C ++：函数左值或右值 [英] c++: function lvalue or rvalue

问题描述

我刚刚通过阅读此页面，开始学习c ++ 11中的右值引用。 ，但我陷入了第一页。这是我从该页面获取的代码。

I just started learning about rvalue references in c++11 by reading this page, but I got stuck into the very first page. Here is the code I took from that page.

  int& foo();
  foo() = 42; // ok, foo() is an lvalue
  int* p1 = &foo(); // ok, foo() is an lvalue

  int foobar();
  j = foobar(); // ok, foobar() is an rvalue
  int* p2 = &foobar(); // error, cannot take the address of an rvalue

1. 为什么是 foo（）一个左值？是因为 foo（）返回 int& 这基本上是一个左值吗？


2. 为什么 foobar（）是右值？是因为 foobar（）返回 int 吗？


3. 通常，为什么您会在乎函数是否为右值？我想，如果我读完该文章的其余部分，就会得到答案。

1. why is foo() an lvalue? is it because foo() returns int& which is basically an lvalue?
2. why is foobar() an rvalue? is it because foobar() returns int?
3. In general, why would you care if a function is an rvalue or not? I think if I read the rest of that article, I'll get my answer to this.

推荐答案

L值是位置，R值是实际值。

L-Values are locations, R-Values are actual values.

所以：

1. 因为 foo（）返回一个引用（ int& ），这使其本身成为左值。


2. 正确。 foobar（）是一个右值，因为 foobar（）返回 int 。


3. 我们并不在乎函数是否为R值。令我们兴奋的是R值引用。

1. since foo() returns a reference(int&), that makes it an lvalue itself.
2. Correct. foobar() is an rvalue because foobar() returns int.
3. We don't care that much if a function is an R-Value or not. What we are getting excited about is R-Value references.

您指向的文章很有趣，我没有考虑过转发或在工厂使用之前。我对R值引用感到兴奋的原因是移动语义，例如：

The article you pointed to is interesting and I had not considered forwarding or the use in factories before. The reason I was excited about R-Value references was the move semantics, such as this:

BigClass my_function (const int& val, const OtherClass & valb);

BigClass x;
x = my_function(5, other_class_instance);

在该示例中，x被销毁，然后使用复制构造函数将my_function的返回值复制到x中。要在历史上解决该问题，您可以这样写：

In that example, x is destroyed, then the return of my_function is copied into x with a copy constructor. To get around that historically, you would write:

void my_function (BigClass *ret, const int& val, const OtherClass & valb);

BigClass x;
my_function(&x, 5, other_class_instance);

这意味着现在 my_function 具有副作用，而且阅读起来不那么简单。现在，使用C ++ 11，我们可以改为：

which means that now my_function has side effects, plus it isn't as plain to read. Now, with C++11, we can instead write:

BigClass & my_function (const int& val, const OtherClass & valb);

BigClass x;
x = my_function(5, other_class_instance);

使其运行效率与第二个示例一样。

And have it operate as efficiently as the second example.

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