c++ - 为什么覆盖全局新运算符和类特定运算符不是模棱两可的行为？

Question

考虑以下代码:

class Foo 
{
public:
    //class-specific
    Foo operator+(Foo& rhs)
    {
       return Foo(); //Just return a temporary
    }

    void* operator new(size_t sd)
    {
        return malloc(sd);
    }
};

//global
Foo operator+(Foo& lhs, Foo& rhs)
{
    return Foo();
}

void* operator new(size_t sd)
{
    return malloc(sd);
}

此代码无法编译，声明调用不明确，因为它匹配两个运算符:

Foo a, b;
a + b;

但是这个带有 new 运算符的编译得很好，并且会调用特定于类的那个。

Foo* a = new Foo();

为什么它不会导致编译错误？编译器是否以不同方式对待 new 运算符？ (对标准的任何引用将不胜感激。)

Answer 1

Why doesn't it result in a compile error? Does the compiler treat the new operator differently? (Any citation to the standard would be appreciated)

关于全局 new 和类特定 new 之间的优先级，引用文献说 this :

As described in allocation function, the C++ program may provide global and class-specific replacements for these functions. If the new-expression begins with the optional :: operator, as in ::new T or ::new T[n], class-specific replacements will be ignored (the function is looked up in global scope). Otherwise, if T is a class type, lookup begins in the class scope of T.

因此类特定的 new 具有优先权。

关于+的重载，可以是成员重载或者全局重载(通常作为类的friend)但不是两者都是因为它产生的歧义。