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How to initialize C++ reference with different constructors based on condition?

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1

1

The reference variable foo below is initialized with either an instance of Foo or its derived class Bar based on condition. Strangely enough, based on the output of the say() method, foo seems to be an instance of Foo rather than an instance of Bar — why?

#include <iostream>

class Foo 
public:
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 const Foo& foo = condition ? Foo() : Bar();
 foo.say(); // outputs "Foo” ??? 
 return 0;

If I annotate each constructor I can see that the Bar constructor is getting invoked when evaluating the ternary expression. If I annotate each destructor I see that Bar destructor is invoked before foo is initialized — this tells me that a temporary Bar object is created by the ternary operator, but is destroyed before initialization — why?

Compiled with (Apple LLVM version 9.0.0 (clang-900.0.39.2))

clang++ -Wall -std=c++11 foo.cpp -o foo

c++ reference

share|improve this question

asked Mar 28 at 14:16

wcochranwcochran

5,69744 gold badges3939 silver badges5151 bronze badges

add a comment
 | 

1

1

The reference variable foo below is initialized with either an instance of Foo or its derived class Bar based on condition. Strangely enough, based on the output of the say() method, foo seems to be an instance of Foo rather than an instance of Bar — why?

#include <iostream>

class Foo 
public:
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 const Foo& foo = condition ? Foo() : Bar();
 foo.say(); // outputs "Foo” ??? 
 return 0;

If I annotate each constructor I can see that the Bar constructor is getting invoked when evaluating the ternary expression. If I annotate each destructor I see that Bar destructor is invoked before foo is initialized — this tells me that a temporary Bar object is created by the ternary operator, but is destroyed before initialization — why?

Compiled with (Apple LLVM version 9.0.0 (clang-900.0.39.2))

clang++ -Wall -std=c++11 foo.cpp -o foo

c++ reference

share|improve this question

asked Mar 28 at 14:16

wcochranwcochran

5,69744 gold badges3939 silver badges5151 bronze badges

add a comment
 | 

The reference variable foo below is initialized with either an instance of Foo or its derived class Bar based on condition. Strangely enough, based on the output of the say() method, foo seems to be an instance of Foo rather than an instance of Bar — why?

#include <iostream>

class Foo 
public:
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 const Foo& foo = condition ? Foo() : Bar();
 foo.say(); // outputs "Foo” ??? 
 return 0;

If I annotate each constructor I can see that the Bar constructor is getting invoked when evaluating the ternary expression. If I annotate each destructor I see that Bar destructor is invoked before foo is initialized — this tells me that a temporary Bar object is created by the ternary operator, but is destroyed before initialization — why?

Compiled with (Apple LLVM version 9.0.0 (clang-900.0.39.2))

clang++ -Wall -std=c++11 foo.cpp -o foo

c++ reference

share|improve this question

asked Mar 28 at 14:16

wcochranwcochran

5,69744 gold badges3939 silver badges5151 bronze badges

#include <iostream>

class Foo 
public:
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 const Foo& foo = condition ? Foo() : Bar();
 foo.say(); // outputs "Foo” ??? 
 return 0;

clang++ -Wall -std=c++11 foo.cpp -o foo

share|improve this question

asked Mar 28 at 14:16

wcochranwcochran

5,69744 gold badges3939 silver badges5151 bronze badges

share|improve this question

asked Mar 28 at 14:16

wcochranwcochran

5,69744 gold badges3939 silver badges5151 bronze badges

asked Mar 28 at 14:16

wcochranwcochran

5,69744 gold badges3939 silver badges5151 bronze badges

asked Mar 28 at 14:16

wcochranwcochran

5,69744 gold badges3939 silver badges5151 bronze badges

1 Answer
1

active

oldest

votes

5

The problem with

const Foo& foo = condition ? Foo() : Bar();

is that both parts need to return the same type. Since Foo() and Bar() aren't the same type, the compiler tries to convert them. The only valid conversion it can do is to slice Bar() into its Foo part. This means no matter what you get, you'll be binding a reference to a Foo and the Bar part disappears.

To fix this you'll need to use pointers like

#include <iostream>
#include <memory>

class Foo 
public:
 virtual ~Foo() = default; // don't forget to add this when using polymorphism
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 auto foo = condition ? std::make_unique<Foo>() : std::make_unique<Bar>();
 foo->say(); // outputs "Bar" now 
 return 0;

share|improve this answer

answered Mar 28 at 14:26

NathanOliverNathanOliver

115k1919 gold badges182182 silver badges263263 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Excellent. Bonus: using unique ptrs saves me from having to explicitly delete as well.
  
