c,pointers,memory-management,global-variables , munmap_chunck(): invalid pointer in c


munmap_chunck(): invalid pointer in c

Question:

Tag: c,pointers,memory-management,global-variables

I have a program that I've written to do some data analysis, and this data is stored in a global structure called P. I allocate the memory for this structure in one function and then, since I need it for the entire program, don't call free until the very end of main. Malloc is not called in main because the size of the array is obtained from a file, so I figured that it made the most sense to read in the file and allocate the memory there, rather than doing all of it in main.

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

typedef struct DATA
{
    /* variables*/
} DATA;

DATA *P;
void function1(void);

int main(int argc, char **argv)
{
    function1();
    /*do some stuff*/
    free(P);
    return 0;
}

void function1(void)
{
    if(!(P = (DATA *)malloc(nums * sizeof(DATA))))
    {
        printf("Error!\n");
        exit(EXIT_FAILURE);
    }
    /*do stuff*/
}

Essentially, when I run my code, I get the error: munmap_chunck(): invalid pointer. I've done a bit of reading and it seems to be related to the function free. I've also read that malloc and free should be called in the same function. If this is the case, then my question is: since P is a global variable, why does it matter which function malloc and free are called in for this particular variable? If, in fact, the problem is not caused by malloc and free being called in different functions, does anyone have any advice on what it might be instead? Thank you very much!


Answer:

This is a common problem which occurs when you provide free a pointer which doesn't point to the start of an allocated memory block. For example, the code

int* something = malloc(sizeof(int)); // Allocate space for 1 int
...
free(something);

will work fine as you are providing free with the original pointer returned by malloc. However, if you do something like this:

int* something = malloc(sizeof(int) * 5); // Allocate space for 5 ints
...
something += sizeof(int); // Shift the allocated memory by 1 int
...
free(something);

Then free has been provided a pointer to somewhere in the middle of the allocated memory block. This means that essentially, free has no idea where the block starts or ends, therefore throwing an error. This can be fixed by storing the original pointer value in another pointer:

int* something = malloc(sizeof(int) * 5);
int* another_something = something; // `another_something` contains the same pointer value as `something`
...
something += sizeof(int); // `something` changes, `another_something` doesn't
...
free(another_something); // Free the original pointer

So, in your case, something like this:

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

typedef struct DATA
{
    /* variables*/
} DATA;

/* Declare another DATA*, P1 */
DATA *P, *P1;
void function1(void);

int main(int argc, char **argv)
{
    function1();
    /*do some stuff*/

    /* Free the original pointer, in P1 */
    free(P1);
    return 0;
}

void function1(void)
{
    if(!(P = (DATA *)malloc(nums * sizeof(DATA))))
    {
        printf("Error!\n");
        exit(EXIT_FAILURE);
    }

    /* Store the original value of `P` in P1 */
    P1 = P;

    /*do stuff*/
}

Lastly, please fix your naming conventions. P and function1 are awful names.


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