Here is my code for converting int up to 15 into hex char:
static char intToHex(int i)
{
switch (i)
{
case 0:
return '0';
case 1:
return '1';
case 2:
return '2';
case 3:
return '3';
case 4:
return '4';
case 5:
return '5';
case 6:
return '6';
case 7:
return '7';
case 8:
return '8';
case 9:
return '9';
case 10:
return 'a';
case 11:
return 'b';
case 12:
return 'c';
case 13:
return 'd';
case 14:
return 'e';
case 15:
return 'f';
default:
break;
}
}
Is there a way to write this nicer, without so many switch cases?
What I tried:
char * returnHex(int i) {
char * hex = malloc(5);
sprintf(hex, "%x", i);
puts(hex);
return hex;
}
but this returns a char array not a char that I need.
Thank you!
"0123456789abcdef"[i]i["0123456789abcdef"]:-)return (unsigned)i < 16 ? "0123456789abcdef"[i] : '\0';