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
– pmacfarlane, Commented Apr 13, 2023 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
– Eric Postpischil, Commented Apr 13, 2023 at 23:50

0___________ · Accepted Answer · 2023-04-14 00:44:36Z

0

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

answered Apr 14, 2023 at 0:44

0___________

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boreddad420 · Accepted Answer · 2023-04-14 13:25:57Z

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, 2023 at 13:25

answered Apr 13, 2023 at 23:39

boreddad420

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5 Comments

JohnScott Over a year ago

Why not just use memcpy()?

boreddad420 Over a year ago

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

ikegami Over a year ago

@JohnScott, memcpy, byte order would vary by system.

ikegami Over a year ago

No need to make data 5 bytes long. This isn't a string.

0___________ Over a year ago

what about endianess

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