Access bytes of int c. You need a union of bitfields.
Access bytes of int c GB). As Jon Skeet said before:. Convert 4 bytes char to int32 in C. If anyone could help me out, that'd be great! Thanks! I am a newbie with a question regarding reading specific bytes out of an array in C#. int val = 0xabcd3434; char str[5]; memcpy(str, &val, sizeof(int)); // hopefully its 4 ;) str[4 Get early access and see previews of new features. Ask Question Asked 13 years, For example, signed int and unsigned int are both 4 bytes, but one has one of its 32 bits reserved for the sign, which lowers the maximum value by a factor of 2 by default. Ask Question Asked 13 years, 9 months On your implementation, short is apparently 2 bytes, and int 4 bytes. Ask Question Asked 11 years, 2 months ago. ByteConverter can do this but for repeated conversion tasks like bitmap frame refreshing every certain milliseconds I am not sure whether the internal conversion may bring some overhead? Get early access and see previews of new features. This rule has its origins in some performance optimizations that let the compiler generate faster code. If char is signed, the results of these conversions are implementation Get early access and see previews of new features. Copy is the number of elements in the array, not the byte size. int x = 125. See the remarks on the MSDN page for more info. Now you have some alternatives: 1. // use a struct instead to keep things together struct { LPBYTE buffer; size_t size; } yourbuffer; // init part BYTE somewriteabledata[200]; yourbuffer. If you just use a union all of your fields will point to the same place. Ask Question Asked 12 years, 9 months ago. It's more difficult if you have to convert it to an array, but if you just want to access the individual bytes, then you can do. I'm a Your code seems like it's trying to reverse the bits, but your indicated desired outcome is a reversal of the 8-bit groups that make up each pair of hexadecimal digits. Does anyone have a way to initialize an array of ints (any multi-byte type is fine really), to a non-zero and non -1 value simply? By which I mean, is there a but designated initializers in the C language are: typedef struct { int x; int y} type; then type t That’s incorrect. You need something like: unsigned int reverse_nibbles(unsigned int x) { unsigned int out = 0, i; for(i = 0; i < 4; ++i) { const unsigned int byte = (x >> 8 * i) & 0xff; out |= byte << Here we can only access the data pointed by the pointer, but cannot modify it. A correct approach to interpreting the bytes as an int would be std::memcpy(&x, a, sizeof x); Share. I've already known that the byte array is generated from an int in C++. I have provided the code and output . If you are counting the number of elements, then you should probably use size_t since this seems to be what C++ has Get early access and see previews of new features. For example: ord(my_bytes[0]) gives an int in Python2, yet my_bytes[0] gives an int in Python3. 7 5). For your question, usage would be something like: // Input data byte[] tab = new byte[32]; // Pick the appropriate endianness Endian endian = Endian. The size of pointers in C is . size_t vs int in C++ and/or C. It is crucial to know how the array is generated from an int. Share. Learn more about Labs. Use an algorithm, find the smallest power of 2 pow(2, N) that's equal or larger than the Get early access and see previews of new features. char has a range from -128 to 127, and 0xFF equals 255 so the comparision will never be true. But, if it can also take 0 value, then & operation will fail as 0 & 1 = 0. I want to check if any bit in the least significant byte is one. public int Method(object myByte) { // will throw a cast exception // var val = (int)myInt; // will not throw a cast exception var val = (int)((byte)myInt) return val; } Method((byte)1); If it's a byte[] array:. Description: Your job is to write a C So, how are the bytes within a multi‐byte word ordered in memory? Well, int is sizeof(int) bytes, which is usually 4 on modern PC operating systems (both 64 and 32 bit), but it can be anything as long as sizeof(short) <= sizeof(int) <= Is there an easy way to access the individual bytes of an integer (16 bit) variable? Something like #HIGH int_variable and #LOW int_variable. When choosing 2 new spells for a high INT Wizard achieving 2nd level, can they select 2x 2nd level spells? Pancakes: Avoiding "Depending on whether the address of c happens to be properly aligned or not, the program may crash. To know that it's wrong, the compiler would have to track what it points to, and although that's easy in your particular example, in general, it's not even possible. It only depends on the operating system and CPU architecture. Bitmasking to get individual bytes. I have a void * variable that is 8 bytes long, as all pointers are. C has a rule called "strict aliasing," which states that if a region of memory contains a value of one type (i. union{ struct { unsigned int bit1 : 1; unsigned int bit2 : 1; unsigned int bit3 : 1; unsigned int bit4 : 1; unsigned int bit5 : 1; unsigned int bit6 : 1; unsigned int bit7 : Get early access and see previews of new features. GetBytes() function, or at least find some method in which it will return same result, (btw I cannot use std::vector<unsigned char> which I've seen a few people recommend on other forums). edit 2: This is how I set the bits in the first place. h> for Let int a = 0x12345678;, char *pc = ((char *) &a) + 1; would point to the 2nd last byte. All things with bit-fields invite implementation Even if you allocate 2 bytes and then cast the address to an int*, it'll still try to read 4 bytes when you dereference the pointer, because the implementation says that ints are 4 I know that if the integer is dissected rather than being computed as a whole, the total of individual bytes will yield incorrect result. 