返回类型扣除与私有成员变量 [英] Return type deduction with a private member variable

问题描述

正如此 Q& A昨天，g ++ 4.8和Clang 3.3都正确地抱怨下面的代码，如b_没有在此范围内声明

  #include< iostream> 
 
 class Test 
 {
 public：
 Test（）：b_（0）{} 
 
 auto foo（）const  - > ; decltype（b_）//只是省略 - > decltype（b_）使用c ++ 1y 
 {
 return b_; 
} 
 private：
 int b_; 
}; 
 
 int main（）
 {
 Test t; 
 std :: cout<< t.foo（）; 
} 
  

移动 private 部分到类定义的顶部消除了错误并输出0.

我的问题是，这个错误也会在C ++ 14中消失，返回类型扣除，以便我可以省略 decltype 并在我的 private

：

解决方案

没有，但是没有了，我们已经开始使用这个工具了。msgctxt@info：whatsthis这是因为你可以说

  decltype（auto）foo（）const {
 return b_; 
} 
  

这将从其正文自动推导返回类型。

As was explained in this Q&A yesterday, both g++ 4.8 and Clang 3.3 correctly complain about the code below with an error like "'b_' was not declared in this scope"

#include <iostream>

class Test
{
public:
    Test(): b_(0) {}

    auto foo() const -> decltype(b_) // just leave out the -> decltype(b_) works with c++1y
    { 
        return b_;    
    }    
private:
    int b_;
};

int main() 
{
    Test t;
    std::cout << t.foo();
}

Moving the private section to the top of the class definition eliminates the error and prints 0.

My question is, will this error also go away in C++14 with return type deduction, so that I can leave out the decltype and have my private section at the end of the class definition?

NOTE: It actually works based on @JesseGood 's answer.

解决方案

No, but there not anymore is a need for this because you can say

decltype(auto) foo() const { 
    return b_;    
}

This will deduce the return type automatically from its body.

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