[c] How can I convert an integer to a hexadecimal string in C?

How can I convert an integer to a hexadecimal string in C?

Example: The integer 50 would be converted to the hexadecimal string "32" or "0x32".

Interesting that these answers utilize printf like it is a given. printf converts the integer to a Hexadecimal string value.

//*************************************************************
// void prntnum(unsigned long n, int base, char sign, char *outbuf)
// unsigned long num = number to be printed
// int base        = number base for conversion;  decimal=10,hex=16
// char sign       = signed or unsigned output
// char *outbuf   = buffer to hold the output number
//*************************************************************

void prntnum(unsigned long n, int base, char sign, char *outbuf)
{

    int i = 12;
    int j = 0;

    do{
        outbuf[i] = "0123456789ABCDEF"[num % base];
        i--;
        n = num/base;
    }while( num > 0);

    if(sign != ' '){
        outbuf[0] = sign;
        ++j;
    }

    while( ++i < 13){
       outbuf[j++] = outbuf[i];
    }

    outbuf[j] = 0;

}

The following code takes an integer and makes a string out of it in hex format:

int  num = 32424;
char hex[5];

sprintf(hex, "%x", num);
puts(hex);

gives

7ea8

I made a librairy to make Hexadecimal / Decimal conversion without the use of stdio.h. Very simple to use :

char* dechex (int dec);

This will use calloc() to to return a pointer to an hexadecimal string, this way the quantity of memory used is optimized, so don't forget to use free()

Here the link on github : https://github.com/kevmuret/libhex/

This code

int a = 5;
printf("%x\n", a);

prints

5

This code

int a = 5; 
printf("0x%x\n", a);

prints

0x5

This code

int a = 89778116;
printf("%x\n", a);

prints

559e7c4

If you capitalize the x in the format it capitalizes the hex value:

int a = 89778116;
printf("%X\n", a);

prints

559E7C4

If you want to print pointers you use the p format specifier:

char* str = "foo";
printf("0x%p\n", str);

prints

0x01275744

Usually with printf (or one of its cousins) using the %x format specifier.

To convert an integer to a string also involves char array or memory management.

To handle that part for such short arrays, code could use a compound literal, since C99, to create array space, on the fly. The string is valid until the end of the block.

#define UNS_HEX_STR_SIZE ((sizeof (unsigned)*CHAR_BIT + 3)/4 + 1)
//                         compound literal v--------------------------v
#define U2HS(x) unsigned_to_hex_string((x), (char[UNS_HEX_STR_SIZE]) {0}, UNS_HEX_STR_SIZE)

char *unsigned_to_hex_string(unsigned x, char *dest, size_t size) {
  snprintf(dest, size, "%X", x);
  return dest;
}

int main(void) {
  // 3 array are formed v               v        v
  printf("%s %s %s\n", U2HS(UINT_MAX), U2HS(0), U2HS(0x12345678));
  char *hs = U2HS(rand());
  puts(hs);
  // `hs` is valid until the end of the block
}

Output

FFFFFFFF 0 12345678
5851F42D