C ++传递Const引用和返回由Const引用 [英] C++ Pass By Const Reference and Return By Const Reference

问题描述

我试图理解返回 const 引用是否有任何好处。我有一个阶乘函数，通常看起来像这样：

 无符号长阶乘（无符号长整数）
 {
 return（n == 0）？ 1：n *阶乘（n-1）; 
} 
  

我假设当我们传递 const 引用，我们返回一个 const 引用...但 const -correctness 总是困惑我。

  const unsigned long& factorial（const unsigned long& n）
 {
 return（n == 0）？ 1：n *阶乘（n-1）; 
} 
  

是否有效返回 const 参考？

解决方案

这是无效的。您不能返回对本地变量的引用。

MSVS C ++编译器甚至会发出以下警告：

  main.cc：warning C4172：返回局部变量或临时变量的地址
  pre> 
 
 
 不太确定GCC，但结果可能是一样的。 
 
I'm trying to understand if there is any benefit to returning a const reference. I have a factorial function that normally looks like this:
unsigned long factorial(unsigned long n)
{
    return (n == 0) ? 1 : n * factorial(n - 1);
}
I'm assuming that there will be a performance increase when we pass by const reference and we return a const reference... but const-correctness always confuses me.
const unsigned long & factorial(const unsigned long& n)
{
    return (n == 0) ? 1 : n * factorial(n - 1);
}
Is it valid to return a const reference? Furthermore, could somebody please tell me: is it beneficial?
解决方案
This is invalid. You can't return reference to a local variable.

MSVS C++ compiler even gives the following warning:
main.cc : warning C4172: returning address of local variable or temporary
Not quite sure about GCC, but probably the result would be the same.

                        
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