Is `&array[5]` a Valid Pointer in C Beyond Array Bounds?

Can Array Subscripting Point Beyond Array Bounds?

The question arises whether the following code is compliant with the C Standard:

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

Specifically, the validity of &array[5] in this context is under scrutiny.

C Standard

According to the C99 draft standard:

• §6.5.2.1, Paragraph 2: "E1[E2] is identical to (*((E1) (E2)))"
• §6.5.3.2, Paragraph 3 (Emphasized): "If the operand 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 one past the last element of an array object, the expression (P) 1 points one past the last element of the array object, and if the expression Q points one past the last element of an array object, the expression (Q) - 1 points to the last element of the array object."

Conclusion

These provisions indicate that:

• Pointers can legitimately point one element past the end of an array (without being dereferenced).
• &array[5] is equivalent to &*(array 5), which in turn is equivalent to (array 5).
• (array 5) points one past the end of the array.
• Since &array[5] does not result in a dereference, it is legal according to the C Standard.

The above is the detailed content of Is `&array[5]` a Valid Pointer in C Beyond Array Bounds?. For more information, please follow other related articles on the PHP Chinese website!