Thread: Convert string to char

It's a little more work, but you break up the string into individual chunks that you can work with and use strtol to convert to hexadecimal:

Code:

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

int main ( void )
{
  char buff[3] = {0};
  char *p = "D7E37B7BF";
  char *pos;
  int  i;

  for ( pos = p; *pos != '\0'; pos += i ) {
    i = ( pos[1] == '\0' ) ? 1 : 2;
    strncpy ( buff, pos, 2 );
    puts ( buff );
    printf ( "%d\n", (int)strtol ( buff, NULL, 16 ) );
  }

  return 0;
}

> have noticed this alot in these forums why do people assign arrays to 0??
It initializes every element in the array to 0. That way I can work with the array without worrying about a nul terminator for the string. I could get the same effect with

Code:

char buff[3];
buff[2] = '\0';

But that is too much like actual work.

or malloc the array with calloc and let it zero it out for you.

Then you'd have to worry about freeing it