如果临时隐含不可修改，这是如何工作的？ [英] If temporaries are implicitly non-modifiable, how does this work?

问题描述

我是告诉，在C ++ 03中，临时

I'm told that, in C++03, temporaries are implicitly non-modifiable.

但是，以下在GCC 4.3上编译了我。 4 （在C ++ 03模式下）：

However, the following compiles for me on GCC 4.3.4 (in C++03 mode):

cout << static_cast<stringstream&>(stringstream() << 3).str();

如何编译？

推荐答案

（不

有人告诉我，在C ++ 03中，暂时是不可修改的。

I'm told that, in C++03, temporaries are implicitly non-modifiable.

这不正确。在其他情况下，通过计算右值创建临时值，并且有非常量值和常量值。表达式的值类别和它表示的对象的常数大多数是正交的 ¹ 。观察：

That is not correct. Temporaries are created, among other circumstances, by evaluating rvalues, and there are both non-const rvalues and const rvalues. The value category of an expression and the constness of the object it denotes are mostly orthogonal ¹. Observe:

      std::string foo();
const std::string bar();

给定上述函数声明，表达式 foo（）是一个非常量的值，它的求值创建一个非常量的临时值， bar（）是一个常量值，创建一个常量临时值。

Given the above function declarations, the expression foo() is a non-const rvalue whose evaluation creates a non-const temporary, and bar() is a const rvalue that creates a const temporary.

请注意，您可以在非常量值上调用任何成员函数，允许您修改对象：

Note that you can call any member function on a non-const rvalue, allowing you to modify the object:

foo().append(" was created by foo")   // okay, modifying a non-const temporary
bar().append(" was created by bar")   // error, modifying a const temporary

/ code>是一个成员函数，你甚至可以分配到非const的值：

Since operator= is a member function, you can even assign to non-const rvalues:

std::string("hello") = "world";

这应该足以证明你说临时是不是 。

This should be enough evidence to convince you that temporaries are not implicitly const.

_{1：例外是标量右值，例如42. 总是非常数。}

这篇关于如果临时隐含不可修改，这是如何工作的？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IT屋！