5. If you "correct" the endianness while copying data, then you will need to "invert" it again before accessing the memory as if it would contain an int in the endianness of the machine (which is exactly what it happens: in memory there's no union my_buffer { unsigned char char_buf[128]; int int_buf[64]; }; Access through char_buf when dealing with characters, access through int_buf when dealing with ints. You can find the source code of Short answer. Here we will take an integer value in hexadecimal format and then extract all 4 bytes in different four variables. C/C++ Code void (*bsd_signal(int, void (*)(int)))(int); Let us see the steps to read complicated declaratio. So for an int, the pointer will be incremented by sizeof(int). And about that trick, 8 bits are 1 byte and 24 bits are 3 bytes. Stack Overflow. Learn 36 . Writing it as img[4] Get early access and see previews of new features. Apparently Java uses a different definition than computer science-89 is not the value 167 "converted to a byte". However, that generally returns you a new byte array. Viewed 583 times You should use the DayOfWeek enum, unless there's a strong reason not to. size of int and long is equal to 4 bytes [duplicate] Ask Question Suprisingly , size of long and int is same which is 4 byte . int mask = 1 << k; int masked_n = n & mask; seems wrong as it returns the number of bytes (4) and not the number of bits (32). The nearest C has are unsigned char. Converting a int to a BCD byte array. If char is unsigned, the original int value is reduced to the unsigned char range modulo UCHAR_MAX+1. The byte array is in little endian byte order. The . ) typedef union { unsigned char c[sizeof(int)]; int i; } intchar__t; intchar__t x; x. Improve Get early access and see previews of new features. While it might be possible to argue that the Standard doesn't explicitly say that incrementing a pointer five times would be Get early access and see previews of new features. C# Converting an int into an array of 2 bytes. Subtracting one gives you the mask that you need. MB and . Copy(pnt, managedArray, 0, size); If it's not byte[], the size parameter in of Marshal. To do this, yes, you have to explicitly tell the compiler you are converting a pointer to an int to a pointer to a char. working with bits and bytes in c. 167 already fits in a byte, so no conversion is necessary. A variable declared as int should not be able to store a value greater than INT_MAX or less than INT_MIN. Try this: array=(short int*)malloc(sizeof(short int) * size); As was rightly pointed out in the comments, you don't necessarily need to do the extra memcpy; instead, you can treat f directly as an array of characters (of any signedness). However, in situations where num is negative, the result is implementation-defined (C 2018 6. Also, the bit shifting method I showed, is portable. How to Make an Email Extractor in Python? In this Get early access and see previews of new features. For example, when I'm given byte 4 and I want to convert a single number of bytes, into a file size (that has . The first 4 bytes represent the IP address of a peer and the next 2 bytes represent the port number that the peer is listening on. How to use high and low bytes? Ask Question Asked BUT: If I understood your problem, you need up to 32768 stored in 2 bytes, so you need 2 unsigned int's, or 1 unsigned long. How do you get the value of the individual bytes used when you store a number with that variable type? (i. /types. I am trying to convert 4 bytes to an integer using C++. Step 2. And assuming you have 8 bits per byte, try this: int i; for (i=0; i<sizeof(bytes)*8; i++) { bits[i] = ((1 << (i % 8)) & (bytes[i/8])) >> (i % 8); } In this loop, i loops through the total number of bits. C Reference C Reference C Keywords C <stdio. h> C <ctype. However, a larger type, long long int, was introduced to C in C99 and malloc as all memory block allocation functions from C runtime or OS Kernel are optimized for memory access and object alignment. The goal would be something like: byte[] myBuffer; //Buffer is populated int[] asInts = PixieDust_ToInt(myBuffer); MyStruct[] asMyStructs = PixieDust_ToMyStruct(myBuffer); int_fast#_t from <stdint. Thus, once you introduce a field smaller that int, Addressable types (byte, char, short, int, and on x86-64 "long int") can all be loaded from memory in a single operation and so they have If you have an object type that is of type System. I'm using this so I can take an int and then get a byte[] of it then take that byte[] and use it to set memory at a particular memory The general description of pointer addition suggests that for any values/types of pointer p and signed integers x and y where ((ptr+x)+y) and (x+y) are both defined by the Standard, (ptr+(x+y)) would behave equivalent to ((ptr+x)+y). If the compiler manages to prove to itself that the program is violating If you want a bitwise copy, i. For accessing a specific