Constructor taking Base& is not called

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I'm working on a simple program for Boolean algebra, but the double negation does not work as expected.

I have the following classes:

Operator:

#ifndef OPERATOR_H
#define OPERATOR_H

class Operator 
public:
 virtual int getArity(void) const = 0;
 virtual bool calc(void) const = 0;
;

#endif // OPERATOR_H

False:

#ifndef FALSE_H
#define FALSE_H

#include "operator.h"

class False : public Operator 
public:
 int getArity() const 
 return 0;
 

 bool calc(void) const 
 return false;
 
;

#endif // FALSE_H

Not:

#ifndef NOT_H
#define NOT_H

#include "operator.h"

class Not : public Operator 
public:
 Not(Operator& child) : m_child(child) 
 std::cout << "not constructor called" << std::endl;
 

 int getArity(void) const 
 return 1;
 

 bool calc(void) const 
 return !m_child.calc();
 

private:
 Operator& m_child;
;

#endif // NOT_H

My main.cpp:

#include <iostream>
#include "operator.h"
#include "not.h"
#include "false.h"

using namespace std;

int main(int argc, char *argv) 

 False f;
 Not n = Not(f);
 Not d = Not(n);

 cout << "n.calc(): " << n.calc() <<endl;
 cout << "d.calc(): " << d.calc() <<endl;
 return 0;

Since d = Not(Not(False())) I expect it to be false.

The output is:

not constructor called
n.calc(): 1
d.calc(): 1 <== should be 0

Why is the constructor of the class Not not called with an object of type Not as child?

edited 11 mins ago

Peter Mortensen

13.2k1983111

asked 5 hours ago

Core

543

New contributor

The best way I've found to figure out what happens in cases like this is to implement every constructor and add some debug statements, explicitly logging every constructor called. It can be very informative.
â€“Â Andrew Henle
5 hours ago

Aside: I would highlight this en.cppreference.com/w/cpp/language/operator_alternative and beg you to reconsider what you're doing. (And at last checking, false is not an operator, but a binary value)
â€“Â UKMonkey
5 hours ago

note that T x = T(a); is defined as being the same as T x(a);, since C++17.
â€“Â M.M
5 hours ago

4

"class False : public Operator" Reminds me of a "famous" german author who managed to let his class Wurstbrot inherit from Wurst and Brot which in turn inherit Supermarkt ...
â€“Â J. Doe
4 hours ago

2

@UKMonkey "false" is usually the name of a boolean value, but it is also the name of a nullary operator (meaning it doesn't take any argument) that returns that boolean value.
â€“Â Nelfeal
4 hours ago

Â |Â
show 1 more comment

up vote
9
down vote

favorite

I'm working on a simple program for Boolean algebra, but the double negation does not work as expected.

I have the following classes:

Operator:

#ifndef OPERATOR_H
#define OPERATOR_H

class Operator 
public:
 virtual int getArity(void) const = 0;
 virtual bool calc(void) const = 0;
;

#endif // OPERATOR_H

False:

#ifndef FALSE_H
#define FALSE_H

#include "operator.h"

class False : public Operator 
public:
 int getArity() const 
 return 0;
 

 bool calc(void) const 
 return false;
 
;

#endif // FALSE_H

Not:

#ifndef NOT_H
#define NOT_H

#include "operator.h"

class Not : public Operator 
public:
 Not(Operator& child) : m_child(child) 
 std::cout << "not constructor called" << std::endl;
 

 int getArity(void) const 
 return 1;
 

 bool calc(void) const 
 return !m_child.calc();
 

private:
 Operator& m_child;
;

#endif // NOT_H

My main.cpp:

#include <iostream>
#include "operator.h"
#include "not.h"
#include "false.h"

using namespace std;

int main(int argc, char *argv) 

 False f;
 Not n = Not(f);
 Not d = Not(n);

 cout << "n.calc(): " << n.calc() <<endl;
 cout << "d.calc(): " << d.calc() <<endl;
 return 0;

Since d = Not(Not(False())) I expect it to be false.

