什么是c ++ 98中的move()? [英] What is move() in c++98?

问题描述

  #include< iostream>#include< vector>使用命名空间std;int main(void){向量< int>一种;a.push_back(3);向量< int>b = move(a);cout ＜b:＜b.data()＜endl；cout ＜a:＜a.data()＜endl；返回0;} 

输出(在c ++ 98中):

 <代码> b:0x7f9a82405730a:0x7f9a82405720 

输出(在c ++ 11中):

 <代码> b:0x7f9a82405730a:0x0 

我正在使用Apple clang 11.0.3.

第一个输出不使用编译器标志.

-std = c ++ 11 标志用于第二个输出.

我知道move()在c ++ 11(及更高版本)中的作用.但是正如我所看到的，在c ++ 98中使用move()对传递的对象没有任何作用，只会进行深度复制.

然后为什么在c ++ 98中有一个move()?

解决方案

在C ++ 11之前的所有版本中都可以调用 std :: move 的原因是libc ++不会包装<在中使用href ="https://github.com/llvm/llvm-project/blob/6d3b81664a4b79b32ed2c2f46b21ab0dca9029cc/libcxx/include/type_traits#L2613-L2622" rel ="noreferrer">其实现 #if _LIBCPP_STD_VER> = 11 .这在libstdc ++(Linux默认使用的)中不起作用，因为b: 0x7f9a82405730 a: 0x7f9a82405720

Output (in c++11):

b: 0x7f9a82405730
a: 0x0

I am using Apple clang 11.0.3.

No compiler flags are used for the first output.

-std=c++11 flag for the second output.

I know what move() does in c++11 (and higher versions). But as I can see using move() in c++98 does nothing to the object passed and just deep copy happens.

Then why is there a move() in c++98??

解决方案

The reason that you can call std::move at all pre-C++11 is that libc++ doesn't wrap its implementation of it in #if _LIBCPP_STD_VER >= 11. This doesn't work in libstdc++ (which Linux uses by default) because it guards std::move with #if __cplusplus >= 201103L.

As for why using it doesn't make a.data() be null, it's because libc++ does wrap the move constructor in #ifndef _LIBCPP_CXX03_LANG, so it falls back to the copy constructor.

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