Is Obtaining the Address One Past the End of a C Array via Subscripting Legal?

Querying the Legality of One-Past-the-End Array Address Acquisition

The Question

In C , can the address of an array element one past its end be obtained through subscripting, as shown in the following code snippet?

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

Specifically, is &array[5] compliant with the C Standard?

The Answer

Yes, the code is compliant with the C Standard.

Explanation

According to the C99 draft standard:

§6.5.2.1, paragraph 2:

A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object.

§6.5.3.2, paragraph 3:

If the operand [of the & operator] is the result of a [] operator, neither the & operator nor the unary * that is implied by the [] is evaluated and the result is as if the & operator were removed and the [] operator were changed to a operator.

§6.5.6, paragraph 8:

If the expression P points to the last element of an array object, the expression (P) 1 points one past the last element of the array object.

Based on these excerpts:

• The & operator yields the address of a pointer's operand.
• When applied to the result of a [] operator, the & operator can be removed and the [] operator becomes a operator.
• An expression pointing to the last element of an array can legally point one past the end of the array.

Therefore, the expression &array[5] is equivalent to *(array 5), which points one past the last element of array. This is legal according to the Standard, as long as the pointer is not dereferenced.

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