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C++ 将 operator=() 赋值转换为构造的具体规则是什么？例如Foo foo = bar实际上会调用 Foo 的构造函数，接受 bar 作为参数，如果它存在的话。我已经用谷歌搜索了它是如何工作的，但似乎找不到任何东西。

我无法弄清楚为什么下面的赋值尝试采用构造函数但没有采用明显正确的构造函数：HandlePtr(TYPE&resource)。使用实际构造语法的构造工作正常，但不适用于赋值运算符。

代码（显然为简洁而编辑）：

template< typename TYPE >
class HandlePtr {
public:
    HandlePtr( void ) = default;
    HandlePtr( HandlePtr< TYPE >& other ) = default;
    HandlePtr( TYPE& resource ) {} // generally I would make this explicit, but for testing purposes I took it out
    ~HandlePtr( void ) = default;

public:
    HandlePtr<TYPE>& operator=( TYPE& resource ) { return *this; }
    HandlePtr<TYPE>& operator=( HandlePtr<TYPE>& other ) { return *this; }
};

int main ( void ) {
    int x = 5;
    HandlePtr< int > g( x ); // works
    HandlePtr< int > i;i = x; // works
    HandlePtr< int > h = x; // doesn't work

            // also tried this just out of curiosity:
    HandlePtr< int > h = HandlePtr< int >( x ); // also does not work

    return 0;
}

错误：

shit.cpp: In function ‘int main()’:
try.cpp:19:24: error: no matching function for call to ‘HandlePtr<int>::HandlePtr(HandlePtr<int>)’
   HandlePtr< int > h = x; // doesn't work
                        ^
try.cpp:19:24: note: candidates are:
try.cpp:7:3: note: HandlePtr<TYPE>::HandlePtr(TYPE&) [with TYPE = int]
   HandlePtr( TYPE& resource ) {} // generally I would make this explicit, but for testing purposes I took it out
   ^
try.cpp:7:3: note:   no known conversion for argument 1 from ‘HandlePtr<int>’ to ‘int&’
try.cpp:6:3: note: HandlePtr<TYPE>::HandlePtr(HandlePtr<TYPE>&) [with TYPE = int]
   HandlePtr( HandlePtr< TYPE >& other ) = default;
   ^
try.cpp:6:3: note:   no known conversion for argument 1 from ‘HandlePtr<int>’ to ‘HandlePtr<int>&’
try.cpp:5:3: note: HandlePtr<TYPE>::HandlePtr() [with TYPE = int]
   HandlePtr( void ) = default;
   ^
try.cpp:5:3: note:   candidate expects 0 arguments, 1 provided
try.cpp:20:20: error: redeclaration of ‘HandlePtr<int> h’
   HandlePtr< int > h = HandlePtr< int >( x ); // also does not work
                    ^
try.cpp:19:20: error: ‘HandlePtr<int> h’ previously declared here
   HandlePtr< int > h = x; // doesn't work
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1 回答 1

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您在声明中忽略了这一点：

T t = u;

这不是赋值运算符。t = u;不是声明的子表达式。这里唯一的表达是u; 并且计算表达式的结果u用作t被声明对象的初始化程序。

如果u有 type T，t则从 复制构造u。

如果u没有 type T，则u首先需要转换为 type T。这将创建一个类型的右值T。

您没有任何接受右值的构造函数，所以T t = u;和相同的T t = T(u);都失败。但是，T t(u)成功是因为没有创建右值；该值u用作构造函数的参数T(U &)。

简化代码示例：

struct T
{
    T(int &);
    T(T&);
    T();
    T &operator=(int &);
};

int main()
{
    int x = 5;
    T g(x);   // OK, T(int &)
    T g2(5);   // fail, looks for T(int const &)
    T i;      // OK, T()
    i = x;    // OK, T::operator=(int&)
    T h3 = i; // OK, T(T&)
    T h1 = T(x);    // fail, looks for T(T const &)
    T h2 = x;       // fail, identical to previous line 
}

通常，您应该将const &其用作复制构造函数和赋值运算符的参数；那么所有这些“失败”的情况都会变成“OK”，因为右值可以绑定到 const 引用。

于 2014-04-12T04:10:16.043 回答