is int[pointer-to-array] in the C++ - standard? [duplicate]

问题

As I have learned, one can write the following code:

char *a = new char[50];
for (int i = 0; i < 50; ++i) {
    i[a] = '5';
}

It compiles. It works. It does exactly the same as

char *a = new char[50];
for (int i = 0; i < 50; ++i) {
    a[i] = '5';
}

Is it just because:

a[b] is implemented as a macro *(a + b) by default and the fact that both code samples are valid is just an accident/compiler specific
it's standardized somewhere and the outcome of such algorithms should be the same on every platform

It is reasonable to assume that addition should be commutative, but if we implement operator[] in that way, we have made something else commutative, what might not be what we wanted.

The interesting fact is that there is no pointer[pointer] operator, so operator[] is not a macro.

I know it's bad. I know it's confusing the people who read the code. But I want to know if it's just an accident and it will not work in a distant land where unicorns have seven legs and horns are on their left cheek.

回答1:

C++ standard, § 8.3.4, note 7 (page 185) (emphasis mine).

Except where it has been declared for a class (13.5.5), the subscript operator [] is interpreted in such a way that E1[E2] is identical to *((E1)+(E2)). Because of the conversion rules that apply to +, if E1 is an array and E2 an integer, then E1[E2] refers to the E2-th member of E1. Therefore, despite its asymmetric appearance, subscripting is a commutative operation.

回答2:

Here is what C++11 standard has to say:

Note: Except where it has been declared for a class (13.5.5), the subscript operator [] is interpreted in such a way that E1[E2] is identical to *((E1)+(E2)). Because of the conversion rules that apply to +, if E1 is an array and E2 an integer, then E1[E2] refers to the E2-th member of E1. Therefore, despite its asymmetric appearance, subscripting is a commutative operation. (emphasis is added).

So your assumption that a[b] is implemented as *(a + b) is correct, except that it is implemented directly in the compiler, not as a macro.

回答3:

The expression E1[E2] is identical (by definition) to *((E1)+(E2))

...and then commutativity of index and pointer takes hold. See your friendly neighbourhood C++ standard, section 5.2.1 in this version: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2012/n3485.pdf

来源：https://stackoverflow.com/questions/23872692/is-intpointer-to-array-in-the-c-standard

标签

c++

arrays

pointers

syntax

subscript-operator