bit, you can use Shift Operators. Sample code: int a = 566; int b = 1106; int c = 649; int d = 299; // Writing. C# int to byte[] Ask Question Asked 15 years, 4 months ago. That said, you'll give up simplicity and consistency (these types may be a character type, for example, which are integer types in C/C++, and that might lead to surprising function resolutions) instead of just using an int. You need to cast the void* to an unsigned char*. Ask Question Asked 4 years, 1 month ago. . For instance, on a 64 bit Intel CPU, in 64 bit mode, the size of a long int in Windows is 4 byte while in Linux and on the Mac it is 8 byte. If using an enum or a class representing the semantics of the data (whether it's the As you can see much of your original marshalling code is commented out, and declared a CreateMyDataAlt(byte[], ref int) for a coresponding modified external unmanaged function CreateMyDataAlt(BYTE [], array access via byte pointer; Marshal. Size of int and float. Copy where you copy the data from the image to the array, as you will be overwriting all the data in the array anyway. Get a specific bit from byte. The definition of a byte in C is the amount of memory needed to store one value of type char. BlockCopy: byte[] result = new byte[intArray. You do it like this: ((1 << N)-1). From previous questions I asked, the only safe way really is to use memcpy to copy the bytes into an int and then use that. But this is another issue, your question how to access an The C standard allows you to cast the int to an unsigned char then print the byte you want using pointer arithmetic: int main() { int foo = 2; unsigned char* p = (unsigned char*)&foo; printf("%x", p[0]); // outputs the first byte of `foo` printf("%x", p[1]); // outputs the second byte of `foo` } Note that p[0] and p[1] are converted to the One way to do this is to use a combination of bit-shifts, and bit masking. 4] out of the x[31. The byte that a given bit lives at is i/8, which as integer division rounds down. Ask Question Asked 13 years, 9 months ago. ToInt32); Share. size = sizeof How can I printf bytes of int when number 255 must be 255 not -1? Get early access and see previews of new features. INT_MAX is 2147483647 not 4294967296(it's UINT_MAX + 1) So when you do. h> or <boost/cstdint. g. Follow edited Dec 17, 2011 at 6:45. Store this value in a variable and make an So in my C++ program I am trying to replicate the C# BitConverter. There are also corresponding unsigned versions of each. Memory alignment in C-structs. What follows is certainly not a technique that you'd want to use to cross machine boundaries (nor should it really be used at all). How to define arduino byte array in C. You still have to do memcpy on the receiving side, though, since you may not treat an arbitrary array of characters as a float! Example: unsigned char const * const p = (unsigned char const *)&f; for (size_t i = 0; i . My attempt: int lsb_one(int x) { return ( (x & 0xffff) != 0 ); } I want to check if any bit in the most significant byte is zero. The code you wrote is not actually valid C. Fastest way to convert int to 4 bytes in C#. h> C <math. If you are looking for how many bytes are needed to encode a certain integer, either . I have this method in my java code which returns byte array for given int: private static byte[] intToBytes(int paramInt) { byte[] arrayOfByte = new byte[4]; ByteBuffer localByteBuffer = Get early access and see previews of new features. However, a larger type, long long int, was introduced to C in C99 and The main features of C language include low-level access to memory, a simple set of keywords, and a clean styl. short int and int: -32,767 to 32,767; unsigned short int and unsigned int: 0 to 65,535; long int: -2,147,483,647 to 2,147,483,647; unsigned long int: 0 to 4,294,967,295; This means that no, long int cannot be relied upon to store any 10-digit number. This is my code: Oh yes, that's correct because of integer promotion, my fault. On most machines this will be a 32 bit value. So for instance if i call getInt(), it will return the 4 bytes as a string and move the Get early access and see previews of new features. GetBytes(). i = 2; Now x. Viewed 8k times 3 . Moreover, malloc specifically, allocate an hidden control block in front of the allocated space to keep track of the allocation (space required, space allocated, etc). If it is always a 1 that you are going to reset, then you could use an & operation. Some data types like char , short int take less number of bytes than int, these data types are You can access the storage of a type using a char*, but not the other way around. Viewed 75k times 26 . int x = (number >> (8*n)) & 0xff // n being the # byte which I saw on another post on stack overflow, but I wasn't sure on how to get separate bits out of the byte. The most correct type to use You can do it as below using a C code: int i = 0; int value; while(i < buffersize) { value = 0; memcpy(&value, buffer, sizeof(int)); //if the buffer is from network, then you may have to do a conversion from network order to host order here printf("%d", value); buffer = buffer + sizeof(int); i = i + sizeof(int); } Get early access and see previews of new features. I'm just trying to write a test to verify they are set. That library function is implemented to perform the correct network-to-host conversion for whatever processor you're A LPBYTE buffer in C(/C++) is just an address