特别地,
cout << getIsTrue< isX<int8>::ikIsX >() << endl;
cout << getIsTrue< isX<uint8>::ikIsX >() << endl;
cout << getIsTrue< isX<char>::ikIsX >() << endl;
这将导致三个类型的模板被实例化:int8、uint8和char。这是什么意思?
但对于int型,情况并非如此:int和uint32会导致相同的模板实例化,signed int则另外一种。
原因似乎在于C++将char、signed char和unsigned char视为三种不同类型,而int则等同于signed int。我理解的对吗?或者我漏掉了些什么?
#include <iostream>
using namespace std;
typedef signed char int8;
typedef unsigned char uint8;
typedef signed short int16;
typedef unsigned short uint16;
typedef signed int int32;
typedef unsigned int uint32;
typedef signed long long int64;
typedef unsigned long long uint64;
struct TrueType {};
struct FalseType {};
template <typename T>
struct isX
{
typedef typename T::ikIsX ikIsX;
};
// This int==int32 is ambiguous
//template <> struct isX<int > { typedef FalseType ikIsX; }; // Fails
template <> struct isX<int32 > { typedef FalseType ikIsX; };
template <> struct isX<uint32 > { typedef FalseType ikIsX; };
// Whay isn't this ambiguous? char==int8
template <> struct isX<char > { typedef FalseType ikIsX; };
template <> struct isX<int8 > { typedef FalseType ikIsX; };
template <> struct isX<uint8 > { typedef FalseType ikIsX; };
template <typename T> bool getIsTrue();
template <> bool getIsTrue<TrueType>() { return true; }
template <> bool getIsTrue<FalseType>() { return false; }
int main(int, char **t )
{
cout << sizeof(int8) << endl; // 1
cout << sizeof(uint8) << endl; // 1
cout << sizeof(char) << endl; // 1
cout << getIsTrue< isX<int8>::ikIsX >() << endl;
cout << getIsTrue< isX<uint8>::ikIsX >() << endl;
cout << getIsTrue< isX<char>::ikIsX >() << endl;
cout << getIsTrue< isX<int32>::ikIsX >() << endl;
cout << getIsTrue< isX<uint32>::ikIsX >() << endl;
cout << getIsTrue< isX<int>::ikIsX >() << endl;
}
我正在使用 g++ 4.x 版本。
int8_t
不一定是signed char
,而uint8_t
也不一定是unsigned char
。特别是在 Solaris 上,如果char
是有符号的,则int8_t
就是char
。换句话说,在那里编写的代码将无法编译通过。 - Michał Górny