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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. The code should look something like this:

#include <stdio.h>
#include <stdlib.h>

typedef char BYTE;

int main() {
    // Create an array of integers
    int num = 2147483647;
    BYTE *b = &num;
    
    // Loop over bytes
    for (int i = 0; i < sizeof(int); i++)
    {
        // Print value of i-th byte
        printf("%i\n", b[i]);
    }

    return 0;
}

While this works for finding the value of a byte within a 4-byte integer, I am looking for a way to find the same value only by utilizing an algorithm. My understanding is that the code above cannot be used for memory-safe languages, such as C#.

The prototype for a function that uses this algorithm might look something like this (assuming we have a type defined that is exactly one byte long):

// n is the integer from which we are getting the byte, index is the position of that byte
BYTE get_byte_from_int(int n, int index);

I assume that this algorithm would require a bit of division and modulus, but I'm struggling a lot with actually figuring out what to do.

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5
  • 1
    Use bitwise shifting and masking for more portable code. But must be careful with signed integers, better switch to unsigned if possible. Commented Oct 28, 2024 at 20:43
  • 4
    Endianness will also be an issue. Commented Oct 28, 2024 at 20:44
  • @Artem Panfilov What output do you want [-128 to 127], [0 - 255], [0-0xFF], or what? Commented Oct 28, 2024 at 23:01
  • 2
    typedef char BYTE; is bad practice. Instead you should be using uint8_t from stdint.h. Commented Oct 29, 2024 at 7:25
  • 1
    "only by utilizing an algorithm" What does that even mean? Everyone seems to be guessing. Does it mean get the nth addressable byte of a data type, or does it mean get the nth value byte of an int, or do you actually want a generic function and int is just a placeholder? Commented Oct 29, 2024 at 7:40

4 Answers 4

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2

you can use memcpy, unions or pointer punning if you want n'th byte from the binary representation of an integer.

typedef unsigned char BYTE;

BYTE get_byte_from_int(int n, int index)
{
    union
    {
        int i;
        BYTE b[sizeof(int)];
    }ui = {.i = n};

    return ui.b[index];
}

BYTE get_byte_from_int(int n, int index)
{
    unsigned char *pb = (unsigned char *)&n;
    return pb[index];
}

BYTE get_byte_from_int(int n, int index)
{
    unsigned char pb[sizeof(n)];

    memcpy(pb, &n, sizeof(pb));
    return pb[index];
}

If you want n'th "arithmetic" byte use bit shift operations:

BYTE get_byte_from_int(int n, int index)
{
    return (((unsigned)n) >> (index * 8)) & 0xff;
}
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5 Comments

@EugeneSh. Yes you right - the second one can't be used in this case
@0___________ I think it is ok to include that one too, as it is not clear what the OP requirements are. Perhaps with some clarification
Note that the last one can return different results than the others. It returns bytes based on their significance, whereas the others give the bytes based on their address. On a little-endian machine, these two are the same. But the result will vary on big-endian machines and machines with are neither LE nor BE.
@ikegami They are separated for a reason.
Perhaps, but the reason was unstated until I addressed that omission.
0

For those who like to avoid a function call: compound literal

for (unsigned index = 0; index < sizeof num; index++) {
  // Print value of i-th byte of num.
  printf("%hhu\n", (union { int i; unsigned char b[sizeof num]; }) {.i = num}.b[index]);
}

Well defined since C99.

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Comments

0

I am looking for a way to find the same value only by utilizing an algorithm. My understanding is that the code above cannot be used for memory-safe languages ... [emphasis mine]

Since you actually want the same endianess sensitive n th byte, like you get with this:

typedef unsigned char BYTE;

BYTE get_byte_from_int(unsigned int num, unsigned index) {
    BYTE *b = (BYTE*)&num;
    return b[index];
}

... but without the cast, you could check what endianess your platform has and bitshift accordingly:

#include <endian.h> // POSIX standard, not C standard
#include <limits.h>

BYTE get_byte_from_int(unsigned int num, unsigned index) {
#if BYTE_ORDER == LITTLE_ENDIAN
    return num >> index * CHAR_BIT;
#elif BYTE_ORDER == BIG_ENDIAN
    return num >> (sizeof num - index - 1) * CHAR_BIT;
#else
#error "unsupported endianess"
#endif
}

These functions gives the same result (on systems where the second function compiles, which will be "most of them"). If you compile them on a big endian machine, the order of the output from both functions will be reversed compared to that on your average PC which is most likely little endian. If you are on a non-POSIX system or using a different programming language, you'll have to check the documentation for how to check for endianess.

If you actually want the same order of the bytes on all platforms, you need to decide what order you want. We can't tell from your program if your current approach prints the least significant byte (LSB) first or the most significant byte (MSB) first on the system you are running it on.

