比较typedef是相同类型 [英] Compare typedef is same type

问题描述

我使用C ++（不是11）和使用一些库具有不同的typedefs整数数据类型。有什么办法，我可以断言两个typedef是相同的类型？我自己想出了以下解决方案..它是安全吗？
感谢

I am using C++ (not 11) and using some libraries which have different typedefs for integer data types. Is there any way I can assert that two typedefs are the same type? I've come up with the following solution myself.. is it safe? Thanks

template<typename T>
struct TypeTest
{
    static void Compare(const TypeTest& other) {}
};

typedef unsigned long long UINT64;
typedef unsigned long long UINT_64;
typedef unsigned int UINT_32;

int main()
{
    TypeTest<UINT64>::Compare(TypeTest<UINT64>()); // pass
    TypeTest<UINT64>::Compare(TypeTest<UINT_64>()); // pass
    TypeTest<UINT64>::Compare(TypeTest<UINT_32>()); // fail
}

推荐答案

+11，您可以使用 std :: is_same< T，U> :: value 。

In C++11, you could use std::is_same<T,U>::value.

没有C ++ 11，你可以自己实现这个功能：

Since you don't have C++11, you could implement this functionality yourself as:

template<typename T, typename U>
struct is_same 
{
    static const bool value = false; 
};

template<typename T>
struct is_same<T,T>  //specialization
{ 
   static const bool value = true; 
};

完成！

static_assert ¹ as：

template<bool> struct static_assert;
template<> struct static_assert<true> {};  //specialization

现在您可以使用它们：

static_assert<is_same<UINT64,UINT64>::value>(); //pass
static_assert<is_same<UINT64,UINT32>::value>(); //fail

或者您可以将此包装在宏中：

Or you could wrap this in a macro as:

#define STATIC_ASSERT(x)  { static_assert<x> static_assert_failed; (void) static_assert_failed; }

然后使用：

STATIC_ASSERT(is_same<UINT64,UINT64>::value); //pass
STATIC_ASSERT(is_same<UINT64,UINT32>::value); //pass

如果使用宏，则会在编译器生成的消息中看到以下字符串if断言失败：

If you use macro, then you would see the following string in the compiler generated message if the assert fails:

static_assert_failed

这是有帮助的。使用错误消息中的其他信息，您将能够找出它失败的原因。

which is helpful. With the other information in the error message, you would be able to figure out why it failed.

希望有帮助。

^{1。注意，在C ++ 11中， static_assert 是一个操作符（在编译期操作），而不是类模板。在上面的代码中， static_assert 是一个类模板。}

^{1. Note that in C++11, static_assert is an operator (which operates at compile-time), not a class template. In the above code, static_assert is a class template.}

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