Populating an Array in C

Question

I'm new to programming and am having difficulty with arrays. I've declared the array but when I go to print an element of the array I get a very different number (possibly a memory address?).

#include <stdio.h>

int main()
{
    int array[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};

    printf("%d", &array[3]);
    return 0
}

Then it prints 6356748 instead of "3." What am I doing wrong?

if one of the answers has been useful, please convalid it

EmLe49
– EmLe49

2017-05-23 11:17:52 +00:00
Commented May 23, 2017 at 11:17 — EmLe49
– EmLe49, Commented May 23, 2017 at 11:17

Bishakh Ghosh · Accepted Answer · 2017-05-22 05:33:49Z

2

Use:

printf("%d", array[3]);

When you use &variable, it returns the address of the variable.

answered May 22, 2017 at 5:33

Bishakh Ghosh

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CinCout · Accepted Answer · 2017-05-22 05:34:06Z

2

That's because you are printing the address:

printf("%d", &array[3]); // note the use of &

Use this to print the value:

printf("%d", array[3]); // no & used

answered May 22, 2017 at 5:34

CinCout

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Comments

EmLe49 · Accepted Answer · 2017-05-22 05:42:31Z

2

using &array[3] you are not refearing to the value array[3] (you are using a pointer, that is a reference to an address of the memory, you will know it later). You simply need to remove the & :

printf("%d", array[3]);

answered May 22, 2017 at 5:42

EmLe49

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