C - Segmentation fault with pointer to array

Question

i'm hoping someone can help me with this little piece of code. It's a stupid test, but i dont know what is it that i am doing wrong. It's like this:

#include <stdio.h>

int **ipp;

int ventas [3][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};

int main(void){

    ipp = (int **)ventas;

    printf("%d\n", **ipp);

    return 0;
}

It compiles (I'm using GCC), but when I execute it I keep getting a segmentation fault. What am I doing wrong? I think it has something to do with an un-initalized pointer, but 'ventas' is an array so it is already initialized, and it's assigned to **ipp.

Hint: Check the types.

Sourav Ghosh
– Sourav Ghosh

2016-11-10 14:13:58 +00:00
Commented Nov 10, 2016 at 14:13 — Sourav Ghosh
– Sourav Ghosh, Commented Nov 10, 2016 at 14:13

Vlad from Moscow · Accepted Answer · 2016-11-10 15:04:56Z

1

When you have such casting like

ipp = (int **)ventas;

then the value of the variable is the address of the first element of the array ventas. In this case after dereferencing the pointer

*ipp

you get the value stored at this address. If to assume that sizeof( int ) is equal to sizeof( int * ) than the first element of the array equal to 1 is considered as a memory address. After applying second dereferencing

**ipp

you get memory access violation.

It will be correct to write either like

int ( *ipp )[4] = ventas;

and then

printf("%d\n", **ipp);

or like

int *ipp = ( int * )ventas;

and then

printf("%d\n", *ipp);

edited Nov 10, 2016 at 15:04

answered Nov 10, 2016 at 14:22

Vlad from Moscow

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Comments

Lundin · Accepted Answer · 2016-11-10 14:18:12Z

1

A pointer-to-pointer is not an array. Nor is it a pointer to a 2D array. They aren't the slightest compatible. Just forget about pointer-to-pointers in this case.
If you want a pointer to a 2D array, you must write int (*ipp)[3][4] = &ventas;
You can't print a pointer using %d format specifier. You should use %p.

Corrected code for printing the address of the 2D array:

#include <stdio.h>

int main(void){
  int ventas [3][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
  int (*ipp)[3][4];

  ipp = &ventas;
  printf("%p\n", ipp);

  return 0;
}

answered Nov 10, 2016 at 14:18

Lundin

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3 Comments

Michi Over a year ago

printf("%p\n", ipp); should not be printf("%p\n", (void*)ipp); ?

Lundin Over a year ago

@Michi Yes indeed, if we are picky.

Elias Van Ootegem Over a year ago

strictly speaking, printf("%p\n", ipp); should be printf("%p\n", (void *) ipp);

David Ranieri · Accepted Answer · 2016-11-10 14:29:16Z

1

A pointer to pointer and a 2D array are not interchangeable, change to:

int (*ipp)[4]; /* A pointer to an array of 4 ints */
...
ipp = ventas;

edited Nov 10, 2016 at 14:29

answered Nov 10, 2016 at 14:14

David Ranieri

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1 Comment

Lundin Over a year ago

It appears that the OP wants a pointer to a 2D array. Anyway, for the purpose of printing the address, a pointer to a 2D array or a pointer to the first element will give the same result.

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