The output is:

not constructor called
n.calc(): 1
d.calc(): 1 <== should be 0

Why is the constructor of the class Not not called with an object of type Not as child?

edited 11 mins ago

Peter Mortensen

13.2k1983111

asked 5 hours ago

Core

543

New contributor

The best way I've found to figure out what happens in cases like this is to implement every constructor and add some debug statements, explicitly logging every constructor called. It can be very informative.
â€“Â Andrew Henle
5 hours ago

Aside: I would highlight this en.cppreference.com/w/cpp/language/operator_alternative and beg you to reconsider what you're doing. (And at last checking, false is not an operator, but a binary value)
â€“Â UKMonkey
5 hours ago

note that T x = T(a); is defined as being the same as T x(a);, since C++17.
â€“Â M.M
5 hours ago

4

"class False : public Operator" Reminds me of a "famous" german author who managed to let his class Wurstbrot inherit from Wurst and Brot which in turn inherit Supermarkt ...
â€“Â J. Doe
4 hours ago

2

@UKMonkey "false" is usually the name of a boolean value, but it is also the name of a nullary operator (meaning it doesn't take any argument) that returns that boolean value.
â€“Â Nelfeal
4 hours ago

Â |Â
show 1 more comment

up vote
9
down vote

favorite

I'm working on a simple program for Boolean algebra, but the double negation does not work as expected.

I have the following classes:

Operator:

#ifndef OPERATOR_H
#define OPERATOR_H

class Operator 
public:
 virtual int getArity(void) const = 0;
 virtual bool calc(void) const = 0;
;

#endif // OPERATOR_H

False:

#ifndef FALSE_H
#define FALSE_H

#include "operator.h"

class False : public Operator 
public:
 int getArity() const 
 return 0;
 

 bool calc(void) const 
 return false;
 
;

#endif // FALSE_H

Not:

#ifndef NOT_H
#define NOT_H

#include "operator.h"

class Not : public Operator 
public:
 Not(Operator& child) : m_child(child) 
 std::cout << "not constructor called" << std::endl;
 

 int getArity(void) const 
 return 1;
 

 bool calc(void) const 
 return !m_child.calc();
 

private:
 Operator& m_child;
;

#endif // NOT_H

My main.cpp:

#include <iostream>
#include "operator.h"
#include "not.h"
#include "false.h"

using namespace std;

int main(int argc, char *argv) 

 False f;
 Not n = Not(f);
 Not d = Not(n);

 cout << "n.calc(): " << n.calc() <<endl;
 cout << "d.calc(): " << d.calc() <<endl;
 return 0;

Since d = Not(Not(False())) I expect it to be false.

The output is:

not constructor called
n.calc(): 1
d.calc(): 1 <== should be 0

Why is the constructor of the class Not not called with an object of type Not as child?

edited 11 mins ago

Peter Mortensen

13.2k1983111

asked 5 hours ago

Core

543

New contributor

I'm working on a simple program for Boolean algebra, but the double negation does not work as expected.

I have the following classes:

Operator:

#ifndef OPERATOR_H
#define OPERATOR_H

class Operator 
public:
 virtual int getArity(void) const = 0;
 virtual bool calc(void) const = 0;
;

#endif // OPERATOR_H

False:

#ifndef FALSE_H
#define FALSE_H

#include "operator.h"

class False : public Operator 
public:
 int getArity() const 
 return 0;
 

 bool calc(void) const 
 return false;
 
;

#endif // FALSE_H

Not:

#ifndef NOT_H
#define NOT_H

#include "operator.h"

class Not : public Operator 
public:
 Not(Operator& child) : m_child(child) 
 std::cout << "not constructor called" << std::endl;
 

 int getArity(void) const 
 return 1;
 

 bool calc(void) const 
 return !m_child.calc();
 

private:
 Operator& m_child;
;

#endif // NOT_H

My main.cpp:

#include <iostream>
#include "operator.h"
#include "not.h"
#include "false.h"

using namespace std;

int main(int argc, char *argv) 

 False f;
 Not n = Not(f);
 Not d = Not(n);

 cout << "n.calc(): " << n.calc() <<endl;
 cout << "d.calc(): " << d.calc() <<endl;
 return 0;

Since d = Not(Not(False())) I expect it to be false.