Here's a version that can be used to print the bytes in the same order on all platforms. It defaults to returning the least significant byte at index 0. If you #define MSB_AT_INDEX0 it will instead return the most significant byte at index 0:

BYTE get_byte_from_int(unsigned int num, unsigned index) {
#ifdef MSB_AT_INDEX0
    return num >> (sizeof num - index - 1) * CHAR_BIT;
#else
    return num >> index * CHAR_BIT;
#endif
}
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3 Comments

I think you have forgotten to apply a mask after shifting the value.
@IanAbbott There's no need. The CHAR_BIT least significant bits in the result of the bitshift will go into the returned value.
I forgot it was a function, so will convert to the return type.
-1

Certainly, you can cast the int * of &some_int as eluded to by others. Here's a strongly nonportable macro along those lines that'll work for any type of integer (must be stored in an lvalue): #define nth_byte(lvalue, n) ((char *) &(lvalue))[n]

On a sidenote, sometimes bytes are larger than 8 bits. We'll come back to that. Imagine using union to type-pun an int and assuming any resulting padding bytes to be "well defined since C99"... my non-portable crud works for any type and doesn't use any function calls either (not that this matters since we have good compilers that can quickly perform complex optimisations like hoisting and dead code elimination)

The AI raises a good point however. We shouldn't be relying upon machine endianness and then codifying translation rules per machine when we can use shift operations to construct and deconstruct the values instead, thus choosing to explicitly encode our own way. You need to know minimums of short, int, long and long long, and the maximums of their unsigned equivalents. I find it's best to stick to the minimums from the stardard. The maximum int is 32 bits in my code, just like in the standard and so my code will work on any system that uses C, albeit a little inefficiently until it's tuned (just like all...). Anyway, see if you can spot the main pro before I mention it; here's the equivalent (nonportable) macro: #define nth_byte(value, n) (value >> (CHAR_BIT * n)) & UCHAR_MAX

Idk what the AI is saying about C# memory safe because casting pointer types wildly? (Irrelevant sidenote: C# actually memory unsafe when you use unsafe qualifier, which introduces this kind of wild pointer casting).

On a more related sidenote, the error modes of right shifting are (erroneous and) less serious than those of relying upon internal representations to match between machines; right-shifting a negative value is implementation-defined behaviour, meaning they're required to document it in the manual. This explains why we should read a book to learn C then read the manuals for our compilers to explore these other neat things like UBSan.

Shifting left out of bounds is more like what you get if you try to access the 8th byte of a 4-byte int... well, it plays nice if you use int unsigned. Negative values cause the problem, which is lack of documentation. When code needs porting, these are the issues we need documentation for.

One of the early standard rationale documents makes the reasoning behind this kind of behaviour clear: C was intended to be a highly efficient language, and so we have undefined behaviour where a class of error was too computationally difficult to correct, or where an opportunity was present to hand-optimise your code and you were doing something out of the spirit of those keywords (const, volatile, register, restrict and for chux there's inline). The result of UB isn't required to be documented let alone stay consistent. To give you some idea of the threshold here, it was deemed too costly to map the side-effects in the first argument of a function call to the correct machine order, so because this error was so taxing printf("%d %d", x++, x++) is undefined behaviour. So in reality different architectures print different results and it's not required to be documented... some day an implementation might crash, we call that UBsan. Nowadays our compilers do PGO so ... what is with those classes of errors that WERE once computationally difficult? Type punning is still possibly implememtation-defined behaviour because AFAIK the method for determining those implementation-defined padding bits isn't required to be documented in the compilers manuals.

Anyway the pros of using the shift operator here is that it's much more clean to express and supports passing in constants as well as lvalues, so you can do nth_byte(42,0) for example. Also, coming back to it: the macro using the right shift contains the implementation defined CHAR_BIT and UCHAR_MAX. See if your manual contains instructions telling you how to do that, and then never do that, because pretty sure that's listed as UB. Just shift 8 bits at a time instead of CHAR_BIT, assume a maximum of 255 instead of using UCHAR_MAX when reading from or writing to disk or the wire. Trust me, you want a portable prototype then optimise it per platform.

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3 Comments

What's up with all the AI refereces? Did it generate much of this answer? You need to fact check whatever the AI tells you, like "The maximum int is 32 bits in my code, just like in the standard..." , which is nonsense.
@TedLyngmo as I understand there's some expectation that those who ask questions (as you did in that comment) take steps to find the answers before asking. Just wondering if you read all of the answers for this question before you asked about the AI? Look, the C standard is pretty clear about maximal and minimal values for INT_MIN and INT_MAX. If you want to write portable code, you should assume those values, or else there will be configurations your code won't work for. You wanna act like an int is 64 bits? Go ahead, then you get signed integer overflow when int is 32 bits.
@TedLygmo you wanna know what is nonsense? Assuming unsigned int has no padding bits in order to imply that sizeof num can be used to determine how many value bits it has. Don't tell me my understanding of C is wrong, unless you're further into writing a C compiler than I am I can prove you wrong.

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