我发现很难理解为什么某些参数是允许的,而其他参数是不允许的.

例如,如果一个类的静态成员,const char *可以作为模板参数给出,如果在外部定义,则不能.

我做了一个小程序来测试所有这些,评论不编译的行.我也根据编译器输出做了几个假设,但是它们可能是错误的.

模板参数值的规则是什么？我看到对象需要外部链接,但是bool是授权的,虽然它显然没有任何形式的联动.

#include <iostream>
using namespace std;

struct tag {
    static char array[];
    static const char carray[];
    static char *ptr;
    static const char *cptr;
    static const char *const cptrc;
    static string str;
    static const string cstr;
};
char tag::array[] = "array";
const char tag::carray[] = "carray";
char *tag::ptr = (char*)"ptr"; // cast because deprecated conversion
const char *tag::cptr = "cptr";
const char *const tag::cptrc = "cptrc";
string tag::str = "str";
const string tag::cstr = "cstr";


namespace ntag {
    char array[] = "array";
    const char carray[] = "carray";
    char *ptr = (char *)"ptr"; // cast because deprecated conversion
    const char *cptr = "cptr";
    const char *const cptrc = "cptrc";
    string str = "str";
    const string cstr = "cstr";
};

template <class T,T t>
void print() { cout << t << endl; };

int main()
{
    cout << "-- class --" << endl;
    // Works
    print<char *,tag::array>();
    print<const char *,tag::carray>();

    // Does not work because it is a lvalue ?
    // print<char *,tag::ptr>();
    // print<const char *,tag::cptr>();
    // print<const char *const,tag::cptrc>();

    // Template type param must be a basic type ?
    // print<string,tag::str>();
    // print<const string*,tag::cstr>();

    cout << "-- namespace --" << endl;
    // Works
    print<char *,ntag::array>();

    // No external linkage ?
    // print<const char *,ntag::carray>();

    // Does not work because it is an lvalue ?
    // print<char *,ntag::ptr>();
    // print<const char *,ntag::cptr>();
    // print<const char *const,ntag::cptrc>();

    // The type of a template value param must a basic type
    // print<string,ntag::str>();
    // print<const string*,ntag::cstr>();
}