将函数的返回值分配给引用 C++?

问题描述

这是一个两部分的问题.可以将函数的返回值分配给引用吗?比如

This is a two part question. Is it ok to assign the return value of a function to a reference? Such as

Foo FuncBar()
{
    return Foo();
}

// some where else
Foo &myFoo = FuncBar();

这样好吗?我的理解是 FuncBar() 返回一个 Foo 对象，现在 myFoo 是对它的引用.

Is this ok? Its my understanding that FuncBar() returns a Foo object and now myFoo is a reference to it.

问题的第二部分.这是优化吗?因此，如果您经常在循环中执行此操作，最好这样做

Second part of the question. Is this an optimization? So if your doing it in a loop a lot of the time is it better to do

Foo &myFoo = FuncBar();

或

Foo myFoo = FuncBar();

考虑到变量的使用，使用 ref 会不会需要更慢的解引用?

And take into account the variables use, won't using the ref require slower dereferences?

推荐答案

Foo &myFoo = FuncBar();

不会编译.应该是:

const Foo &myFoo = FuncBar();

因为 FuncBar() 返回一个临时对象(即右值)并且只有左值可以绑定到对非常量的引用.

because FuncBar() returns a temporary object (i.e., rvalue) and only lvalues can be bound to references to non-const.

安全吗?

是的，它是安全的.

C++ 标准规定，将临时对象绑定到对 const 的引用可将临时对象的生命周期延长至引用本身的生命周期，从而避免常见的悬空引用错误.

C++ standard specifies that binding a temporary object to a reference to const lengthens the lifetime of the temporary to the lifetime of the reference itself, and thus avoids what would otherwise be a common dangling-reference error.

Foo myFoo = FuncBar();      

是复制初始化.
它创建由 FuncBar() 返回的对象的副本，然后使用该副本来初始化 myFoo.myFoo 是语句执行后一个单独的对象.

Is Copy Initialization.
It creates a copy of the object returned by FuncBar() and then uses that copy to initalize myFoo. myFoo is an separate object after the statement is executed.

const Foo &myFoo = FuncBar();

将 FuncBar() 返回的临时对象绑定到引用 myFoo，注意 myFoo 只是返回的临时对象的别名，而不是一个单独的对象.

Binds the temporary returned by FuncBar() to the reference myFoo, note that myFoo is just an alias to the returned temporary and not a separate object.

这篇关于将函数的返回值分配给引用 C++?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！