为什么左值转换工作? [英] Why does an lvalue cast work?
问题描述
我今天第一次看到这种演员,我很好奇为什么这个演出。我认为以这种方式投入将分配给临时,而不是类成员。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
使用cast符号的 float
的显式类型转换将是一个prvalue(§5.4):
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 的结果类型为T
。如果T是一个左值引用类型或对函数类型的右值引用,并且如果T是对象类型的右值引用,则结果为左值;
The result of the expression
(T)
cast-expression is of typeT
. 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.
我强调的是
赋值运算符需要左值作为其左操作数(§5.17):
The assignment operator requires an lvalue as its left operand (§5.17):
所有都需要可修改的左值左操作数并返回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.
prvalue不是左值。
A prvalue is not an lvalue.
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