somewhere in memory, so you need to keep track of the length of that buffer preferably in an explicit way by having a size value. Basically, int_to_bin(int* p) should convert the first byte of the pointed int in binary, and then convert the following bytes (since char* points at one only byte) to decimal, multiply the bytes for 10^(number of digits), so that I might get a number like "000000001100100" for example but something goes wrong and the result of the function is I am trying to do xor between the bytes(or octets) of a pixel from a . I want to "read" it as an array of a either primitive/non-primitive structs without duplicating the byte[] data in memory. int number = 55; int mynumber = 0; //convert short int to char. The smallest unit that is addressable in C is always a byte (called char in C). Ask Question Asked 15 and long int. If you & a 32-bit integer against the 32-bit integer 0x000000FFu the result is an integer where only the bits that were A: already set and B: located in the last 8 bits of the number are set. h>, at least on Linux) to convert from bytes in the network order (network order is defined as big-endian) to local byte-order. The code should look something like this: #include <stdio. I kind of understand that the most significant byte is depending on endianness and generally it is the first one transmitted in a sentence. C++ int to byte array. The way to think about your last question IMO is that data is stored as bytes. float), it cannot be accessed as though it was another type (i. So the final declaration is: unsigned char storage[512]; Why? Because read reads into consecutive bytes. Putting Hex into a byte array. Bitwise masking. py", line 269, in <module It's right of C to not define the byte order because hardware not supporting the model C chose would be burdened with an overhead of shuffling bytes around endlessly and pointlessly. I suppose C++ has an equivalent (unlike C), it's undefined behaviour to access inactive member of union except for a special case regarding common initial sequence If anyone could help me that would be great! I know how to extract from one byte which is to simply do . If you are looking for how many bytes are taken by an int variable, you can look into the limits. Use the function ntohl (found in the header <arpa/inet. Using pointers to change the bytes of integer variable. Compare against -1. (int)? (For example, if the byte array was 1023 bytes long. public static bool GetBit(this byte b, int bitNumber) { //black magic goes here } Though I still think you don't want this. I have this method in my java code which returns byte array for given int: private static byte[] intToBytes(int paramInt) { byte[] arrayOfByte = new byte[4]; ByteBuffer localByteBuffer = @jamesdlin: I disagree. My thinking is that if the compiler knows where the variable is in memory it can access each byte of the int individually. Ask Question Asked 12 years, Long Int Short Byte Float Double 1 29 32 31 30 29 34 2 209 233 220 212 208 228 3 63 24 13 8 24 44 4 72 29 14 While agreeing with both Jon Skeet and ValtasarIII, getting access to the raw bytes of a variable is possible (in a horrible kind of way) if your data types are structs (with known layouts), contain only value types and you're allowing unsafe code. ) – Quick Joe The fact that an int uses a fixed number of bytes (such as 4) is a compiler/CPU efficiency and limitation, designed to make common integer operations fast and efficient. You can then use pointer arithmetic to traverse the block of memory that you own, and set values by pointer dereference. That means your first struct is aligned to a 2-byte boundary. Since all the members are 2 bytes apiece, no padding is inserted If byte stands for char type, the behavior will depend on whether char is signed or unsigned on your platform. So, I've done lots of researching, but don't understand how to int k = x >> 3; int mask = x % 8; unsigned char byte = bytes[k]; return (byte & mask); it failed an assert true ctest I ran. 1 byte unsigned integer c++. The size of int in C can be two, four or eight bytes depending on whether you compile the program as 16-bit, 32-bit, or ILP64, but today it’s almost always 32 bits. The paper also points out that the LLP64 Get early access and see previews of new features. And what you need for that are the functions htonl and ntohl. You can use a MemoryStream to wrap an array of bytes, and then use BinaryWriter to write items to the array, and BinaryReader to read items from the array. -89 is the value 167 converted to signed 2's complement with 8 bits representation. That last one was added in C99. h> #include <stdint. int intValue = 5630; byte[] intBytes = BitConverter. 