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Comments on Is it undefined behaviour to just make a pointer point outside boundaries of an array without dereferencing it?

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Is it undefined behaviour to just make a pointer point outside boundaries of an array without dereferencing it?

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I have heard that it is undefined behaviour to make a pointer point outside boundaries of an array even without dereferencing it. Can that really be true? Consider this code:

int main(void) 
{
    char arr[10];
    char *ptr = &arr[-1];
    char c = *ptr;
}

The line char c = *ptr is obviously bad, because it's accessing out of bounds. But I heard something that even the second line char *ptr = &arr[-1] invokes undefined behaviour? Is this true? What does the standard say?

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1 comment thread

General comments (5 comments)
General comments
Olin Lathrop‭ wrote over 4 years ago

Your example might be better if you referenced arr[10] instead of arr[-1]. That eliminates issues of signed versus unsigned arithmetic of array subscripts, which it sounds like is not what you are trying to ask about.

klutt‭ wrote over 4 years ago

@olinlathrop Except that arr[10] is legal to point at, but not arr[11] ;) I mostly posted this question to get this site going.

Skipping 1 deleted comment.

klutt‭ wrote over 4 years ago

@kami Nope, that line is exactly as it should be.

Lundin‭ wrote over 4 years ago

Actually, char *ptr = arr[-1]; is not valid, there needs to be an & or it's a constraint violation of simple assignment. I didn't think of it when I originally answered the question.

Kami‭ wrote over 4 years ago

@klutt sneaky edit is sneaky :P I really miss a note when the last edit was...