C - 如何将int转换为uint8_t？

Question

I have this struct:

struct block{
    uint8_t *tBlock;
}

This struct will have 1024 bytes so tBlock = malloc(1024) .

I have an integer that I want to write in 4 bytes so tBlock[0] to tBlock[3] in little endian. I have this :

uint8_t little[4];

void inttolitend(uint32_t x, uint8_t* lit_int){
   lit_int[3] = (uint8_t)x / (256*256*256);
   lit_int[2] = (uint8_t)(x % (256*256*256)) / (256*256);
   lit_int[1] = (uint8_t)((x % (256*256*256)) % (256*256)) / 256;
   lit_int[0] = (uint8_t)((x % (256*256*256)) % (256*256)) % 256;
}

But when I do:

int x = 7;
inttolitend(x, little);

I got little[0] = 7, little[1] = 0, little[2] = 0 and little[3] = 0 so I totally fail my converter. How could I get 7 in uint8_t in 4 bytes?

Answer 1

Here is the standard way to do it - nice and concise:

void inttolitend(uint32_t x, uint8_t *lit_int) {
    lit_int[0] = (uint8_t)(x >>  0);
    lit_int[1] = (uint8_t)(x >>  8);
    lit_int[2] = (uint8_t)(x >> 16);
    lit_int[3] = (uint8_t)(x >> 24);
}

Or using arithmetic similar to your question:

void inttolitend(uint32_t x, uint8_t *lit_int) {
    lit_int[0] = (uint8_t)(x % 256);
    lit_int[1] = (uint8_t)(x / 256 % 256);
    lit_int[2] = (uint8_t)(x / 256 / 256 % 256);
    lit_int[3] = (uint8_t)(x / 256 / 256 / 256 % 256);
}

---

Addendum:

The reverse conversion - idiomatic:

uint32_t litendtoint(uint8_t *lit_int) {
    return (uint32_t)lit_int[0] <<  0
         | (uint32_t)lit_int[1] <<  8
         | (uint32_t)lit_int[2] << 16
         | (uint32_t)lit_int[3] << 24;
}

Or using arithmetic similar to your question:

uint32_t litendtoint(uint8_t *lit_int) {
    return (uint32_t)lit_int[0]
         + (uint32_t)lit_int[1] * 256
         + (uint32_t)lit_int[2] * 256 * 256
         + (uint32_t)lit_int[3] * 256 * 256 * 256;
}

Answer 2

void inttolitend(uint32_t x, uint8_t* lit_int){

    lit_int[0] = x & 0xff;
    lit_int[1] = (x>> 8) & 0xff;
    lit_int[2] = (x>> 16) & 0xff;
    lit_int[3] = (x>> 24) & 0xff;
}

I got little[0] = 7, little[1] = 0, little[2] = 0 and little[3] = 0

Btw, this is Little Endian for 7.

Answer 3

OP uses of lit_int[3] = (uint8_t)x / (256*256*256); mistakenly did the cast before the division.

void inttolitend(uint32_t x, uint8_t* lit_int){
   lit_int[3] = (uint8_t) (x / 16777216);
   lit_int[2] = (uint8_t) (x / 65536);
   lit_int[1] = (uint8_t) (x / 256);
   lit_int[0] = (uint8_t) x;
}

---

Calling int x = 7; inttolitend(x, little); int x = 7; inttolitend(x, little); is a problem if int is not the same as int32_t .

256*256*256 overflow on 16-bit systems.