Class template having pointers arguments with default value

Question

I am trying to understand the below code snippet

template <typename T>
class testopaque {
public:
    void test(T var = T()) {}
};

How does the default argument work when called with a pointer type example int *

int main() {
    testopaque<int *> obj1;
    obj1.test();
}

what would the compiler generate when obj1.test() is called. I get a compiler error when I try

int main() {
    int * var = int *();
}

error: expected primary-expression before ‘int’
int * ptr = int *();

Answer 1

Suppose you have x=1+2. Would you expect x*3, which is 9, to equal 1+2*3, which is 7?

A similar problem is happening here. int*() isn't the same as T=int* then T().

Try (int*){}, which solves the combined parsing and precident problems. Or using T=int*; int* x=T();, or even int*x={};

(Simply using (int*)() doesn't solve the problem due to how types are parsed, which is probably something you don't want to go too deeply into if you value your sanity.)

Answer 2

This is an example of how C++'s complicated syntax and grammar just produces unexpected results:

int *();

Your C++ compiler is very tempted to interpret this construct as a "function returning a pointer to an int". Your C++ compiler gives in to this temptation, with the observed results.

You need to teach your C++ compiler what you're trying to accomplish here:

typedef int *intp;

int main()
{
    int * var = intp();
}

Answer 3

When you use an expression like:

T var = T();

Where T is a pointer type then var will be assigned nullptr