在S s = S（）是否保证不会创建临时？ [英] In S s = S() is it guaranteed that no temporary will be created?

问题描述

在下面的代码中， pS 和 s.pS 保证在最后一行相等？换句话说，在语句 S s = S（）; 中，我可以确定临时 S 不建构？

In the following code, are pS and s.pS guaranteed to be equal in the final line? In other words, in the statement S s = S();, can I be sure that a temporary S will not be constructed?

#include <iostream>
using namespace std;

struct S
{
  S() { pS = this; }
  S* pS;
};

int main()
{
  S s = S();
  S* pS = &s;
  cout << pS << " " << s.pS << endl;
}

在每个编译器中，我都在 pS == s.pS ，但我不太熟悉标准，以便能够满足自己，这是保证。

In every compiler I've tested this in pS == s.pS, but I'm not sufficiently familiar with the standard to be able to satisfy myself that this is guaranteed.

推荐答案

否

编译器不是该标准简单地指定了[class.copy]：

The compiler isn't obligated to do copy elision. The standard simply specifies that, [class.copy]:

当满足某些标准时，允许< ...>

When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object [...]

我可以通过<$ c $禁用复制elision c> -fno-elide-constructors ，然后两个指针肯定会不同。例如：

I can disable copy elision via -fno-elide-constructors, and then the two pointers will definitely be different. For example:

$g++ -std=c++11 -Wall -pedantic -fno-elide-constructors -Wall -Wextra main.cpp && ./a.out
0x7fff5a598920 0x7fff5a598930

在一般情况下， code> S（S&&）= delete ，那么上面的代码甚至不会编译。

And in the general case, if we add S(S&& ) = delete, then the above code wouldn't even compile.

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