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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) { ...
#3: Post edited
by
klutt
·
2020-08-19T10:49:23Z (about 4 years ago)
- 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 invokes undefined behaviour? Is this true? What does the standard say?
- 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?
#2: Post edited
by
klutt
·
2020-08-18T09:36:25Z (about 4 years ago)
- 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:
char arr[10];char *ptr = arr[-1];char c = *ptr;The third line is obviously bad, because it's accessing out of bounds. But I heard something that even the second line invokes undefined behaviour? Is this true? What does the standard say?
- 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 invokes undefined behaviour? Is this true? What does the standard say?
#1: Initial revision
by
klutt
·
2020-08-11T08:45:25Z (about 4 years ago)
Is it undefined behaviour to just make a pointer point outside boundaries of an array without dereferencing it?
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:
char arr[10];
char *ptr = arr[-1];
char c = *ptr;
The third line is obviously bad, because it's accessing out of bounds. But I heard something that even the second line invokes undefined behaviour? Is this true? What does the standard say?