but how do the constructor work?

Question

#include <iostream>
using namespace std;
class A{
public:
    A(){cout << "A()" << endl; }
    A(string s): str(s){ cout << "A(" << str << ")" << endl; }
    ~A(){ cout << "delete!" << endl;}
    string str = "000";
};
int main(int argc, char** argv)
{
    //A a0;    // it will call A() by default
    A a1();    //even if i comment A(), the compiler will no warn me anything
    A a2("123");
    cout << a2.x << endl;
    return 0;
}

output:

A(123)
123
delete!

so, i just want to know why "A a1();" can't works? can someone helps me ? thank you!

You misspelled ``, and `A(string s)` refers to a nonexistent `x` variable. Please post your actual code. — jwodder, Nov 21 '14 at 17:26
This is probably a duplicate, but that's not really the right question to link to. — James Curran, Nov 21 '14 at 17:28
@JamesCurran If you've got a better one you can reopen it (and then close again, I think). — jrok, Nov 21 '14 at 17:29
[Turn up your warning level](http://coliru.stacked-crooked.com/a/2823639134892991) (I fixed `iostram`). — chris, Nov 21 '14 at 17:37

score 3 · Accepted Answer · answered Nov 21 '14 at 17:26

3

A a1(); is a function prototype, taking no parameters and returning an A

answered Nov 21 '14 at 17:26

Bathsheba

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Whoever downvoted this: Most Vexing Parse. Look it up. – Borgleader Nov 21 '14 at 17:26
How about int a (5); – ventsyv Nov 21 '14 at 17:29
`int a(5);` cannot be a function declaration, and therefore must be a variable definition, specifically the equivalent of `int a = 5;` – James Curran Nov 21 '14 at 17:31
Oh, I see now. But is declaring functions inside other functions even legal? Excluding lambdas obviously... – ventsyv Nov 21 '14 at 17:35
@ventsyv Declaring is legal, defining isn't. – molbdnilo Nov 21 '14 at 17:46

but how do the constructor work?

1 Answers1