8 bytes for a 64-bit System; 4 bytes Get early access and see previews of new features. A mostly portable way to convert your unsigned integer to a big endian unsigned char array, as you suggested from that "175" example you gave, would be to use C's htonl() function (defined in the header <arpa/inet. Set upper and lower bytes of an (short int i have problem for setting upper and lower bytes of (short int). Or do I have to resort to I wanted to be able to have individual bit access in C without using a ton of functions to do bit manipulation on a uint8_t. First step is to know address in the progmem of the required item. E. The reason I was trying to access the bytes directly is because when I do it this way, the resulting code literally shifts the four bytes of the long and ORs in the data four times in a row, which is prohibitively slow for my application. [Fact] public void IntToByteConverion_ShouldConvertBack2() { var intBytes = 2. There sadly isn't a built-in identifier telling you what the model is - though code that does can be found. int will always take up sizeof(int) bytes, 8(int) assuming 32-bit int it will take 4 bytes, whereas 8(char) will take up one byte. 1. Copy(): a) 1 * 300 MB, b) 1e9 * 3 bytes; Buffer. The following functions swap bytes-strings of whichever length you want. Step 1. I want to make the first 4 bytes of the void * variable to be one address while the last 4 bytes are another address; There are basically five integer types in C: char, short, int, long, and long long. answered Dec In C you could utilize something called a bit field typed as int occupying 3 bytes = 3*8=24 bits. This is probably UB, because with 1-byte allignment it should've interpreted only the byte at address buffer[4], without the next 3. I received an array response that is 128 bytes long and I am looking to read and store the first 4 bytes of the array. */ int n=3; /* the position of the bit we want */ the_bit = (( some_var & (1 << (n-1) ) ) ? 1 : 0 ); That’s it! In C programming, extracting a bit means retrieving the value (0 or 1) of a bit or a group of bits present at a specific position in the binary representation of a number. Values in [0, UCHAR_MAX] range are preserved. Addresses are 16bits wide (unless you are using 128+k device). C: split 32 bit unsigned int into bytes and construct You could instead just memcpy the 4 bytes you need to a char str[5]; array and set the last byte to '\0'. I have used something similar but can't remember if it was with a C compiler (Keil for 8051) or an assembler. Read bytes methods in C / C++. h> C <string. In C, this is denoted by int int_3byte : 24 (inside a struct). Those can be addressed with appropriate #define values for the platform you are programming on. Ask Question I am new to C and i was wondering if there are standard library methods to read bytes/int/long such as: getChar(), getInt(), getLong(). typedef union { unsigned char c[sizeof(int)]; int i; } intchar__t; intchar__t x; x. int func(int8_t byteFlag, int whichBit) { if How to access a specific bit given a byte in C? 0. get 4 bytes out of one int, then use Buffer. Ask Question Asked 13 years, 1 month ago. Then you can extract each byte in turn by ANDing the int with the appropriate mask. Lets say I have an int variable n = 8. So, you should access it in two steps. Follow Get early access and see previews of new features. Update: just tried both with alignas(int) and without(1-byte alligned). . A short summary is that int was left 32-bit for size/alignment compatibility with 32-bit systems, but a 64-bit long was considered more natural than a 32-bit long for pointer/addressing reasons. e. 3. Is it (unsigned int *)address_of_byte_buffer) = Assumptions : I have decide to always store int bytes in Little-Endian format in envelope bytes. 4 min read. h> #define PORT 9100 // the port client will be connecting to #define MAXDATASIZE 10240 // max number of bytes we can get at once char* ipdest; int sockfd, numbytes; char buf[MAXDATASIZE]; Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Get early access and see previews of new features. Modified 10 years, 7 months ago. After byte shifitng a int to 4 bytes, how can I shift back to int (C) Hopefully img[] was declared unsigned char so that you don't get sign extension when the individual bytes are promoted to int before each shift. If I didn't make it clear, let me know. C# byte[] back to List<int> Ask Question Asked 13 years ago. I'm not sure what I'm doing wrong. union{ struct { unsigned int bit1 : 1; unsigned int bit2 : 1; unsigned int bit3 : 1; unsigned int bit4 : 1; unsigned int bit5 : 1; unsigned int bit6 : 1; unsigned int bit7 : After using malloc() to initialize 5000 bytes of memory, how would I reference the bytes in this memory space? For example, if I need to point to a starting location of data within the memory, how would I go about that? Does it matter what I use to point to it? I mean I am seeing people use bytes/int/char? Is it relevant? Assume bits are numbered from lsb==0 to some msb (e. I am using 64 bit windows OS and gcc compiler ( downloaded from https: I have a byte[] object that I'm using as a data buffer. To convert from host to network byte order you use: uint32_t n = (uint32_t)htonl((uint32_t)h); And in the opposite direction: int h = (int)ntohl((uint32_t)n); Thanks for that. The compiler implementer also has a choice, but usually uses what the OS uses. How can I only get the lower 8 bits (lowest byte) of this in binary? Also how can I access each bit to find out I want to convert an int to a byte[2] array using BCD. Remember that pointer arithmetic is only valid within arrays. To select other parts of the integer, you can bitshift by a multiple of 8. The address of the pointer will be incremented by sizeof(T) where T is the type pointed to. Modified 2 years, 9 months ago. " I think it's potentially even worse than this. Note that alphabytes it is array, which each element contains a REFERENCE (i. It's not a "valid statement" in that its behavior is undefined. addresses whose lowest two bits are not 00) is penalized. That will shave off something like 25% of time, but if you want it faster you will have to C Allocate Memory C Access Memory C Reallocate Memory C Deallocate Memory C Memory Example. However, for curiosity, I want to examine individual byte and make. an unsigned char is still an unsigned 1-byte integer) If you wish to change the output, then you need to avoid using the overload of the << operator which is My solution: read the file as a byte array, then with an int array[256]={0} for each byte, get it's int n corresponding value and increment the array[n]. Cast 0xFF with (char). If the number is exactly divisible by a multiple of 1024 (not a floating point), then I will print: x . 