How to generate an 8-bit random number in C>

Question

int main(){

    uint8_t *wdata = NULL;
    wdata = calloc(5, sizeof(uint8_t));

    for (int j =0;j<5;j++){

        wdata[j] = rand();

    }

}

The rand() function generates 16 bits of data. How do I generate 8 bits of random values? Do I need to use a custom function for the same?

Answer 1

How to generate an 8 bit random number in C

Given code such as

uint8_t *wdata = calloc( 5, sizeof( uint8_t ) );

something like

for ( int j = 0;j < 5; j++ )
{
    wdata[ j ] = rand();
}

will work fine.

Integer assignment to an unsigned value truncates, which is exactly what you want. See 6.5.16.1 Simple assignment of the C Standard .

Note that RAND_MAX is guaranteed by 7.22.2 Pseudo-random sequence generation functions , paragraph 5 to be at least 32767, which is "wider" than 8 bits, so the result is guaranteed to "fill up" an 8-bit variable.

Answer 2

Because 8 bit number can contain maximum 0 - 255 . so if i divide rand()%256 we get 8 bit random number.

  int main(){
    uint8_t *wdata = NULL;
    wdata = calloc(5, sizeof(uint8_t));
    for (int j =0;j<5;j++){
        wdata[j] = rand()%256;
   }
}

for more randomness you use some prime number add or subtract difference after dividing from 256 .