Is &array[5] a Valid C Expression for Accessing the One-Past-the-End Array Element?

Accessing One-Past-the-End Array Elements with &array[5]

Question:

Is &array[5] considered a valid C code expression that references the element one past the end of an array?

Answer:

Yes, &array[5] is considered legal C code according to the C Standard.

Detailed Explanation:

C99 Standard:

• 6.5.2.1, Paragraph 2: Defines the subscript operator [] as equivalent to (((E1) (E2))), where E1 is an array object and E2 is an integer.
• 6.5.3.2, Paragraph 3: States that if the operand of the & operator is the result of a [] operator, the & operator and the implied unary * are not evaluated, resulting in the & operator being removed and the [] operator replaced with a operator.

C Standard:

• 6.5.6, Paragraph 8: Allows pointers to point one past the last element of an array provided they are not dereferenced.

Based on these standards, the expression &array[5] is evaluated as follows:

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

Since array 5 points one past the end of the array and is not dereferenced, &array[5] is a valid expression.

Comparison with C Standard:

The C standard matches the C standard in this regard.

Reason for Treating It Differently from array 5 or &array[4] 1:

The main difference between &array[5] and expressions like array 5 or &array[4] 1 is their intended use. While array 5 and &array[4] 1 explicitly perform pointer arithmetic to obtain a pointer that is offset from the start of the array, &array[5] uses the [] operator to directly access the element at that offset. This distinction allows programmers to more easily reference the end of the array without having to perform explicit pointer arithmetic.

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