Dynamic Memory Allocation in C ( return value 3221225477 & Segmentation fault: 11 )

Question

Can anyone explain what is the wrong of the code below? İf I give the "line" variable as 1,2,3 or 4, it runs correctly. But after 4 it just shows a count of row correctly, the count of columns is not correct.

output of dev-c/c++ on windows >> "return value 3221225477"

output of gcc on mac >> "Segmentation fault: 11"

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

int main(){

int **matris;
int size;
int i, j;

printf( "Size >> " );
scanf( "%d", &size );

matris = (int **)malloc( size * sizeof(int) );

if( matris == NULL )
    printf( "It's required more memory'!" );

for( i = 0; i < size; i++ ) {
    matris[i] = malloc( size * sizeof(int) );
    if( matris[i] == NULL )
        printf( "It's required more memory'!!" );
}

for( i = 0; i < size; i++ ) {
    for( j = 0; j < size; j++ ){    
        matris[i][j]=i+1;
        printf( "%d ", matris[i][j] );
    }
    printf( "\n" );
}


for( i = 0; i < size; i++ ) {
    free( matris[i] );
}

free( matris );

return 0;
}

`(int **)malloc( size * sizeof(int) );` looks wrong, it should be `sizeof(int*)` — UnholySheep, Jun 26 '20 at 11:08
`matris = (int **)malloc( size * sizeof(int) );` You should not cast the return value of `malloc` in C. And you do not allocate correct size. `matrix` points to `int*`, not `int`. — Gerhardh, Jun 26 '20 at 11:08
Run it in gdb and it will tell you exactly what the issue is. — lostbard, Jun 26 '20 at 11:08
Others have already posted the solution. Try printing the sizes of ints vs int pointer with `printf("int: %zu, ptr: %zu", sizeof(int), sizeof(int*));`. On an x86_64/AMD64 machine, that gives me `int: 4, ptr: 8`. That's why your code crashes. — Benjamin Maurer, Jun 26 '20 at 11:15
Just replace all of the icky, segmented bugginess with `int (*matris)[j] = malloc(sizeof(int[size][size]));` ... `free(matris);`. — Lundin, Jun 26 '20 at 11:41

RobertS supports Monica Cellio · Answer 1 · 2020-06-26T11:37:21.680

matris = (int **)malloc( size * sizeof(int) );

with especially ( size * sizeof(int) ) is wrong. You need to allocate memory for objects of type int*(pointer to int), not int.

This is a problem if sizeof(int) < sizeof(int*) on your implementation -> Translated: The size of an object of type pointer to int is bigger than the size of an object of type int, which is quite common.

Then you get a segmentation fault, because you access memory beyond the bounds of the pointer array matris in the following code.

Use:

matris = malloc ( sizeof(int*) * size );

or even better

matris = malloc ( sizeof(*matris) * size );

Side Notes:

You don't need to cast the return of malloc(). The returned pointer is automatically aligned. Do I cast the result of malloc?
sizeof(*matris) requires less maintenance and is more safe if the type of matris and the pointed objects changes.
Set the sizeof operation first to ensure size_t arithmetic, since size is of type int.

Deepak Patankar · Answer 2 · 2020-06-26T14:24:58.333

0

The line matris = (int **)malloc( size * sizeof(int) ); is incorrect as you want to create an array of int pointers and not of int. Changing the int to int* works.

edited Jun 26 '20 at 14:24

answered Jun 26 '20 at 11:14

Deepak Patankar

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`malloc( size * sizeof(int) );` actually creates an array of `int` and the OP wants an array of int pointers. You probably meant the other way around? – Lundin Jun 26 '20 at 11:44
Hey @Lundin, I meant the other way, thanks for correcting. – Deepak Patankar Jun 26 '20 at 14:25

Dynamic Memory Allocation in C ( return value 3221225477 & Segmentation fault: 11 )

2 Answers2