I have to ask about aborted(core dumped) error i get from this code-CodePudding

i got that error while compiling this code. It prints maximum number between 5 numbers using dynamic memory allocation.

--------------------------------------------------------------------c

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

int main() {
    int* ptr = (int*)malloc(5*sizeof(int));
    *ptr = 5;
    *(ptr 1) = 7;
    *(ptr 2) = 2;
    *(ptr 3) = 9;
    *(ptr 4) = 8;

    int* ptr_max = (int*)malloc(sizeof(int));
    *ptr_max = 0;

    for(int i = 0; i < 5; i  ){
        if(*ptr_max < *ptr){
            *ptr_max = *ptr;
            ptr  ;
        }
        else
            ptr  ;
    }   
    printf("%d\n", *ptr_max);
    free(ptr);
    free(ptr_max);
    return 0;
}

I want to know why exactly i get this error from this code please can you explain me ?

CodePudding user response：

The real problem lies when you are free()-ing the ptr. Once, you increment a pointer its address jumps to next address allocated by malloc(). Always make sure that *ptr is same as ptr[0]. So, to fix this issue, you can decrement the ptr by 5, or create a copied pointer.

Example of address given to free(), they are not pointing to the same memory block:

Before decrementing 0x213f2b4
After decrementing 0x213f2a0

The reason for decrementing by 5, is the difference between these two hexadecimal values which is 20, same as sizeof(int) * 5.

ptr -= 5;

You can create a copy of your pointer and then perform operations on that copied one:

int *my_copied_ptr = ptr; // you don't need to free this pointer

Then, free() them:

free(ptr);
free(ptr_max);

Now, to avoid these mistakes further in a large code bases, try using [] operator like this:

ptr[0] = 5;
ptr[1] = 7;
ptr[2] = 2;
ptr[3] = 9;
ptr[4] = 8;