The output is:

not constructor called
n.calc(): 1
d.calc(): 1 <== should be 0

Why is the constructor of the class Not not called with an object of type Not as child?

c++ inheritance constructor

edited 11 mins ago

Peter Mortensen

13.2k1983111

asked 5 hours ago

Core

543

New contributor

edited 11 mins ago

Peter Mortensen

13.2k1983111

asked 5 hours ago

Core

543

New contributor

edited 11 mins ago

Peter Mortensen

13.2k1983111

edited 11 mins ago

Peter Mortensen

13.2k1983111

edited 11 mins ago

Peter Mortensen

13.2k1983111

asked 5 hours ago

Core

543

New contributor

asked 5 hours ago

Core

543

asked 5 hours ago

Core

543

New contributor

Core is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

The best way I've found to figure out what happens in cases like this is to implement every constructor and add some debug statements, explicitly logging every constructor called. It can be very informative.
â€“Â Andrew Henle
5 hours ago

Aside: I would highlight this en.cppreference.com/w/cpp/language/operator_alternative and beg you to reconsider what you're doing. (And at last checking, false is not an operator, but a binary value)
â€“Â UKMonkey
5 hours ago

note that T x = T(a); is defined as being the same as T x(a);, since C++17.
â€“Â M.M
5 hours ago

4

"class False : public Operator" Reminds me of a "famous" german author who managed to let his class Wurstbrot inherit from Wurst and Brot which in turn inherit Supermarkt ...
â€“Â J. Doe
4 hours ago

2

@UKMonkey "false" is usually the name of a boolean value, but it is also the name of a nullary operator (meaning it doesn't take any argument) that returns that boolean value.
â€“Â Nelfeal
4 hours ago

Â |Â
show 1 more comment

The best way I've found to figure out what happens in cases like this is to implement every constructor and add some debug statements, explicitly logging every constructor called. It can be very informative.
â€“Â Andrew Henle
5 hours ago

Aside: I would highlight this en.cppreference.com/w/cpp/language/operator_alternative and beg you to reconsider what you're doing. (And at last checking, false is not an operator, but a binary value)
â€“Â UKMonkey
5 hours ago

note that T x = T(a); is defined as being the same as T x(a);, since C++17.
â€“Â M.M
5 hours ago

4

"class False : public Operator" Reminds me of a "famous" german author who managed to let his class Wurstbrot inherit from Wurst and Brot which in turn inherit Supermarkt ...
â€“Â J. Doe
4 hours ago

2

@UKMonkey "false" is usually the name of a boolean value, but it is also the name of a nullary operator (meaning it doesn't take any argument) that returns that boolean value.
â€“Â Nelfeal
4 hours ago

The best way I've found to figure out what happens in cases like this is to implement every constructor and add some debug statements, explicitly logging every constructor called. It can be very informative.
â€“Â Andrew Henle
5 hours ago

Aside: I would highlight this en.cppreference.com/w/cpp/language/operator_alternative and beg you to reconsider what you're doing. (And at last checking, false is not an operator, but a binary value)
â€“Â UKMonkey
5 hours ago

note that T x = T(a); is defined as being the same as T x(a);, since C++17.
â€“Â M.M
5 hours ago

"class False : public Operator" Reminds me of a "famous" german author who managed to let his class Wurstbrot inherit from Wurst and Brot which in turn inherit Supermarkt ...
â€“Â J. Doe
4 hours ago

@UKMonkey "false" is usually the name of a boolean value, but it is also the name of a nullary operator (meaning it doesn't take any argument) that returns that boolean value.
â€“Â Nelfeal
4 hours ago

