Seg Fault trying to assign a value to an array

**Vespasian_2** · 04-05-2016

I create a function called define_array which takes a pointer as a parameter and makes it point to a new array allocated in the heap whilst also returning the new arrays size (in this case hard coded)

Line 11 however gives a Seg Fault when trying to assign a value specifically to any element after 2. It works just fine on the first two elements. Any idea why?

Code:

#include <stdio.h>

#define TEST_ARRAY_SIZE 100

int define_array(int **new_ptr)
{
    *new_ptr = malloc(sizeof(int) * TEST_ARRAY_SIZE);
    int array_size = 3;
    *new_ptr[0] = 40;
    *new_ptr[1] = 50;
    *new_ptr[2] = 20; // seg fault
    printf("%d\n%d", *new_ptr[0], *new_ptr[1]);
    return array_size;
}

int main ()
{
    int *local_ptr = NULL;
    define_array(&local_ptr);
    return 0;
}

**jimblumberg** · 04-05-2016

I can't duplicate your problem, probably because the code doesn't compile on my machine.

main.c||In function ‘define_array’:|
main.c|7|error: implicit declaration of function ‘malloc’ [-Wimplicit-function-declaration]|
main.c|7|warning: incompatible implicit declaration of built-in function ‘malloc’|

It looks like you forgot to include a required file.

However the problem is probably that you need to use parentheses when accessing the array elements.

Code:

    (*new_ptr)[0] = 40;

Jim

**Vespasian_2** · 04-05-2016

Originally Posted by jimblumberg

I can't duplicate your problem, probably because the code doesn't compile on my machine.

It looks like you forgot to include a required file.

However the problem is probably that you need to use parentheses when accessing the array elements.

Code:

    (*new_ptr)[0] = 40;

Jim

Tried your second solution and although there is no compile error, its printing garbage for elements 2 and 3.

For me to try out your first proposed solution, what do I need to do? What do you mean forgot to include a required file?

**jimblumberg** · 04-05-2016

Tried your second solution and although there is no compile error, its printing garbage for elements 2 and 3.

Show your current code.

For me to try out your first proposed solution, what do I need to do? What do you mean forgot to include a required file?

Look up the documentation for the functions you're using and be sure to include the proper include files.

Jim

**Vespasian_2** · 04-05-2016

Originally Posted by jimblumberg

Show your current code.

Look up the documentation for the functions you're using and be sure to include the proper include files.

Jim

Code:

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


#define TEST_ARRAY_SIZE 100


int define_array(int **new_ptr)
{
    *new_ptr = malloc(sizeof(int) * TEST_ARRAY_SIZE);
    int array_size = 3;
    (*new_ptr)[0] = 40;
    (*new_ptr)[1] = 50; // prints garbage
    (*new_ptr)[2] = 20; // prints garbage
    printf("%d\n%d\n%d", *new_ptr[0], *new_ptr[1], *new_ptr[2]);
    return array_size;
}


int main ()
{
    int *local_ptr = NULL;
    define_array(&local_ptr);
    return 0;
}

Ive included stdlib and stdio. Those are the only two libraries I need

**Salem** · 04-05-2016

Originally Posted by Vespasian_2

Tried your second solution and although there is no compile error, its printing garbage for elements 2 and 3.

Did you also apply the () to the expressions in the printf call, as in the assignments?

> *new_ptr[0]
This is index, then dereference.

> (*new_ptr)[0]
This is dereference, then index.

**algorism** · 04-05-2016

Why do you think you need the parentheses when assigning but not when printing? You need them in both places. Or, better yet, print the values in main.

**jimblumberg** · 04-05-2016

Ive included stdlib and stdio. Those are the only two libraries I need

You mean "I've included stdlib.h and stdio.h, those are the only two include files I need." Include files are not libraries.

That is correct you need stdlib.h for malloc() and stdio.h for printf().

Your printing "garbage" because you missed the parentheses in your printf() statement. The problem is being caused by operator precedence, so you need the parentheses for every occurrence.