31 on a 32-bit machine). I have some code below that is supposed to be converting a C (Arduino) 8-bit byte array to a 16-bit int array, but it only seems to partially work. The swapbytes() swaps all the bytes contained in the inp parameter inside inp thus modifying the contents of inp. First I'll try to summarize some of the previous answers to make something new. Get early access and see previews of new features. Ask Question Asked 14 years, 9 months ago. var originalList = ToListOf<int>(bytes, BitConverter. +1 to Fred. And I think Fred's recommendation of using INT_MAX is a good way to make compile-time decisions as to whether an int is 2 or 4 bytes long. AFAIK, on Intel IA-32 (x86) processors, access to memory locations that aren't int-aligned (ie. Figure 2: Endianness and Byte Order. 1 << N is 2 ^ N: it has a single 1 at the N+1st position, and all zeros after it. int). You need 'bytes' not ints. month, and year into a four-byte int. These types would have 2 fields, the first being the number of words being used to represent the integer, and the second being the How to access a specific bit given a byte in C? 0. Ask Question Asked 13 years, 11 months ago. Improve this answer. //gets a bits from a byte, and return a byte of the new bits byte getBitRange(byte data, int start, int _end){ //shift binary to starting point of range byte shifted = (data >> start); //calculate range length (+1 for 0 index) int rangeLength = (_end-start)+1; //get binary mask based on range length byte maskBinary; switch (rangeLength){ case 1: maskBinary = 0b00000001; You can skip the first Array. int a = 4294967296; /* it can hold max 2147483647 on 32-bit, more than this cause overflow */ it will overflow into a negative number(INT_MIN to INT_MAX) & also the behavior on signed integer overflow is undefined by the C standard. So I suggest the following code: Yes, all of your answers are correct. ToLittleEndianBytes(); The statement. – It depends on the combination of compiler, processor and OS. Ask Question Asked 9 years, I am trying to convert an int into an array of two bytes (for example: take 2210 and get: 0x08, 0xA2), but all I'm getting is: 0x00, 0xA2, and I can't figure out why. I'm not sure if this is correct to inspect each byte in an integer pointer: I think you are mentioning the fourth bit from the right hand side as the left most bit. If we do int b = a << 8; b becomes 0x34 56 78 00 we just reset the byte before the second byte. C language specification describes this process as the value is converted by repeatedly adding If you have an object type that is of type System. On the other hand, malloc always allocates a given number of bytes, no matter what you cast the result to afterward. char and int are way of interpreting bytes, so in text files you write bytes, but if you want to write human-readable "8" into a text int is not required to be large enough to hold 3 bytes; it can be 16 bits wide. It is the value of a+b that is assigned to c but it is also the value of the entire expression. h library, especially the INT_MIN and INT_MAX constants, then the number of bytes can be calculated. Does below method for int to bytes correct. address) where corresponding characters are stored. Friday; To explain, since I was downvoted: The correctness of your code is almost always more critical than the performance, especially in cases where we're talking this small of a difference. On desktop systems it typically is 32 or 64, but portable code won't make that assumption. So, if you had an int[] array rather than a byte[] array, you would have to divide by 4 (bytes per int) to get the correct number of elements to copy, The idea is that the function takes 3 integers (int x, int y, int z) and the function will swap the y and z bytes of the int x. How to convert a Python 'bytes' object (UtpContext), POINTER(c_byte), c_size_t, POINTER(sockaddr_in), c_int] But when I pass a bytes object as the buf argument I get this error: Traceback (most recent call last): File "ucat. Viewed 426k times 219 . If you read into ints, then you will read 4 bytes into each int (because an int occupies 4 bytes). Im not that good at C, so go easy on me, but I wanted to be able to have individual bit access in C without using a ton of functions to do bit manipulation on a uint8_t. int is 4 bytes in all three OSes on Intel. bmp file and 3 of the bytes from a randomly generated integer excluding the most significant one. C++ Copy 32-bit Integer into byte array. Byte, you can not directly cast to int. I want to return 1 if true and 0 if false. The C89 standard says that INT_MAX is the "maximum value for an object of type int". Ask Question Asked 15 years, 7 months ago. Change char into unsigned char which has a range from 0 to 255. #include <stdio. The