How is this array declaration in C possible?

Question

int arr[5]={1,2,3,4,5,6,7,8,9};

This way of declaration is not giving an error and the array is stored up till the 4th index, if I try to output arr[5] it will give garbage value. Can anyone explain how this way is possible?

Edit: I was trying to run the following code in an online compiler:

#include <stdio.h>
int main() {
    int arr[5]={1,2,3,4,5,6,7,8,9};
    int i;
    for(int i=0; i<6;i++){
        printf("arr[%d]=%d\n", i,arr[i]);
    }
    return 0;
}

You might tell us what compiler you are using. You should get some diagnostic message about providing excess elements in array initializer — Gerhardh
– Gerhardh, Commented Aug 24, 2021 at 12:33
It's possible because the compiler is not required to refuse it. Hmm, perhaps it is required to refuse it. Add -pedantic-errors if you're using gcc or clang, /permissive- if you are using MSVC. The program has undefined behavior anyway though. — Ted Lyngmo
– Ted Lyngmo, Commented Aug 24, 2021 at 12:33

Andreas Wenzel · Accepted Answer · 2021-08-24 13:10:33Z

6

It will initialize only 5 elements of the array. The remaining initializers will be ignored. The compiler will emit a diagnostic message (warning).

<source>:11:23: note: (near initialization for 'arr')
<source>:11:25: warning: excess elements in array initializer

This way of declaration is not giving an error and the array is stored up till the 4th index

This is correct - you need to read the warning messages and notes. Do not ignore warnings!!

if I try to output arr[5] it will give garbage value

You invoke Undefined Bahavour by reading outside the array bounds.

edited Aug 24, 2021 at 13:10

Andreas Wenzel

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answered Aug 24, 2021 at 12:33

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2 Comments

Jonathan Leffler Over a year ago

The standard requires a diagnostic for a constraint violation.

0___________ Over a year ago

@JonathanLeffler ammended

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