Â |Â
show 1 more comment

1 Answer
1

active

oldest

votes

up vote
18
down vote

accepted

Not d = Not(n); invokes the copy constructor of Not, because the argument is also of type Not. The copy constructor's signature matches better and it's therefore selected.

answered 5 hours ago

Angew

129k11240335

3

More detail: Not(Operator& child) is not a copy-constructor, despite the fact that it can be called with an argument of the same type. The implicitly-defined copy-constructor also exists.
â€“Â M.M
5 hours ago

3

As a possible solution, you could take Operator* child, or pass the value with static_const<Operator&>. To avoid the silent copy, you can delete the copy-constructor and use uniform initialization syntax.
â€“Â JosÃ© Manuel
4 hours ago

add a commentÂ |Â

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
18
down vote

accepted

Not d = Not(n); invokes the copy constructor of Not, because the argument is also of type Not. The copy constructor's signature matches better and it's therefore selected.

answered 5 hours ago

Angew

129k11240335

3

More detail: Not(Operator& child) is not a copy-constructor, despite the fact that it can be called with an argument of the same type. The implicitly-defined copy-constructor also exists.
â€“Â M.M
5 hours ago

3

As a possible solution, you could take Operator* child, or pass the value with static_const<Operator&>. To avoid the silent copy, you can delete the copy-constructor and use uniform initialization syntax.
â€“Â JosÃ© Manuel
4 hours ago

add a commentÂ |Â

up vote
18
down vote

accepted

Not d = Not(n); invokes the copy constructor of Not, because the argument is also of type Not. The copy constructor's signature matches better and it's therefore selected.

answered 5 hours ago

Angew

129k11240335

3

More detail: Not(Operator& child) is not a copy-constructor, despite the fact that it can be called with an argument of the same type. The implicitly-defined copy-constructor also exists.
â€“Â M.M
5 hours ago

3

As a possible solution, you could take Operator* child, or pass the value with static_const<Operator&>. To avoid the silent copy, you can delete the copy-constructor and use uniform initialization syntax.
â€“Â JosÃ© Manuel
4 hours ago

add a commentÂ |Â

up vote
18
down vote

accepted

Not d = Not(n); invokes the copy constructor of Not, because the argument is also of type Not. The copy constructor's signature matches better and it's therefore selected.

answered 5 hours ago

Angew

129k11240335

Not d = Not(n); invokes the copy constructor of Not, because the argument is also of type Not. The copy constructor's signature matches better and it's therefore selected.

answered 5 hours ago

Angew

129k11240335

answered 5 hours ago

Angew

129k11240335

answered 5 hours ago

Angew

129k11240335

answered 5 hours ago

Angew

129k11240335

3

More detail: Not(Operator& child) is not a copy-constructor, despite the fact that it can be called with an argument of the same type. The implicitly-defined copy-constructor also exists.
â€“Â M.M
5 hours ago

3

As a possible solution, you could take Operator* child, or pass the value with static_const<Operator&>. To avoid the silent copy, you can delete the copy-constructor and use uniform initialization syntax.
â€“Â JosÃ© Manuel
4 hours ago

add a commentÂ |Â

3

More detail: Not(Operator& child) is not a copy-constructor, despite the fact that it can be called with an argument of the same type. The implicitly-defined copy-constructor also exists.
â€“Â M.M
5 hours ago

3

As a possible solution, you could take Operator* child, or pass the value with static_const<Operator&>. To avoid the silent copy, you can delete the copy-constructor and use uniform initialization syntax.
â€“Â JosÃ© Manuel
4 hours ago

More detail: Not(Operator& child) is not a copy-constructor, despite the fact that it can be called with an argument of the same type. The implicitly-defined copy-constructor also exists.
â€“Â M.M
5 hours ago

As a possible solution, you could take Operator* child, or pass the value with static_const<Operator&>. To avoid the silent copy, you can delete the copy-constructor and use uniform initialization syntax.
â€“Â JosÃ© Manuel
4 hours ago

add a commentÂ |Â

Core is a new contributor. Be nice, and check out our Code of Conduct.

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