  – wcochran
  Mar 28 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @wcochran Yeah, not having to write new and delete is really nice.
  
  – NathanOliver
  Mar 28 at 15:26
  

  
  
  
  

add a comment
 | 

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5

The problem with

const Foo& foo = condition ? Foo() : Bar();

is that both parts need to return the same type. Since Foo() and Bar() aren't the same type, the compiler tries to convert them. The only valid conversion it can do is to slice Bar() into its Foo part. This means no matter what you get, you'll be binding a reference to a Foo and the Bar part disappears.

To fix this you'll need to use pointers like

#include <iostream>
#include <memory>

class Foo 
public:
 virtual ~Foo() = default; // don't forget to add this when using polymorphism
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 auto foo = condition ? std::make_unique<Foo>() : std::make_unique<Bar>();
 foo->say(); // outputs "Bar" now 
 return 0;

share|improve this answer

answered Mar 28 at 14:26

NathanOliverNathanOliver

115k1919 gold badges182182 silver badges263263 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Excellent. Bonus: using unique ptrs saves me from having to explicitly delete as well.
  
  – wcochran
  Mar 28 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @wcochran Yeah, not having to write new and delete is really nice.
  
  – NathanOliver
  Mar 28 at 15:26
  

  
  
  
  

add a comment
 | 

5

The problem with

const Foo& foo = condition ? Foo() : Bar();

is that both parts need to return the same type. Since Foo() and Bar() aren't the same type, the compiler tries to convert them. The only valid conversion it can do is to slice Bar() into its Foo part. This means no matter what you get, you'll be binding a reference to a Foo and the Bar part disappears.

To fix this you'll need to use pointers like

#include <iostream>
#include <memory>

class Foo 
public:
 virtual ~Foo() = default; // don't forget to add this when using polymorphism
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 auto foo = condition ? std::make_unique<Foo>() : std::make_unique<Bar>();
 foo->say(); // outputs "Bar" now 
 return 0;

share|improve this answer

answered Mar 28 at 14:26

NathanOliverNathanOliver

115k1919 gold badges182182 silver badges263263 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Excellent. Bonus: using unique ptrs saves me from having to explicitly delete as well.
  
  – wcochran
  Mar 28 at 14:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @wcochran Yeah, not having to write new and delete is really nice.
  
  – NathanOliver
  Mar 28 at 15:26
  

  
  
  
  

add a comment
 | 

The problem with

const Foo& foo = condition ? Foo() : Bar();

is that both parts need to return the same type. Since Foo() and Bar() aren't the same type, the compiler tries to convert them. The only valid conversion it can do is to slice Bar() into its Foo part. This means no matter what you get, you'll be binding a reference to a Foo and the Bar part disappears.

To fix this you'll need to use pointers like

#include <iostream>
#include <memory>

class Foo 
public:
 virtual ~Foo() = default; // don't forget to add this when using polymorphism
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 auto foo = condition ? std::make_unique<Foo>() : std::make_unique<Bar>();
 foo->say(); // outputs "Bar" now 
 return 0;

share|improve this answer

answered Mar 28 at 14:26

NathanOliverNathanOliver

115k1919 gold badges182182 silver badges263263 bronze badges

const Foo& foo = condition ? Foo() : Bar();

#include <iostream>
#include <memory>

class Foo 
public:
 virtual ~Foo() = default; // don't forget to add this when using polymorphism
 virtual void say() const 
 std::cout << "Foon";
 
;

class Bar : public Foo 
public:
 virtual void say() const 
 std::cout << "Barn";
 
;

int main() 
 constexpr bool condition = false;
 auto foo = condition ? std::make_unique<Foo>() : std::make_unique<Bar>();
 foo->say(); // outputs "Bar" now 
 return 0;

share|improve this answer

answered Mar 28 at 14:26

NathanOliverNathanOliver

115k1919 gold badges182182 silver badges263263 bronze badges

answered Mar 28 at 14:26

NathanOliverNathanOliver

115k1919 gold badges182182 silver badges263263 bronze badges

answered Mar 28 at 14:26

NathanOliverNathanOliver

115k1919 gold badges182182 silver badges263263 bronze badges