I created a class which basically serves as a public structure in this example, and let's say the class name is X. I want to declare a local object in main function. The short version of my question is: I know we can do X foo;, but I think X foo(); (attach a pair of parenthesis) should work, and I thought the first usage is actually a shorthand of the second usage. The entire code is provided below:
#include <iostream>
using namespace std;
class X {
public:
int val1;
int val2;
};
int main() {
X a;
X b(); // A warning here
X *c = new X;
X *d = new X();
cout << "val of a: " << a.val1 << " " << a.val2 << endl;
cout << "val of b: " << b.val1 << " " << b.val2 << endl; // Compile error
cout << "val of c: " << c->val1 << " " << c->val2 << endl;
cout << "val of d: " << d->val1 << " " << d->val2 << endl;
return 0;
}
And the compiler complains:
11_stack.cpp:16:6: warning: empty parentheses interpreted as a function declaration [-Wvexing-parse]
X ab();
^~
11_stack.cpp:16:6: note: replace parentheses with an initializer to declare a variable
X ab();
^~
{}
11_stack.cpp:22:26: error: use of undeclared identifier 'b'
cout << "val of b: " << b.val1 << " " << b.val2 << endl;
^
11_stack.cpp:22:43: error: use of undeclared identifier 'b'
cout << "val of b: " << b.val1 << " " << b.val2 << endl;
^
1 warning and 2 errors generated.
My initial guesses are the following:
- In any case, we shouldn't put an empty parenthesis after a declared variable.
- It triggers
operator().
But later on I disproved both hypotheses. We can see the first disproof in the code: both X *c = new X; and X *d = new X(); work. For the second one, I put additional code like this:
a();
Then I got compilation error messages:
11_stack.cpp:26:2: error: type 'X' does not provide a call operator
a();
^
So what exactly cause the error?
Working environment:
- Mac OS 10.11.3
- g++ with flag c++11
- Xcode Version 7.2 (7C68)
P.S. Please also help me think a better and descriptive post title if it's too ambiguous...
