How to convert 32-bit binary value to int8_t array

Question

I'm trying to covert a 32 bit binary value into int8_t array. I'm not sure how to do this, and I'm struggling to find any documentation explaining the process.

I was thinking that each 8 bits represents an integer, and then that's the array, but I'm not sure.

Edit: The number I'm trying to represent is 01000011 01010000 00000000 00000000.

What do you mean by a "binary value"? All numbers are stored, in the end, in binary, in a computer. Is this a string of `0` and `1`? Your question is very unclear. Do you have some code to share for your attempt? — pmacfarlane, Apr 13 '23 at 23:21
Usually when disassembling an object into the bytes that represent it, unsigned types (`unsigned char` or `uint8_t`) are used, not a signed type, due to issues with sign bits and incomplete specifications of the behavior of signed types in the C standard. Are you sure you want `int8_t`, not `uint8_t`? — Eric Postpischil, Apr 13 '23 at 23:50

boreddad420 · Answer 1 · 2023-04-14T13:25:57.240

0

I'm assuming that your "32 bit binary" data is a signed 32 bit integer. If that is the case, try this:

uint32_t some_number = <some 32 bit signed or unsigned int>;
int8_t data[4] = { 0 };
data[0] = (int8_t)(some_number>>24);
data[1] = (int8_t)(some_number>>16);
data[2] = (int8_t)(some_number>>8);
data[3] = (int8_t)(some_number);

edited Apr 14 '23 at 13:25

answered Apr 13 '23 at 23:39

boreddad420

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Why not just use `memcpy()`? – JohnScott Apr 13 '23 at 23:56
valid point, memcpy would likely be more efficient in terms of compiler output, but for simplicity and educational sake, I just assigned them straight up like that that @JohnScott – boreddad420 Apr 14 '23 at 00:03
@JohnScott, `memcpy`, byte order would vary by system. – ikegami Apr 14 '23 at 00:38
1

No need to make `data` 5 bytes long. This isn't a string. – ikegami Apr 14 '23 at 00:43
what about endianess – 0___________ Apr 14 '23 at 00:44

score 0 · Answer 2 · answered Apr 14 '23 at 00:44

int8_t *conv(uint32_t num, void *buff, int endianess)
{
    uint8_t *arr = buff;

    if(endianess)
        for(int i = 0; i < sizeof(num); i++)
        {
            arr[i] = num;
            num >>= 8;
        }
    else
        for(int i = sizeof(num) - 1; i >= 0; i--)
        {
            arr[i] = num;
            num >>= 8;
        }

    return buff;
}

https://godbolt.org/z/M7YqrW1j1

How to convert 32-bit binary value to int8_t array

2 Answers2