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10

1

I have member function (method) which uses

std::enable_shared_from_this::weak_from_this() 

In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.

Is there a way to check against it at compile time?

c++ c++17 shared-ptr weak-ptr

share|improve this question

edited 5 hours ago

armitus

509114

asked 5 hours ago

KorriKorri

33127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.
  
  – rubenvb
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.
  
  – SergeyA
  5 hours ago
  

  
  
  
  

add a comment | 

10

1

I have member function (method) which uses

std::enable_shared_from_this::weak_from_this() 

In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.

Is there a way to check against it at compile time?

c++ c++17 shared-ptr weak-ptr

share|improve this question

edited 5 hours ago

armitus

509114

asked 5 hours ago

KorriKorri

33127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.
  
  – rubenvb
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.
  
  – SergeyA
  5 hours ago
  

  
  
  
  

add a comment | 

I have member function (method) which uses

std::enable_shared_from_this::weak_from_this() 

In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.

Is there a way to check against it at compile time?

c++ c++17 shared-ptr weak-ptr

share|improve this question

edited 5 hours ago

armitus

509114

asked 5 hours ago

KorriKorri

33127

std::enable_shared_from_this::weak_from_this() 

share|improve this question

edited 5 hours ago

armitus

509114

asked 5 hours ago

KorriKorri

33127

share|improve this question

edited 5 hours ago

armitus

509114

asked 5 hours ago

KorriKorri

33127

edited 5 hours ago

armitus

509114

edited 5 hours ago

armitus

509114

asked 5 hours ago

KorriKorri

33127

asked 5 hours ago

KorriKorri

33127

3 Answers
3

active

oldest

votes

7

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered 5 hours ago

AngewAngew

134k11260353

add a comment | 

3

No there is no way. Consider:

void call_me(struct widget*);



struct widget : std::enable_shared_from_this<widget> {

    widget() {

        call_me(this);

    }



    void display() {

        shared_from_this();

    }

};



// later:



void call_me(widget* w) {

    w->display(); // crash

}

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

16.3k53872

add a comment | 

2

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A {



    // ... whatever ...



    A() { 

        // do construction work

        constructed = true;

    }



    foo() {

        if (not constructed)  { 

            throw std::logic_error("Cannot call foo() during construction"); 

        }

        // the rest of foo

    }



    bool constructed { false };

}

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

answered 3 hours ago

einpoklumeinpoklum

37k28132263

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
  
  – Korri
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.
  
  – Jesper Juhl
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

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7

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered 5 hours ago

AngewAngew

134k11260353

add a comment | 

7

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered 5 hours ago

AngewAngew

134k11260353

add a comment | 

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered 5 hours ago

AngewAngew

134k11260353

share|improve this answer

answered 5 hours ago

AngewAngew

134k11260353

answered 5 hours ago

AngewAngew

134k11260353

answered 5 hours ago

AngewAngew

134k11260353

3

No there is no way. Consider:

void call_me(struct widget*);



struct widget : std::enable_shared_from_this<widget> {

    widget() {

        call_me(this);

    }



    void display() {

        shared_from_this();

    }

};



// later:



void call_me(widget* w) {

    w->display(); // crash

}

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

16.3k53872

add a comment | 

3

No there is no way. Consider:

void call_me(struct widget*);



struct widget : std::enable_shared_from_this<widget> {

    widget() {

        call_me(this);

    }



    void display() {

        shared_from_this();

    }

};



// later:



void call_me(widget* w) {

    w->display(); // crash

}

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

16.3k53872

add a comment | 

No there is no way. Consider:

void call_me(struct widget*);



struct widget : std::enable_shared_from_this<widget> {

    widget() {

        call_me(this);

    }



    void display() {

        shared_from_this();

    }

};



// later:



void call_me(widget* w) {

    w->display(); // crash

}

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

16.3k53872

void call_me(struct widget*);



struct widget : std::enable_shared_from_this<widget> {

    widget() {

        call_me(this);

    }



    void display() {

        shared_from_this();

    }

};



// later:



void call_me(widget* w) {

    w->display(); // crash

}

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

16.3k53872

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

16.3k53872

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

16.3k53872

2

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A {



    // ... whatever ...



    A() { 

        // do construction work

        constructed = true;

    }



    foo() {

        if (not constructed)  { 

            throw std::logic_error("Cannot call foo() during construction"); 

        }

        // the rest of foo

    }



    bool constructed { false };

}

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

answered 3 hours ago

einpoklumeinpoklum

37k28132263

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
  
  – Korri
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.
  
  – Jesper Juhl
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

2

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A {



    // ... whatever ...



    A() { 

        // do construction work

        constructed = true;

    }



    foo() {

        if (not constructed)  { 

            throw std::logic_error("Cannot call foo() during construction"); 

        }

        // the rest of foo

    }



    bool constructed { false };

}

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

answered 3 hours ago

einpoklumeinpoklum

37k28132263

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
  
  – Korri
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.
  
  – Jesper Juhl
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A {



    // ... whatever ...



    A() { 

        // do construction work

        constructed = true;

    }



    foo() {

        if (not constructed)  { 

            throw std::logic_error("Cannot call foo() during construction"); 

        }

        // the rest of foo

    }



    bool constructed { false };

}

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

answered 3 hours ago

einpoklumeinpoklum

37k28132263

class A {



    // ... whatever ...



    A() { 

        // do construction work

        constructed = true;

    }



    foo() {

        if (not constructed)  { 

            throw std::logic_error("Cannot call foo() during construction"); 

        }

        // the rest of foo

    }



    bool constructed { false };

}

share|improve this answer

answered 3 hours ago

einpoklumeinpoklum

37k28132263

answered 3 hours ago

einpoklumeinpoklum

37k28132263

answered 3 hours ago

einpoklumeinpoklum

37k28132263