为什么我可以在operator + =的右侧使用初始化列表，但不能使用operator +？

Question

This is a follow-up to an earlier question about why I can't use a brace-enclosed initializer as an argument to operator+ , which was resolved by looking at this earlier question on the subject .

Consider the following C++ code, which you can try live at ideone.com :

#include <iostream>
#include <initializer_list>
using namespace std;

struct AddInitializerList {
    void operator+= (initializer_list<int> values) {
        // Do nothing   
    }

    void operator+ (initializer_list<int> values) {
        // Do nothing
    }
};

int main() {
    AddInitializerList adder;
    adder += {1, 2, 3};  // Totally legit
    adder +  {1, 2, 3};  // Not okay!

    return 0;
}

The line in main that uses operator+ with a brace-enclosed initializer list does not compile (and, after asking that earlier question, I now know why this is). However, I'm confused why the code that uses opeartor+= in main does indeed compile just fine.

I'm confused as to precisely why I can overload += and have it work just fine, while overloading + doesn't seem to work here. Is there a particular provision in the standard that permits brace-enclosed initializers in the context of the += operator but not the + operator? Or is this just a weird compiler quirk?

Answer 1

It is explained in the answer to this question (which is linked from the question you linked to).

The language grammar only allows a braced list in certain grammatical contexts, not in place of an arbitrary expression. That list includes the right-hand side of assignment operators, but NOT the right-hand side of operators in general.

+= is an assignment operator, + is not.

The grammar for assignment expressions is:

\n  assignment-expression: \n     conditional-expression \n     logical-or-expression assignment-operator initializer-clause \n     throw-expression \n  assignment-operator: one of \n      \n  = *= *= /= %= += -= >>= <<= &= ^= |= 

Answer 2

C++14 §5.17/9:

” A braced-init-list may appear on the right-hand side of

• an assignment to a scalar, in which case the initializer list shall have at most a single element. The meaning of x={v} , where T is the scalar type of the expression x , is that of x=T{v} . The meaning of x={} is x=T{} .
• an assignment to an object of class type, in which case the initializer list is passed as the argument to the assignment operator function selected by overload resolution (13.5.3, 13.3).

This applies to a += b via its $5.7/7 equivalence to a = a + b (except that a is evaluated only once for += ). Put another way, due to a comment by MM, because of the equivalence for the built-in operators += is regarded as an assignment operator, and not a special update operator. Hence the quoted text above about “assignment” applies to += .

Answer 3

+= operator is a compound assignment . The standard explicitly permits initializer lists on the right-hand side of assignments:

§8.5.4/1 [...] Note: List-initialization can be used

...

— on the right-hand side of an assignment (5.17)

§5.17 talks about all assignments, including compound ones:

assignment-expression:
- conditional-expression
- logical-or-expression assignment-operator initializer-clause
- throw-expression

assignment-operator: one of
= *= /= %= += -= >>= <<= &= ˆ= |=