I have this program that claims to convert a decimal number to hexadecimal. The issue is that the program throws the above warning and I don't know how to correct it. The return (variable identifier) is supposed to be a %s because what you want to display is a string.
#include <stdio.h>
int main() {
int n = 0;
char hex = '\0';
const char * HEX_DIG = "0123456789ABCDEF";
printf ("Enter a positive integer: ");
scanf ("%d",&n);
do{
hex = HEX_DIG [n % 16 + 1] + hex;
n = (int) n/16;
}while ( n != 0 );
printf ("\nHexadecimal= %s", hex);
return 0;
}
The %s format token expects to be given a string which, in C, is a pointer to a char (or an array of chars). You've provided it a char which, when passed to a function with variadic arguments (in this case, printf), gets promoted to an int. That's what the compiler is complaining about.
Furthermore, even if your code was valid C, you'd be printing the integer backwards.
To print a char, you need to use %c. The best way to incorporate that into your existing code would be to move the printf inside the loop:
for (int k=0; k<sizeof(int)*2; k++) {
hex = HEX_DIG[(n >> (sizeof(int)*8-4)) & 0xf]; // Move the high four bits all the way to the right and then mask them off.
n <<= 4; // Shift the next four bits into place.
if ( hex != '0' ) { // No need to print leading 0's.
printf("%c", hex);
}
}
printf("\n");
To explain the sizeof calculations, let's assume that ints are four-bytes wide. Then sizeof(int)*2 would be eight which accounts for the eight sets of four bits. sizeof(int)*8-4 would be 28 and so we'd be shifting off all 32 bits except for the four highest.
Note that this code does ignore the case when n is zero. In such a case, the loop would print nothing.
Backing up a bit, unless you're doing this as an exercise (perhaps a homework problem), I'd recommend not reinventing the wheel as printf already supports the printing of integers in hex format:
printf("%x\n", n);