Is Taking the Address One Past the End of a C Array Legal?

Is Taking the Address of an Array Element One Past the End Allowed by the C Standard?

The code snippet below has raised discussions about its validity according to the C Standard:

int array[5];
int *array_begin = &array[0];
int *array_end = &array[5];

Specifically, is &array[5] a legal C expression in this context?

Legality According to the C Standard

C Standard defines the legality of this expression:

• C 17 Standard 6.5.2.1, Paragraph 2: Defines E1[E2] as identical to (*((E1) (E2))).
• C 17 Standard 6.5.3.2, Paragraph 3: States that for a [] expression, the & operator and the implied * are not evaluated, resulting in &(E1 E2).
• C 17 Standard 6.5.6, Paragraph 8: Allows pointers to point one element past the end of an array, but prohibits dereferencing them.

Interpretation

By applying these rules to &array[5], we have:

&array[5] = &*(array + 5) = (array + 5)

Since (array 5) points one past the end of the array and is not dereferenced, it is a legal expression according to the C Standard.

Comparison with the C Standard

This behavior is also consistent with the C Standard, which allows taking the address of elements beyond the end of an array, again provided that they are not dereferenced.

Reason for Different Treatment

The decision to treat array 5 and &array[4] 1 differently is rooted in the differing purposes of these expressions:

• array 5 provides a pointer one past the end of the array, allowing for comparisons and traversing beyond the array's bounds.
• &array[4] 1 attempts to create a reference to a nonexistent element, which would lead to undefined behavior.

The above is the detailed content of Is Taking the Address One Past the End of a C Array Legal?. For more information, please follow other related articles on the PHP Chinese website!