为什么左值转换有效? [英] Why does an lvalue cast work?

问题描述

我今天第一次看到这种演员阵容，我很好奇为什么会这样.我认为以这种方式铸造会分配给临时成员，而不是班级成员.使用 VC2010.

I saw this kind of cast for the first time today, and I'm curious as to why this works. I thought casting in this manner would assign to the temporary, and not the class member. Using VC2010.

class A
{
public:

   A() :
      m_value(1.f)
   {
      ((float)m_value) = 10.f;
   }

   const float m_value;
};

推荐答案

它不应该工作.使用强制转换表示法到 float 的显式类型转换将是纯右值(第 5.4 节):

It shouldn't work. An explicit type conversion to float with cast notation will be a prvalue (§5.4):

表达式 (T) cast-expression 的结果是 T 类型.如果 T 是左值引用类型或函数类型的右值引用，则结果是左值；如果 T 是对象类型的右值引用，则结果是 xvalue；否则结果是纯右值.

The result of the expression (T) cast-expression is of type T. The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type and an xvalue if T is an rvalue reference to object type; otherwise the result is a prvalue.

我强调了.

赋值运算符需要一个左值作为其左操作数(第 5.17 节):

The assignment operator requires an lvalue as its left operand (§5.17):

都需要一个可修改的左值作为它们的左操作数，并返回一个引用左操作数的左值.

All require a modifiable lvalue as their left operand and return an lvalue referring to the left operand.

纯右值不是左值.

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