CODEWITHSUNDEEP
cmalloc
Manoj
Manoj

Reputation: 863

Free memory in another function

So I have this code in which allocation is happening in one function and deallocation is being done in the calling function. Getting Segmentation fault or Abort message while trying to free the memory either by ptr or *ptr. Kindly have a look:

#include <stdio.h>

int main()
{

    char *ptr;
    fun(&ptr);
    printf("ptr = %p\n",ptr);
    printf("&ptr = %p\n",&ptr);
    printf("String ptr = %s\n",ptr);
    free (ptr);

    return 0;
}
void fun(char **str)
{
    *str = malloc(10);
    *str = "HELLO";
    printf("str = %p\n",str);
    printf("&str = %p\n",&str);

    printf("String str = %s\n",*str);
}

Following is the output:

str = 0x7ffe63247858                                                                                                                 
&str = 0x7ffe63247838                                                                                                                
String str = HELLO                                                                                                                   
ptr = 0x400764                                                                                                                       
&ptr = 0x7ffe63247858                                                                                                                
String ptr = HELLO                                                                                                                   
*** Error in `/home/a.out': munmap_chunk(): invalid pointer: 0x0000000000400764 ***                                                  
Aborted

Question :

Why can't we free ptr ? And if we can, what is the best way to do it ?

Upvotes: 0

Views: 179

Answers (4)

Paresh Dhandhukiya
Paresh Dhandhukiya

Reputation: 212

you are changing memory address pointed str pointer after malloc in fun. you should do strcpy instead of direct assignment :

#include <stdio.h>

int main()
{

    char *ptr;
    fun(&ptr);
    printf("ptr = %p\n",ptr);
    printf("&ptr = %p\n",&ptr);
    printf("String ptr = %s\n",ptr);
    free (ptr);

    return 0;
}
void fun(char **str)
{
    *str = malloc(10);
    strcpy(*str,"HELLO");//*str = "HELLO";
    printf("str = %p\n",str);
    printf("&str = %p\n",&str);

    printf("String str = %s\n",*str);
}

Upvotes: 0

Stephen Docy
Stephen Docy

Reputation: 4788

To answer the question of how to do this without explicit string copying, instead of

char *p = "hello";

do

char p[] = "hello";

The first points p to a non-modifiable portion of memory. The second allocates an array large enough to hold the literal and copies the literal into it.

Upvotes: 0

Jean-Fran&#231;ois Fabre
Jean-Fran&#231;ois Fabre

Reputation: 140168

you're overwriting the pointer value in the line after, making *str point to a literal.

So you have a memory leak and undefined behaviour by trying to free a literal.

You probably mean

strcpy(*str,"hello");

Note that those 2 lines could be changed into a working single line like this:

*str = strdup("HELLO");

which has the advantage of allocating the exact number of bytes for the string instead of guessing.

Upvotes: 6

Ivaylo Strandjev
Ivaylo Strandjev

Reputation: 70929

When you do *str = "HELLO"; it does not have the effect you expect. Instead of assigning value to the allocated memory, you are re-assigning the pointer to point to the string literal and thus later you try to deallocate it (thus the error). Instead of this you should use strcpy like strcpy(*str, "HELLO").

Upvotes: 1

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