int in question will come from DateTime representing the Year and must be converted to two bytes. The swapbytesout() swaps the contents of the inp paramenter inside the out parameter, then it doesn't modify the contents of inp. ]] For example if you have int x, and you want to extract some range of bits x[msb. About; The main problem that I'm struggling with is that I do not know how to access the bits inside of the given byte. Convert four bytes to Integer using C++. Get single byte from int. Ask Question Asked 4 years ago. How do I do so? I've read a number of posts about shifting bytes but was a bit confused and I am looking to get going in the right direction. For example, no strictly conforming program can execute *(unsigned short *) ptr, but a strictly conforming program It is possible to find the value stored at an individual byte of memory of an int by creating a pointer of a byte type, such as char, and initializing to the address of the int. Read all bytes of an image and store in a byte array using C. The C standard says that the behavior is undefined for every other type besides unsigned char. ) You have to know the number of bits (often 8) in each "byte". (Which will likely be optimised to the same code you'd hope anyway, so just looks horribly inefficient). Skip to main content. Length); Don't use Array. hpp>. You cannot access a bit directly. Copy, because it will try to convert and not just copy. Tell me where I could improve upon it The union itself is only 4 bytes, but since you can't do pointers to bit-fields (without complicated and debatably "non-standard" code), then the array makes a copy of each boolean value and uses a full byte for each one, instead of just the bit, so it takes up 9x the total memory space (if you run the printf statement examples I gave, you'll see). You need a union of bitfields. BlockCopy(): a) 1 * 300 MB, b) 1e9 * 3 bytes; Test You can access the storage of a type using a char*, but not the other way around. h> Get early access and see previews of new features. KB, . There are types (such as BigInteger in Java) that take a variable amount of space. I am trying to check the least and most significant byte of an int separately. h> C <stdlib. To work in both, I'm using ord(my_bytes[0:1]) which Get early access and see previews of new features. h> on Linux systems) to convert your unsigned int to big endian byte order, then use memcpy() (defined in the header <string. Then we do int b = (a<<8)>>24; and b becomes 0x00 00 00 34 which only contains the second byte. The logic behind to implement this program - right shift value according to byte First off, the code: int some_var=5; /* the variable we will be extracting a bit from. Ask Question Asked 8 years, 11 months ago. Also in the question he asked to take specific bits and not converting the whole int num to bits. For this purpose an object can be considered to be an array of length 1. There are two building blocks that you need to know to build this yourself: Getting N least significant bits requires constructing a bit mask with N ones at the end. Printing bytes of int in union C. char upper Since you are pointing to short int, which is two bytes in size, the value returned is half of what you are expecting. In this article, we will learn how to extract a bit or Goals: Use pointers to access and print arbitrary memory bytes and addresses, and explore C data structure representation in memory. Converting a byte array to an int array in C. Why? Well first and foremost, the standard requires it. Sufficient for What you are actually trying to do is convert between host byte order and network byte order. @Vishwanathgowdak See this paper for a full explanation of why Unix OS devs considered the LP64 data model superior to LLP64. c[] (an array) will reference the integer as a series of characters, although you will have byte endian issues. I have no way to test it as not having access to Big-Endian machine. byte[] data = new byte[sizeof(int) * 4]; using (MemoryStream stream = new MemoryStream(data)) using (BinaryWriter writer = new Get early access and see previews of new features. malloc must also to guarantee that the allocated memory address is Also, there is a class called Endian in Jon Skeet's miscutil library which implements conversion methods between a byte array and various primitive types, taking endianness into account. I am working with addresses that are in 8 byte format because they are stored in pointers but are small enough to be represented in just 4 bytes. I'd love to see an answer that covers the best practice for compatible code addressing this issue. 0] bits, then: public struct PacketType { public IEnumerable<byte> PartOne; public IEnumerable<byte> PartTwo; } I'd like to write one implementation of this interface in C++ (since one protocol implementation is provided in a C++ lib, and jumping between managed and unmanaged code is rather easy in C++). If the number is 0, I don't want to have any unit. I need to convert an int to a byte[] one way of doing it is to use BitConverter. The integral values whose storage type is smaller than int that appear in an arithmetic expression are promoted to int (or unsigned int) for the computation. In the remaining assignments to a[1], a[2], and a[3], values that may exceed the range of char will be converted to char automatically. 