#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int get_val(char hex_digit) {
if (hex_digit >= '0' && hex_digit <= '9') {
return hex_digit - '0';
} else {
return hex_digit - 'A' + 10;
}
}
void convert_to_oct(const char* hex, char** res) {
int oct_len = (hex_len/3) * 4;
int i;
// One hex digit left that is 4 bits or 2 oct digits.
if (hex_len%3 == 1) {
oct_len += 2;
} else if (hex_len%3 == 2) { // 2 hex digits map to 3 oct digits
oct_len += 3;
}
(*res
) = malloc((oct_len
+1) * sizeof(char)); (*res)[oct_len] = 0; // don't forget the terminating char.
int oct_index = oct_len - 1; // position we are changing in the oct representation.
for (i = hex_len - 1; i - 3 >= 0; i -= 3) {
(*res)[oct_index] = get_val(hex[i]) % 8 + '0';
(*res)[oct_index - 1] = (get_val(hex[i])/8+ (get_val(hex[i-1])%4) * 2) + '0';
(*res)[oct_index - 2] = get_val(hex[i-1])/4 + (get_val(hex[i-2])%2)*4 + '0';
(*res)[oct_index - 3] = get_val(hex[i-2])/2 + '0';
oct_index -= 4;
}
// if hex_len is not divisible by 4 we have to take care of the extra digits:
if (hex_len%3 == 1) {
(*res)[oct_index] = get_val(hex[0])%8 + '0';
(*res)[oct_index - 1] = get_val(hex[0])/8 + '0';
} else if (hex_len%3 == 2) {
(*res)[oct_index] = get_val(hex[1])%8 + '0';
(*res)[oct_index - 1] = get_val(hex[1])/8 + (get_val(hex[0])%4)*4 + '0';
(*res)[oct_index - 2] = get_val(hex[0])/4 + '0';
}
}
int main() {
const char* hex_val = "A111";
char * res;
convert_to_oct(hex_val, &res);
return 0;
}