相关疑难解决方法(0)

下面的代码编译,但char类型的行为与int类型的行为不同.

特别是

   cout << getIsTrue< isX<int8>::ikIsX  >() << endl;
   cout << getIsTrue< isX<uint8>::ikIsX  >() << endl;
   cout << getIsTrue< isX<char>::ikIsX  >() << endl;

导致三种类型的模板的3个实例化:int8,uint8和char.是什么赋予了？

对于ints来说也是如此:int和uint32导致相同的模板实例化,而signed int则是另一个.

原因似乎是C++将char,signed char和unsigned char视为三种不同的类型.而int与signed int相同.这是对的还是我错过了什么？

#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 …