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); }

There's no problem with free(s). You don't use that memory any more. — Barmar
– Barmar, Commented May 6, 2019 at 16:34
There is a compiler warning for pow(2, j). You don't need maths for this, just (1 << j). — Weather Vane
– Weather Vane, Commented May 6, 2019 at 16:34

bruno · Accepted Answer · 2019-05-06 16:47:44Z

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

@SangminChoi you welcome, happy coding, and use valgrind, that tool is just magic ;-)

Barmar · Accepted Answer · 2019-05-06 16:41:22Z

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';

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