什么是static_cast< T>做一个T& [英] What does static_cast<T> do to a T&?

问题描述

所以我问了这个问题，我正在通过 static_cast 。 （顺便提一句，它确实解决了问题，我只是不知道为什么。）

在代码中：

 矢量< int> foo = {0，42，0，42，0，42}; 
 replace（foo），end（foo），static_cast< int>（foo.front（）），13）; 
  

static_cast 值 int ？它和只是调用有什么区别：

  replace（begin（foo），end（foo），int {foo。 front（）}，13）; 
  

编辑：

根据 static_cast 的回答建构R-Value int ： http://ideone.com/dVPIhD

但此代码不是在Visual Studio 2015上工作。这是一个编译器错误？在此处测试： http://webcompiler.cloudapp.net/

解决方案

1. 是的，它与 int {...} 相同，除非 .front（）返回了需要缩小转换的类型。在这种情况下， int（...）将是相同的。




2. 错误，静态转换稍微不太可能做某事危险，例如将指针转换为 int（...）的int

注意，由于前面的元素被replace操作修改，消除了未定义行为的转换结果，并且可以破坏 std ::替换。

我会使用

  template< class T> 
 std :: decay_t< T> copy_of（T& t）{return std :: forward< T>（t）; } 
 
 
 
 
 
 
为什么这不工作在MSVC ... 
 
 
 
 MSVC有帮助的情况下，你转换类型 T 到 T ，并且不做任何操作。这会中断您的代码。
 
 
 
有一个编译器标志（/ Zc：rvalueCast），您可以使用MSVC不再破坏您的代码。
 
So I asked this question and I was tinkering around with solving it via static_cast. (Incidentally it does solve the problem, I'm just not sure if I understand why.)

In the code:
vector<int> foo = {0, 42, 0, 42, 0, 42};
replace(begin(foo), end(foo), static_cast<int>(foo.front()), 13);
Is the static_cast simply constructing an R-Value int? What's the difference between that and just the call:
replace(begin(foo), end(foo), int{foo.front()}, 13);
EDIT:

As inferred by the answers static_cast does seem to construct an R-Value int: http://ideone.com/dVPIhD

But this code does not work on Visual Studio 2015. Is this a compiler bug? Test here: http://webcompiler.cloudapp.net/
解决方案

Yes, it is the same as int{...}, unless .front() returned a type that required a narrowing conversion.  In that case, int(...) would be identical.
In the case of programmer error, static cast is marginally less likely to do something dangerous, like convert a pointer into an int than int(...).
Note eliminating the cast results in undefined behaviour as the front element is modified by the replace operation, and that could break std::replace.

I would use
template<class T>
std::decay_t<T> copy_of(T&& t){return std::forward<T>(t); }
myself here.

As for why this isn't working in MSVC...

MSVC helpfully takes situations where you cast a variable of type T to a T and proceeds to do nothing.  This breaks your code.

There is a compiler flag (/Zc:rvalueCast) you can use to make MSVC no longer break your code.

                        
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