将const引用传递给原始类型有什么用？ [英] What is the use of passing const references to primitive types?

问题描述

在我维护的项目中，我看到了很多类似这样的代码，它们用于简单的 get / set 方法

In a project I maintain, I see a lot of code like this for simple get/set methods

const int & MyClass::getFoo() { return m_foo; }

void MyClass::setFoo(const int & foo) { m_foo = foo; }

这样做不是什么意思？

int MyClass::getFoo() { return m_foo; }  // Removed 'const' and '&' 

void MyClass::setFoo(const int foo) { m_foo = foo; }  // Removed '&'

将引用传递给原始类型应要求相同（或

毕竟只是一个数字...

这只是一些尝试过的微优化还是有真正的好处？

Passing a reference to a primitive type should require the same (or more) effort as passing the type's value itself, right?
It's just a number after all...
Is this just some attempted micro-optimization or is there a true benefit?

推荐答案

不同之处在于，如果您自己将结果获取到引用中，则可以跟踪整数变量在您的变量中的变化自己的变量名，而无需调用该函数。

The difference is that if you get that result into a reference yourself you can track the changes of the integer member variable in your own variable name without recalling the function.

const &int x = myObject.getFoo();
cout<<x<<endl;
//...
cout<<x<<endl;//x might have changed

这可能不是最佳的设计选择，并且如果返回的变量从作用域中释放出来，则返回一个引用（是否为const）是非常危险的。因此，如果返回引用，请务必确保它不是超出范围的变量。

It's probably not the best design choice, and it's very dangerous to return a reference (const or not), in case a variable that gets freed from scope is returned. So if you return a reference, be careful to be sure it is not a variable that goes out of scope.

修饰符也有细微差别，但同样

There is a slight difference for the modifier too, but again probably not something that is worth doing or that was intended.

void test1(int x)
{
  cout<<x<<endl;//prints 1
}

void test2(const int &x)
{
  cout<<x<<endl;//prints 1 or something else possibly, another thread could have changed x
}

int main(int argc, char**argv)
{
  int x = 1;
  test1(x);
  //...
  test2(x);
  return 0;
}

因此，最终结果是即使传递了参数，您也可以获得更改。

So the end result is that you obtain changes even after the parameters are passed.

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