This question already has answers here:
Default constructor with empty brackets
(9 answers)
Closed 4 years ago.
This code doesn't behave how I expect it to.
#include<iostream>
using namespace std;
class Class
{
Class()
{
cout<<"default constructor called";
}
~Class()
{
cout<<"destrutor called";
}
};
int main()
{
Class object();
}
I expected the output 'default constructor called', but I did not see anything as the output. What is the problem?
Nope. Your line Class object(); Declared a function. What you want to write is Class object;
Try it out.
You may also be interested in the most vexing parse (as others have noted). A great example is in Effective STL Item 6 on page 33. (In 12th printing, September 2009.) Specifically the example at the top of page 35 is what you did, and it explains why the parser handles it as a function declaration.
No call to constructor
Because the constructor never gets called actually.
Class object(); is interpreted as the declaration of a function object taking no argument and returning an object of Class [by value]
Try Class object;
EDIT:
As Mike noticed this is not exactly the same code as what you are feeding to the compiler. Is the constructor/destructor public or is Class a struct?
However google for C++ most vexing parse.
You can use it like this:
Class obj;
//or
Class *obj = new Class(/*constructor arguments*/);
Related
This question already has answers here:
My attempt at value initialization is interpreted as a function declaration, and why doesn't A a(()); solve it?
(5 answers)
Closed 5 years ago.
#include <iostream>
using namespace std;
class A{
public:
A(){
cout << "Class A!";}
};
int main()
{
A a();
}
the above code does not call the constructor A :: A() even though they have the same input parameters (none). however if, in the main function, I remove the parenthesis from
A a();
it calls the constructor.
so what is the difference between A a; and A a();
i believe the question here is very similar maybe even the same however if some could explain in simpler terms I would be very grateful.
Do the parentheses after the type name make a difference with new?
would declaring A a(); ever call a constructor, under any circumstances?
do parameterless constructors exist in c++, or is that the same as a default constructor?
With
A a();
you declare a as a function taking no arguments and returning an A object.
This question already has answers here:
Default constructor with empty brackets
(9 answers)
Closed 4 years ago.
This code doesn't behave how I expect it to.
#include<iostream>
using namespace std;
class Class
{
Class()
{
cout<<"default constructor called";
}
~Class()
{
cout<<"destrutor called";
}
};
int main()
{
Class object();
}
I expected the output 'default constructor called', but I did not see anything as the output. What is the problem?
Nope. Your line Class object(); Declared a function. What you want to write is Class object;
Try it out.
You may also be interested in the most vexing parse (as others have noted). A great example is in Effective STL Item 6 on page 33. (In 12th printing, September 2009.) Specifically the example at the top of page 35 is what you did, and it explains why the parser handles it as a function declaration.
No call to constructor
Because the constructor never gets called actually.
Class object(); is interpreted as the declaration of a function object taking no argument and returning an object of Class [by value]
Try Class object;
EDIT:
As Mike noticed this is not exactly the same code as what you are feeding to the compiler. Is the constructor/destructor public or is Class a struct?
However google for C++ most vexing parse.
You can use it like this:
Class obj;
//or
Class *obj = new Class(/*constructor arguments*/);
This question already has answers here:
Default constructor with empty brackets
(9 answers)
Closed 5 years ago.
Suppose I have the following structure:
struct A {
A() { cout << "Default ctor\n"; }
A(int) { cout << "Ctor with params\n"; }
};
And then I want to create an object. In Java I've got accustomed to use brackets when I create an object, so the first desire is to write something like that:
A a();
The code compiles, but a actually isn't an instance of A, it is something different.
So the question is: what is a and why should I omit the brackets to call the default constructor?
See the most vexing parse, what you are actually doing is declaring a function. The way to alleviate this problem is to either eliminate the () or to use the C++11 uniform initialization syntax,
A a;
A a{};
In C++, A a(); is a forward declaration for a function called a that takes no arguments, and returns an A. It does not create an instance of A using the default constructor.
In Java, there is no need for forward declarations of functions, so A a(); can be read to be equivalent to A a;
This curiousity of C++ even has a name: see https://en.wikipedia.org/wiki/Most_vexing_parse
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Most vexing parse: why doesn't A a(()); work?
I have two classes in file1.h:
class ZoneRecord {
public:
//a lof of stuff here
};
class RegisterRecord {
public:
RegisterRecord(ZoneRecord rec); //this function register object rec in a fabric
};
And file2.cpp has:
#include "file1.h"
class MockZoneRecord: public ZoneRecord {
public:
MockZoneRecord(): ZoneRecord() {}
};
RegisterRecord mockrecord_register(MockZoneRecord());
This code compiles perfectly, except one thing. It says that mockrecord_register is a declaration of a function. But I actually wanted to create an global object of type RegisterRecord with name mockrecord_register. How to explicitly tell to compiler that this is not a function prototype, but an object?
You are experiencing the most vexing parse.
One way to solve this is to use copying, like
RegisterRecord mockrecord_register = RegisterRecord(MockZoneRecord());
Another is the use of parenthesis like in the answer by yuri kilochek.
If your compiler is C++11 compatible, you could use this construct:
RegisterRecord mockrecord_register{MockZoneRecord()};
Place parenthesis around argument:
RegisterRecord mockrecord_register((MockZoneRecord()));
This question already has answers here:
Default constructor with empty brackets
(9 answers)
Closed 4 years ago.
This code doesn't behave how I expect it to.
#include<iostream>
using namespace std;
class Class
{
Class()
{
cout<<"default constructor called";
}
~Class()
{
cout<<"destrutor called";
}
};
int main()
{
Class object();
}
I expected the output 'default constructor called', but I did not see anything as the output. What is the problem?
Nope. Your line Class object(); Declared a function. What you want to write is Class object;
Try it out.
You may also be interested in the most vexing parse (as others have noted). A great example is in Effective STL Item 6 on page 33. (In 12th printing, September 2009.) Specifically the example at the top of page 35 is what you did, and it explains why the parser handles it as a function declaration.
No call to constructor
Because the constructor never gets called actually.
Class object(); is interpreted as the declaration of a function object taking no argument and returning an object of Class [by value]
Try Class object;
EDIT:
As Mike noticed this is not exactly the same code as what you are feeding to the compiler. Is the constructor/destructor public or is Class a struct?
However google for C++ most vexing parse.
You can use it like this:
Class obj;
//or
Class *obj = new Class(/*constructor arguments*/);