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Is there a way to make member function NOT callable from constructor?
What are the differences between a pointer variable and a reference variable in C++?Can I call a constructor from another constructor (do constructor chaining) in C++?Throwing exceptions from constructorsHow do I call ::std::make_shared on a class with only protected or private constructors?Calling a base member in constructor in multiple inheritance in C++Equality-compare std::weak_ptrClass inheritance: Constructor and member functions of class not recognized by compilerHow does shared_ptr<T> detect that T derives from enable_shared_from_this<T>?enable_shared_from_this derived class methods are undefined referenceDefault move constructor with mutex member
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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
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
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
15 hours ago
3
Nice question. One way would be to make a dummy class with pure virtual functionweak_from_this
and inherit yours from it. This will make it a hard compile error.
– SergeyA
15 hours ago
3
@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
– Bakuriu
9 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
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
c++ c++17 shared-ptr weak-ptr
edited 15 hours ago
armitus
524114
524114
asked 16 hours ago
KorriKorri
36129
36129
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
15 hours ago
3
Nice question. One way would be to make a dummy class with pure virtual functionweak_from_this
and inherit yours from it. This will make it a hard compile error.
– SergeyA
15 hours ago
3
@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
– Bakuriu
9 hours ago
add a comment |
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
15 hours ago
3
Nice question. One way would be to make a dummy class with pure virtual functionweak_from_this
and inherit yours from it. This will make it a hard compile error.
– SergeyA
15 hours ago
3
@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
– Bakuriu
9 hours ago
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
15 hours ago
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
15 hours ago
3
3
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
15 hours ago
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
15 hours ago
3
3
@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
– Bakuriu
9 hours ago
@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
– Bakuriu
9 hours ago
add a comment |
3 Answers
3
active
oldest
votes
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.
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.
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 noexcept
s though.
An alternative to throwing could be assert()
'ing.
Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go withassert(!wptr.expired())
. I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
– Korri
13 hours ago
@Korri remember thatassert
s 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; firstassert
thenthrow
, or justthrow
.
– Jesper Juhl
12 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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.
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.
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.
answered 15 hours ago
AngewAngew
135k11261354
135k11261354
add a comment |
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.
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.
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.
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.
answered 15 hours ago
Guillaume RacicotGuillaume Racicot
16.3k53872
16.3k53872
add a comment |
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 noexcept
s though.
An alternative to throwing could be assert()
'ing.
Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go withassert(!wptr.expired())
. I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
– Korri
13 hours ago
@Korri remember thatassert
s 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; firstassert
thenthrow
, or justthrow
.
– Jesper Juhl
12 hours 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 noexcept
s though.
An alternative to throwing could be assert()
'ing.
Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go withassert(!wptr.expired())
. I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
– Korri
13 hours ago
@Korri remember thatassert
s 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; firstassert
thenthrow
, or justthrow
.
– Jesper Juhl
12 hours 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 noexcept
s though.
An alternative to throwing could be assert()
'ing.
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 noexcept
s though.
An alternative to throwing could be assert()
'ing.
answered 13 hours ago
einpoklumeinpoklum
37k28132263
37k28132263
Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go withassert(!wptr.expired())
. I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
– Korri
13 hours ago
@Korri remember thatassert
s 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; firstassert
thenthrow
, or justthrow
.
– Jesper Juhl
12 hours ago
add a comment |
Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go withassert(!wptr.expired())
. I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
– Korri
13 hours ago
@Korri remember thatassert
s 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; firstassert
thenthrow
, or justthrow
.
– Jesper Juhl
12 hours ago
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
13 hours ago
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
13 hours ago
@Korri remember that
assert
s 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
12 hours ago
@Korri remember that
assert
s 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
12 hours ago
add a comment |
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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
15 hours ago
3
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
15 hours ago
3
@SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
– Bakuriu
9 hours ago