Basic array segmentation fault

**Argentius** · 11-03-2007

Hi,

I'm having a problem with entering numbers into an array and then printing them out again. I'm pretty new to programming which is probably what the problem is...

So far, I have the following:

Code:

#include<stdio.h>

int main(){

int array[10];
int i;

for (i = 0;i <= 10;i++)

        {
        printf("\nEnter a number:");
        scanf("%d", array[i]);
        }

for (i = 0;i <= 10;i++)

        {
        printf("\nArray %d: %d", i, array[i]);
        }
}

When I compile and run the program, it asks me to enter a number. I enter the number and then it says 'Segmentation fault' and closes.

Could anyone please advise on where I'm going wrong with this?

Many thanks

**matsp** · 11-03-2007

Code:

for (i = 0;i <= 10;i++)

        {
        printf("\nEnter a number:");
        scanf("%d", &array[i]);

Your loop runs to 10, but you onlyhave a 0..9 array [ten elements]. Remove the equal sign.

You need to pass the address of the data you are filling in with scanf, so you need an "&" in front of your variable.

--
Mats

**Argentius** · 11-03-2007

I can't believe I missed the &...

Thank you!

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