C ++：是一个L值的返回值？ [英] C++: is return value a L-value?

问题描述

请考虑以下代码：

struct foo
{
  int a;
};

foo q() { foo f; f.a =4; return f;}

int main()
{
  foo i;
  i.a = 5;
  q() = i;
}

没有编译器抱怨，甚至Clang。为什么 q（）= ... 行是正确的？

No compiler complains about it, even Clang. Why q() = ... line is correct?

推荐答案

不，当且仅当它是一个引用（C ++ 03）时，函数的返回值是一个l值。 （5.2.2 [expr.call] / 10）

No, the return value of a function is an l-value if and only if it is a reference (C++03). (5.2.2 [expr.call] / 10)

如果返回的类型是基本类型，那么这将是一个编译错误。 （5.17 [expr.ass] / 1）

If the type returned were a basic type then this would be a compile error. (5.17 [expr.ass] / 1)

这是因为你可以调用成员函数（甚至非 - const 成员函数）对类类型的r值和 foo 的赋值是一个实现定义的成员函数： foo& foo :: operator =（const foo&）。第5节中对运算符的限制仅适用于内置运算符（5 [expr] / 3），如果重载解析为运算符选择了重载函数调用，则对该函数调用的限制适用

The reason that this works is that you are allowed to call member functions (even non-const member functions) on r-values of class type and the assignment of foo is an implementation defined member function: foo& foo::operator=(const foo&). The restrictions for operators in clause 5 only apply to built-in operators, (5 [expr] / 3), if overload resolution selects an overloaded function call for an operator then the restrictions for that function call apply instead.

这就是为什么有时候推荐返回类型为 const 对象的对象（例如 const foo q（）; ），但是这可能对C ++ 0x有负面影响，它可以阻止move语义从他们应该工作。

This is why it is sometimes recommended to return objects of class type as const objects (e.g. const foo q();), however this can have a negative impact in C++0x where it can inhibit move semantics from working as they should.

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