Question

In Effective C++, Item 3, Scott Meyers suggests overloading operator* for a class named Rational:

    class Rational { ... };
    const Rational operator*(const Rational& lhs, const Rational& rhs);

收益价值为<代码>const-standard”的原因在以下各行文中作了解释:如果未填写const,方案人员可撰写如下的代码:

    (a * b) = c;

或更可能的话:

     if (a*b = c)

Fair enough. Now I’m confused as I thought that the return value of a function, here operator*, was a rvalue, therefore not assignable. I take it not being assignable because if I had:

    int foo();
    foo() += 3;

that would fail to compile with invalid lvalue in assignment. Why doesn’t that happen here? Can someone shed some light on this?

www.un.org/Depts/DGACM/index_spanish.htm 我在斯科特·迈耶斯的这一项目上看到了许多其他线索,但没有人解决我在这里暴露的珍贵问题。

Answer 1

要点是,对于类别,a = b只是一个短手,可到a.operator=(b),其中operator=为成员职能。成员的职能可以被提升为高价值。

请注意,在C++11中,你可以通过制定<代码>operator=加以抑制。仅限值:

class Rational
{
public:
  Rational& operator=(Rational const& other) &;
  // ...
};

<代码>& 告诉汇编者,这一职能可能不会被提升为高价值。

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