Memory allocation and leaking issue?

Question

I am working on the print_selected(int x) function which, takes x then extracts bits 5, 6, 7, and 8 (starting from bit 0 at the rightmost) then print them in both binaries then in hexadecimal.

There is a helper function, int2bin(int n) within the print_selected(int x) and int2bin(int) function returns a char type array to s .

The problem is that I can print out the binary number of bits of 5, 6, 7, and 8bits correctly, however, the hexadecimal number returns somehow weird number(due to the memory leaking issue?). I strongly doubt that char* s = int2bin(x) and free(s) within print_selected(int) might be the problem but I do not know which part I should make a change to correctly print out the right hexadecimal number.

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

// takes integer and returns char array with 32 binary number.
char* int2bin(int n)  
{
    int nbits = sizeof(n) * 8;
    char *s = malloc(nbits + 1);

    s[nbits] = '\0';

    unsigned int u = *(unsigned int*)&n;
    int i;
    unsigned int mask = 1 << (nbits - 1);
    for (i = 0; i < nbits; i++, mask >>= 1)
        s[i] = ((u & mask) != 0) + '0';
    return s;
}

// takes an integer and print in "binary" and "hexadecimal".
void print_selected(int x) 
{
    int hex[4];  // to store the 4 bits of (5,6,7,8) 
    int i;

    char *s = int2bin(x); //  I think this part makes problem?

    printf("bits 5 to 8 in bin: ");
    for (i = 23; i <= 26; i++)
    {
        printf("%d", s[i] - '0');
        hex[i] = s[i] - '0';
    }

    free(s); // I think this part makes problem?

    printf("\n");

    int j = 3; // since we already know the number of bits,
    int sum = 0;

    for (i = 0; i <= 3; i++)
    {
        sum = sum + (hex[i] * pow(2, j));
        j--;
    }

    printf("in hex: 0x%x\n", sum);
}

int main(void)
{
    int a = 278;
    print_selected(a);
}

Answer 1

The problem is here:

    for (i = 23; i <= 26; i++)
    {
        printf("%d", s[i] - '0');
        hex[i] = s[i] - '0';
    }

hex indexes go from 0 to 3 , so assigning to hex[23] causes undefined behavior. Then you try to print the elements of hex , which you never actually filled in.

You need:

hex[i - 23] = s[i] - '0';

Answer 2

however, the hexadecimal number returns somehow weird number(due to the memory leaking issue?).

no, you access to non initialized memory in

 for (i = 0; i <= 3; i++) { sum = sum + (hex[i] * pow(2, j)); j--; } 

because hex is only initialized in

 for (i = 23; i <= 26; i++) { printf("%d", s[i] - '0'); hex[i] = s[i] - '0'; } 

so for the indexes 23 to 26, not for the indexes 0 to 3

note you also write out of hex , with an undefined bahevior, you need to use the index i -23 to write hex

so the value of sum is undefined

I encourage you to use valgrind to detect your memory problems, I used it to answer you