Can this code create memory leak and how to test it

Question

Hi i have a source code C as below:

#include <stdio.h>
#include <stdlib.h>
void test(int *a,int n)
{
    a=(int*)malloc(n * sizeof(int)); 
    for(int i=0;i<n;i++)
        a[i] = i+1;
}
int main()
{
    int *ptr;
    int n;
    n = 5;
    printf("Enter number of elements: %d\n", n);
    test(ptr,n);
    if (ptr == NULL) { 
        printf("Memory not allocated.\n");
    }
    else { 
     //...
    } 
    return 0; 
}

As my understanding, when we call function test, the program will create a shadow of pointer ptr to put inside test and then when we go out of test the shadow of ptr will be delete so in the main() the ptr still be NULL, but inside the test we have malloca memory for ptr and this memory is in the heap and it is not free when we go out of test. So if i call test many time this will make memory leak is this true ? And how can i free this memory with free() function in the main ?

score 2 · Answer 1 · answered Jun 26 '20 at 09:29

As my understanding, when we call function test, the program will create a shadow of pointer ptr to put inside test and then when we go out of test the shadow of ptr will be delete so in the main() the ptr still be NULL, but inside the test we have malloc a memory for ptr and this memory is in the heap and it is not free when we go out of test. So if i call test many time this will make memory leak is this true ?

Your understanding is correct. It's enough to call the function once to get a memory leak.

And how can i free this memory with free() function in the main ?

You must return the pointer to the caller in order to enable that. See this:
Dynamic memory access only works inside function

Actually, this question is probably a duplicate to the one I linked. — Lundin, Jun 26 '20 at 09:30

Adam · Accepted Answer · 2020-06-26T13:47:38.113

void test(int *a,int n)
{
    a=(int*)malloc(n * sizeof(int)); 
    for(int i=0;i<n;i++)
        a[i] = i+1;
}

this will lead to memory leak for sure!. No call was encountered to free(void* buffer) before existing the test function. you are getting a copy of the user pointer, since evrey thing is passed by value in C, that means each operation on the copy pointer a will not mirrored or affects the calle pointer (ptr). how to fix?

change the function signature to recieve double pointer! And calling free() before rturn from main if malloc succedd to allocate.

Remember: Allways check malloc() system call return !!

see fixes below:

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

void test(int **a,int n)
{
    int* temp=(int*)malloc(n * sizeof(int));
    if(!temp)  /*always check malloc system call return*/
    {
       *a = NULL;
       return;
    }

    *a = temp;  
    for(int i=0;i<n;i++)
        a[i] = i+1;
}
int main()
{
    int* ptr = NULL;
    ...
    
    test(&ptr, n);
    if(!ptr){
        perror("malloc failed");  /*write to standard error*/ 
        exit(EXIT_FAILURE);
    } 
    ... do work ..
    
    /*if we reach this code it means that malloc succedd then free the ptr which points to the previous allocated heap buffer*/   
    free(ptr);
    return 0; 
}

As my understand, you are using the address of pointer ptr to pass in function test, this mean pass-by-reference is this true ? So if i using a pointer-to-pointer **p to point to *ptr and pass-by-value **p to function test(), it will make the same result. But i think two method will have some different points. Can you explain about the difference between two methods ? — Tiến Thành Nguyễn, Jun 26 '20 at 10:17

Can this code create memory leak and how to test it

2 Answers2