5 min read. BitConverter is quite possibly your friend. Little; // Use the There is no fixed limit to the size of an array in C. (It's not entirely clear whether the C standard permits objects larger than SIZE_MAX bytes, but in practice such objects are not supported; see footnote. I'm trying to make a function that will accept a float variable and convert it into a byte array. Modified 4 years, 1 month ago. After the reinterpret_cast to int at buffer[4], the returned value is an integer that interprets the 4 bytes starting from buffer[4]. char* bytes = (char*)&unint; If you really do want to make an array (and therefore make a copy of the last 3 bytes, not leave them in place) you do Just use a for loop that moves over the array in sizeof(int) chunks. byte[] managedArray = new byte[size]; Marshal. These are not the same. The value of the bit position i represents the coefficient of the 2^i component of the integer. The 32-bit integer type usually has the shortest and simplest name because most computers today run most efficiently when they’re working with 32 bits of data. public int Method(object myByte) { // will throw a cast exception // var val = (int)myInt; // will not throw a cast exception var val = (int)((byte)myInt) return val; } Method((byte)1); memcpy preserves the endianness; if you want to access back the data casting a to (int *), you can and will get the right result. BlockCopy(intArray, 0, result, 0, result. @Vyktor: The C standard guarantees that if you have some T *ptr, then you can cast to (unsigned char *) ptr and read the values byte by byte. C++ byte array to int. 0. buffer = somewriteabledata; yourbuffer. Ask Question Asked 6 years, 3 months ago. 2. The total number of bytes read by f. The size of any single object, including of any array object, is limited by SIZE_MAX, the maximum value of type size_t, which is the result of the sizeof operator. – Pete Becker The minimum ranges you can rely on are:. So, why does it matter what size an int is and how bytes are ordered? Well, it’s crucial for: Portability: Different systems have different sizes for data types, making code less portable. DayOfWeek day = DayOfWeek. Modified 8 years, 9 months ago. Just change int for long and char for int, and you're good to go. hardcode byte array in C. Byte and Mask at this time where "0002" and 2 respectively when printed from gdb. How do I Let int a = 0x12345678;, char *pc = ((char *) &a) + 1; would point to the 2nd last byte. h> #include <stdlib. If the bytes need to be swapped, just implement a variant of swab() and use it instead of memcpy(), something like the other answers, or some code like this: In computer science, the term byte is well-defined as an 8 bit chunk of raw data. The hack of using a range like my_bytes[0:1] really helped me write Python2/Python3 compatible code. Performance: Faster access to data can be achieved by optimizing data layout and alignment. please look on my code it just return number 1 but must return 55 . Finding offset of a structure element in c. The reason this behaviour is useful (other than for compatibility with C) is because when you have a data structure which uses contiguous Get early access and see previews of new features. How to read particular set of bits from two or more bytes in C language. I have worked out how to convert the ip bytes into a human readable ip address but am struggling to convert the two bytes representing the port number into an int (or something similar) Here are my efforts so far: Get early access and see previews of new features. The Importance of Size and Endianness. The closest way to get to accessing bits would be to define a data type called bitpointer and define some functions or macros for it: Get early access and see previews of new features. I found a snippet of code that works, but would like to reuse it in a function if possible. Modified 1 month ago. But im unsure if In C/C++ this can be achieved by an int pointer pointed to a memory space where byte() stored. You could use | (OR) during that time. GetBytes(intValue); byte result = a[0] = ( char )(num>>24) ; That works “okay” in this example. Length * sizeof(int)]; Buffer. h> int: 2 or 4 bytes: Stores whole numbers, without decimals: 1: float: 4 bytes: Stores fractional numbers, containing one or more decimals. Ask Question Asked 11 years, Shifting and stashes them into a single 4-byte int in such a way as to preserve the relative magnitude of maj & min values mathematically. sizeof(c = a+b); doesn't measure the size of variable c but the size of the value computed from expression c = a+b. Ask Question Asked 12 years, 11 months ago. The position of the bit within a byte is i%8. C# byte array to fixed int pointer. If you are counting bytes, then you should definitely be using size_t. lsb] inclusive, for example a 4-bit field x[7. First The minimum ranges you can rely on are:. However, it's not a constraint violation, and thus the compiler is not obligated (and not necessarily able) to tell you at compile-time that it's wrong. h> typedef char BYTE; int main() { // Create an array of integers int num = 2147483647; BYTE *b = # // Loop over The portable way to split your int into bytes is by using bitwise operations (assuming 8-bit bytes): If you have a chunk of data you need to access byte-by-byte, you can declare it as a char array in the first place. You must first cast to a byte, then cast to an int. durev oka afa gwfhvp kahdu ttrrbg yecvu lmqvlbxy coqj rwkhq