Question

In C++, these two have no difference.

char *pC;
char* pC2;

I thought that the type char and char* have got to be different types and all the C++ programmers had to use this form char* p instead of char *p, because when you do char *p it doesn’t seem to specify that the programmer is using a pointer to char.

However, when you take a look at this code below, which is the same as nullptr from C++0x, it seems that T doesn’t recognize any pointer type.

const class{
public:
    // T is char, not char* in this case.
    // T* is char*, but I thought it would be char**
    template <class T>
    operator T*() const{   
        cout << "Operator T*() Called" << endl;
        return 0;
    }

    template <class C, class T> 
    operator T C::*() const{
        cout << "Operator C T::*() Called" << endl;
        return 0;
    }

private:
    void operator&() const;
}AAA = {};

int main(){
 // operator T*() const is called, 
 // but it's awkward,
 // because it looks like T will be char* type, so T* will be char**
    char* pC = AAA;
}

Thanks in advance!

Answer 1

You’re correct that char and char* are different types. Whether programmers say char* p or char *p is irrelevant. Those are three separate tokens, and there can be as many or as few spaces between them as you want; it has no effect on the type of p.

When you initialize pC with AAA, the compiler needs to choose a conversion operator from AAA. Since it’s initializing a char* variable, it wants operator char*. To make that work, the compiler needs to choose T = char. Since T is char, there’s no way T* could be char**.

If you want T to be a pointer type, then use AAA in a context where a pointer-to-pointer type is expected, such as char**. Then T will be char*.