在什么情况下，将调用类型自身的转换运算符? [英] Under what circumstances would a type's conversion operator to itself be invoked?

问题描述

考虑类型bar，该类型具有用户定义的转换操作符来引用类型bar的引用:

Consider a type bar which has user-defined conversion operators to references of type bar:

struct bar
{
  operator bar & ();
  operator const bar & () const;
};

何时将应用这些转换?而且，这意味着这些运算符是否为deleted?两种功能都有有趣的用途吗?

When would these conversions be applied? Moreover, what does it imply if these operators were deleted? Is there any interesting use of either feature?

以下程序似乎均未应用任何转换:

The following program does not appear to apply either conversion:

#include <iostream>

struct bar
{
  operator bar & ()
  {
    std::cout << "operator bar &()" << std::endl;
    return *this;
  }

  operator const bar & () const
  {
    std::cout << "operator const bar &() const" << std::endl;
    return *this;
  }
};

void foo(bar x)
{
}

int main()
{
  bar x;

  bar y = x;         // copy, no conversion

  y = x;             // assignment, no conversion

  foo(x);            // copy, no conversion

  y = (bar&)x;       // no output

  y = (const bar&)x; // no output

  return 0;
}

推荐答案

C ++ 11§12.3.2

C++11 §12.3.2

从不使用转换函数将(可能具有cv限定)对象转换为(可能具有cv限定)相同的对象类型(或对其的引用)，转换为该对象的(可能具有cv限定)的基类.类型(或对其的引用)，或(可能具有cv资格)作废

A conversion function is never used to convert a (possibly cv-qualified) object to the (possibly cv-qualified) same object type (or a reference to it), to a (possibly cv-qualified) base class of that type (or a reference to it), or to (possibly cv-qualified) void

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