C++ const重载赋值运算符机制

Question

这是来自 '03 或 '04 的 C++ 筛选测试。

Which of the following declares the assignment operator of MyClass so that the code on Line 3 generates a compilation error while the code on line 2 does not?

Consider the following code fragment:

   (Line 1) MyClass a, b, c; 
   (Line 2) a=b=c; 
   (Line 3) (a=b)=c; 

1) void operator=( const MyClass& X );
2) MyClass operator=( const MyClass& X );
3) const MyClass operator=( const MyClass& X );
4) None of the above

正确答案是 3)。现在，怎么会这样？它是如何工作的？什么叫什么？ parens是否强制创建临时文件？ const 如何参与其中？

Answer 1

将赋值重写为函数调用而不是运算符可能会有所帮助(正如其他人明确指出的那样，没有任何 ()，赋值从右到左进行):

Line 2: a.operator=(b.operator=(c));
Line 3: (a.operator=(b)).operator=(c);

这使得在第 3 行更清楚一点，您正在尝试获取 a=b 的结果并将 c 分配给它。但在情况 3 中，operator= 的输出是一个 const 对象，因此这是不允许的(因为 operator= 是一个非常量函数)。

将每个赋值分解为两个单独的语句可能会更清楚:

// with const MyClass MyClass::operator=( const MyClass& X );

// line 1
MyClass a, b, c; 

//line 2; originally a=b=c
const MyClass d = (b = c);
a = d; // no problem; a is nonconst, and it's ok for d (the parameter to operator=) to be const

//line 3; originally (a=b)=c
const MyClass e = (a = b); // note: the leftmost = here is initialization rather than assignment
e = c; // no good! e is const, and you can't call a nonconst function like operator= on a const object

或者更进一步，通过使用名为 assign 的函数代替 operator= 来消除任何与运算符相关的混淆:

// with const MyClass MyClass::assign( const MyClass& X );

// line 1
MyClass a, b, c; 

//line 2; originally a=b=c
const MyClass d = b.assign(c);
a.assign(d); // ok. assign() is nonconst, but so is a.

//line 3; originally (a=b)=c
const MyClass e = a.assign(b);
e.assign(c); // no good! e is const and assign() is not.