Jim

**Vespasian_2** · 04-05-2016

F sakes, I forgot to apply the parenthesis in the printf.

Thanks all, it works

Originally Posted by jimblumberg

You mean "I've included stdlib.h and stdio.h, those are the only two include files I need." Include files are not libraries.

That is correct you need stdlib.h for malloc() and stdio.h for printf().

Your printing "garbage" because you missed the parentheses in your printf() statement. The problem is being caused by operator precedence, so you need the parentheses for every occurrence.

Jim

Im assuming the difference between the two is explained here:
c++ - What's the difference between a header file and a library? - Stack Overflow

**jimblumberg** · 04-05-2016

Im assuming the difference between the two is explained here:

Pretty much, yes. Or you could go with a compiler's documentation explanation.

Libraries for use by C programs really consist of two parts: header files that define types and macros and declare variables and functions; and the actual library or archive that contains the definitions of the variables and functions.

(Recall that in C, a declaration merely provides information that a function or variable exists and gives its type. For a function declaration, information about the types of its arguments might be provided as well. The purpose of declarations is to allow the compiler to correctly process references to the declared variables and functions. A definition, on the other hand, actually allocates storage for a variable or says what a function does.)

Jim

**Vespasian_2** · 04-08-2016

Originally Posted by Salem

Did you also apply the () to the expressions in the printf call, as in the assignments?

> *new_ptr[0]
This is index, then dereference.

> (*new_ptr)[0]
This is dereference, then index.

I have tried doing the pointer arithmetic version of this code instead of accessing elements through an array notation. I've been bashing my brains out for hours, its just not working. Check this code (I simplified the array to take 2 elements not 3):

Code:

#include <stdio.h>#define TEST_ARRAY_SIZE 100


int define_array(int **new_ptr)
{
    *new_ptr = malloc(sizeof(int) * TEST_ARRAY_SIZE);
    int array_size = 2;
    int **iterator_pointer = new_ptr;
    (*new_ptr)[0] = 40;
    (*new_ptr)++;//this is the pointer arithmetic solution which doesn't work. This instruction moves the pointer to pointer to point to one element down the array
    **new_ptr = 50;//this then should assign the value of 50 to the next element which was "tagged" before.this is the pointer arithmetic solution which doesn't work. This for some reason DOES NOT update the second element but REPLACES the 40 and the second element is left as garbage values
    //(*new_ptr)[1] = 50; //this was the original array'd solution which works perfect. Thanks for that
    return array_size;
}


void print_array (int *pointer_to_first_element, int array_size)
{
    int counter;
    for(counter = 0; counter < array_size; counter++)
    {
        printf("%d\n", *pointer_to_first_element);
        pointer_to_first_element++;
    }
}


int main ()
{
    int *local_ptr = NULL;
    int array_size;
    array_size = define_array(&local_ptr);
    print_array (local_ptr, array_size);
    return 0;
}

You will see that your original array'd solution is in there and it works perfectly. But when I try the pointer arithmetic version it doesnt work:
The statements on 10 and 11 for some reason DOES NOT update the second element but REPLACES the 40 and the second element is left as garbage values. WHY on earth is it doing this???

**Salem** · 04-08-2016

> (*new_ptr)++;//this is the pointer arithmetic solution which doesn't work. This instruction moves the pointer to pointer to point to one element down the array
The difference between what you wrote, and

(*new_ptr)[1] = 50;
is that it does NOT modify new_ptr, or what it points to.

int *temp = *new_ptr;
*temp++ = 40;
*temp++ = 50;
is another way.

**Vespasian_2** · 04-11-2016

Originally Posted by Salem

> (*new_ptr)++;//this is the pointer arithmetic solution which doesn't work. This instruction moves the pointer to pointer to point to one element down the array
The difference between what you wrote, and

(*new_ptr)[1] = 50;
is that it does NOT modify new_ptr, or what it points to.

int *temp = *new_ptr;
*temp++ = 40;
*temp++ = 50;
is another way.

Oh I tried to manipulate with pointer to pointer but actually needed to manipulate with a pointer. Thanks

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