C ++编译器警告 - 返回局部变量 [英] C++ compiler warning - returning local variable

问题描述

 引用本地变量'
 tmp'returned 
  

这是重载的代码

  const Int& Int :: operator +（const Int& p）const 
 {
 Int tmp = value + p.value; 
 return tmp; 
} 
  

以下是类

  class Int {
 int value; 
 public：
 Int（）{} //默认构造函数
 Int（int v）：value（v）{} 
 Int& operator =（const Int&）; 
 const Int& operator +（const Int&）const; 
}; 
  

解决方案

您不能返回对局部变量的引用。在运算符+（）函数中，您正在创建一个局部变量 tmp 。一旦函数退出，它就会被销毁。

将函数的定义更改为：

p>

  const Int运算符+（const Int&）const; 
  

它会构建没有警告，也很好。

I'm simply trying to overload a + operator and I'm getting this compiler warning

reference to local variable 'tmp' returned

Here is the code for the overload

const Int& Int::operator+(const Int& p) const
{
    Int tmp = value + p.value;
    return tmp;
}

Here is the class

class Int{
    int value;
public:
    Int() {}    // default constructor
    Int(int v) : value(v) {}
    Int& operator=(const Int&);
    const Int& operator+(const Int&) const;
};

解决方案

You can't return a reference to a local variable. Inside the operator+() function, you're creating a local variable called tmp. It will get destroyed as soon as the function exits. You can't return a reference to that variable, because it no longer exists when the calling function gets the return value.

Change your definition of the function to:

const Int operator+(const Int&) const;

It would build without warnings and work fine too.

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