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'Segmentation fault' is very familiar error for C developer.

Then, when this runtime error is issued?

At the moment of accessing not-allocated-memory?

This may be right, but this is NOT strict enough.

'Segmentation fault' is issued by kernel through signal to notify that,

"You - user process - try to access not-permitted memory area. And usually, this happens when it tries to access not-allocated-memory - for the user process (In case of ARM architecture, it is triggered by Data Abort cpu exception).

Let's see below code.

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

int
main(int argc, const char *argv[]) {
        int i = 0;
        char *s = malloc(1);
        s += 1;
        for (i = 0; i < 1024 * 1024 * 1024; i++) {
                printf("%d\n", i);
                s[i] = 4;
        }
        return 0;
}

Do you think when this process is stopped by 'Segmentation fault' signal?

"i == 0" ? (that is at the first moment of trying to access not-allocated-memory-by-malloc).

It, is totally up to malloc implementation.

But, at above case, this is definitely NO.

The reason is, kernel(linux) allocate memory for user process at least 1-page.

So, above code will run even if i > 1024 especially on the Linux machine.

And, there are various implementation(algorithm) of malloc.

So, we cannot expect exact value of i for above code.

Important thing to know here is, answer the question - "What exactly does 'Segmentation fault' mean?"

See https://gcc.gnu.org/onlinedocs/gcc-4.4.4/gcc/Designated-Inits.html

And here is one more tips.

Sometimes, you may want to initialize all array element with same value.

Then, you can do it as follows if you are using GCC.

static const int arr[10] = {[0 ... (sizeof(arr)/sizeof(arr[0]) - 1)] = 10};