Free(): invalid pointer in C program

Question

I am executing following C program and getting runtime error as "free(): Invalid Pointer"

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

static void freeArgs(char** args);

int main()
{
    char** argv = calloc(4, 10);
    int  argc = 0;

    argv[argc++]="yogita";
    argv[argc++] ="negi";
    argv[argc] = NULL;
    freeArgs(argv);
    return 0;
}

static void freeArgs(char** args)
{
    char** af = args;
    for (; *af; af++)
        free(*af);
    free(args);
}

Can anyone suggest me the solution?

You never `malloc` (or `callor` or `realloc`) the pointers in `argv`. Yet you pass them to `free`? — Some programmer dude, Feb 12 '21 at 11:17
Also, what do you think will happen if `sizeof(char *) != 4` (like on a typical 64-bit system)? — Some programmer dude, Feb 12 '21 at 11:17
Nevermind my close vote, bad duplicate. I can't be bothered to find a better one. — Lundin, Feb 12 '21 at 11:22

score 4 · Accepted Answer · answered Feb 12 '21 at 11:18

4

free(*af);

tries to free memory that was not allocated through malloc/calloc:

argv[argc++]="yogita";

"yogita" is a string literal, thus not dynamically allocated. You can't free its memory.

answered Feb 12 '21 at 11:18

Jean-Baptiste Yunès

34,548
4
48
69

Vlad from Moscow · Answer 2 · 2021-02-12T11:37:43.023

This call

char** argv = calloc(4, 10);

(Note:it seems you mean

char** argv = calloc( 4, sizeof( char * ) );

end note)

with the magic number 10 allocates dynamically only one extent of memory of the size 4 * 10 that is equal to 40. So you need to call the function free only one time for the allocated extent. That is how many times malloc or calloc was called the same number of times free should be called.

I suspect that you meant something like the following

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

static void freeArgs( char **args )
{
    char **af = args;
    
    while ( *af ) free( *af++ );
    
    free( args );
}

int main(void) 
{
    enum { N = 4, LEN = 10 };
    char **argv = calloc( N, sizeof( char * ) );
    
    size_t  argc = 0;
    
    argv[argc] = malloc( LEN * sizeof( char ) );
    strcpy( argv[argc++], "yogita" );
    
    argv[argc] = malloc( LEN * sizeof( char ) );
    strcpy( argv[argc++], "negi" );
    
    freeArgs( argv );
    
    return 0;
}

Free(): invalid pointer in C program

2 Answers2