转换后是右值还是左值

Question

此处的代码在类型转换后测试左值或右值：

#include <stdio.h>

template <typename T>
T const f1(T  const &t) {
  printf("T const \n");
  return t;
}
template <typename T>
T  f1(T  &t) {
  printf("T\n");
  return t;
}
struct KK {
  int a;
};

int main()
{
  KK kk;
  kk.a=0;

  int ii;
  f1(kk);
  f1((KK)kk);

  f1(ii);
  f1((int)ii);
 return 0;
}

在 gcc link 中，结果是这样的，表明右值在类型转换后产生：

T
T const 
T
T const

但是在VC++2010中，只有当它是class类型时，这是表示右值的结果：

T
T const
T
T

这是编译器错误还是类型转换为 int 时的一些未定义行为？

Answer 1

来自 expr.cast（这适用于 C++11 及更高版本）

The result of the expression (T) cast-expression is of type T. The result is an lvalue if T is an lvalue reference type or an rvalue reference to function type and an xvalue if T is an rvalue reference to object type; otherwise the result is a prvalue. [ Note: If T is a non-class type that is cv-qualified, the cv-qualifiers are discarded when determining the type of the resulting prvalue; see Clause [expr]. — end note ]

对于 C++98：

The result of the expression (T) cast-expression is of type T. The result is an lvalue if T is a reference type, otherwise the result is an rvalue. [ Note: if T is a non-class type that is cv-qualified, the cv-qualifiers are ignored when determining the type of the resulting rvalue; see 3.10. — end note ]

那么，gcc就对了

根据 mkaes 的评论，这似乎是 （可以说是有用的）MSVC 扩展

转换后是右值还是左值

Is it an Rvalue or Lvalue After a Cast

c++

compiler-bug

lvalue-to-rvalue