应该不应该decltype Trigger编译它的参数？ [英] Shouldn't decltype Trigger Compilation of its Argument?

问题描述

所以我对如何工作感到困惑。给定：

 模板< typename T> 
 int foo（T t）{t.foo（）; } 
  

看起来这个调用应该失败：

  decltype（foo（int {13}））fail = 42; 
 
 cout<<失败< endl; 
  

而是只打印：

42

它以这种方式在我可以访问的所有编译器。这是正确的行为吗？

解决方案

在 [dcl.spec] ：

对于表达式e， （e）被定义为
如下：

如果e是未加括号的id-表达式，命名从分解的标识符列表引入的值b $ b声明，decltype（e）是分解声明（[dcl.decomp]）的
规范中给出的引用类型;否则为

，如果e是未加括号的id表达式或未加括号的类成员访问（[expr.ref]），则decltype（e）是由e命名的实体的
类型。如果没有这样的实体，或者如果e
命名一组重载的函数，则程序是错误的;

否则，如果e是x值，decltype（e）是T&& amp;& amp;其中T是e的类型;否则，decltype（e）是e的类型。

decltype说明符的操作数是未求值的操作数
（Clause [expr]）。

（强调我的）

所以你的 foo（int {13}） 从不评估。

So I'm perplexed as to how this works. Given:

template <typename T>
int foo(T t) { t.foo(); }

It seems like this call should fail:

decltype(foo(int{ 13 })) fail = 42;

cout << fail << endl;

Instead it just prints:

42

It works this way on all the compilers I have access to. Is this correct behavior? I request a quote from the C++ Standard.

解决方案

In [dcl.spec] :

For an expression e, the type denoted by decltype(e) is defined as follows:

if e is an unparenthesized id-expression naming an lvalue or reference introduced from the identifier-list of a decomposition declaration, decltype(e) is the referenced type as given in the specification of the decomposition declaration ([dcl.decomp]);

otherwise, if e is an unparenthesized id-expression or an unparenthesized class member access ([expr.ref]), decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;

otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;

otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;

otherwise, decltype(e) is the type of e.

The operand of the decltype specifier is an unevaluated operand (Clause [expr]).

(Emphasis mine)

So your foo(int{ 13 }) is never evaluated.

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