CODEWITHSUNDEEP
cmemory-leaksmalloc
Filipe
Filipe

Reputation: 331

Memory leak | LeakSanitizer: detected memory leaks | C programming

I can't understand why the memory leak below happens, and how I can avoid it.

I am using malloc to allocate memory, and I initiate this memory allocated with a loop, copying value by value from string a to string b.

As far as my understanding goes, I am not leaving any memory whithout use.

my code:

#include <stdlib.h>

void    run_test_one(void)
{
    char    *a;
    char    *b;
    int     i;

    a = "up the irons";
    b = (char *)malloc(13 * sizeof(char));
    i = 0;
    while(a[i] != '\0')
    {
        b[i] = a[i];
        ++i;
    }
    b[i] = '\0';
}

int main(void)
{
    run_test_one();

    return (0);
}

Error:

==9003==ERROR: LeakSanitizer: detected memory leaks

Direct leak of 13 byte(s) in 1 object(s) allocated from:
    #0 0x7ffb28c1db40 in __interceptor_malloc (/usr/lib/x86_64-linux-gnu/libasan.so.4+0xdeb40)
    #1 0x55a9533b29c6 in run_test_one test.c:10
    #2 0x55a9533b2b04 in main test.c:22
    #3 0x7ffb2876fbf6 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x21bf6)

SUMMARY: AddressSanitizer: 13 byte(s) leaked in 1 allocation(s).

Upvotes: 1

Views: 3736

Answers (1)

Maciej Fender
Maciej Fender

Reputation: 331

Memory is allocated by malloc regardless of using it or not. To fix leak you need to free memory.

#include <stdlib.h>

void    run_test_one(void)
{
    char    *a;
    char    *b;
    int     i;

    a = "up the irons";
    b = (char *)malloc(13 * sizeof(char));
    i = 0;
    while(a[i] != '\0')
    {
        b[i] = a[i];
        ++i;
    }
    b[i] = '\0';
    // new code
    free(b)
    // 
}

int main(void)
{
    run_test_one();

    return (0);
}

Second solution is to use automatic allocation. After your program leaves run_test_one function memory is freed:

#include <stdlib.h>

void    run_test_one(void)
{
    char    *a;
    char    *b;
    int     i;

    a = "up the irons";
    //edit
    char b[13];
    //
    i = 0;
    while(a[i] != '\0')
    {
        b[i] = a[i];
        ++i;
    }
    b[i] = '\0';
}

int main(void)
{
    run_test_one();

    return (0);
}

Upvotes: 1

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