differences in behavior between char, signed char and unsigned char

The code below compiles successfully, but char behaves differently than integer types.

cout ::ikIsX  >() ::ikIsX  >() ::ikIsX  >() 
The result is three instantiation patterns for three types: int8, uint8 and char. Why does this happen?
The same is not true for integers: int and uint32 result in one pattern instantiation, and signed int results in another.
The reason is probably because C treats char , signed char and unsigned char as three different types. While int is the same as signed int. Is this true, or am I missing something?
#include 

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 
struct isX
{
   typedef typename T::ikIsX ikIsX;
};


// Это  int==int32 неоднозначно
//template  bool getIsTrue();
template ::ikIsX  >() ::ikIsX  >() ::ikIsX  >() ::ikIsX  >() ::ikIsX  >() ::ikIsX  >() 
I use g 4.something