Mixing
Can I call a virtual destructor using a function pointer?
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
I have class Data which can hold a pointer to an object. I want to be able to call its destructor manually later on, for which I need its address stored in a variable but it seems that taking the address of constructor/destructor is forbidden. Is there any way around this ?
struct Data {
union
long i;
float f;
void* data_ptr;
_data;
std::type_index _typeIndex;
void (*_destructor_ptr)();
template<typename T>
void Init()
if constexpr (std::is_integral<T>::value)
//
else if constexpr (std::is_floating_point<T>::value)
//
else
_data.data_ptr = new T;
_typeIndex = std::type_index(typeid(T));
_destructor_ptr = &T::~T; // << -- can't do this
c++ explicit-destructor-call
asked 4 hours ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
 |Â
show 1 more comment
up vote
6
down vote
favorite
I have class Data which can hold a pointer to an object. I want to be able to call its destructor manually later on, for which I need its address stored in a variable but it seems that taking the address of constructor/destructor is forbidden. Is there any way around this ?
struct Data {
union
long i;
float f;
void* data_ptr;
_data;
std::type_index _typeIndex;
void (*_destructor_ptr)();
template<typename T>
void Init()
if constexpr (std::is_integral<T>::value)
//
else if constexpr (std::is_floating_point<T>::value)
//
else
_data.data_ptr = new T;
_typeIndex = std::type_index(typeid(T));
_destructor_ptr = &T::~T; // << -- can't do this
c++ explicit-destructor-call
asked 4 hours ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
stackoverflow.com/questions/7905272/… ?
– Kamil Cuk
4 hours ago
5
Just to clarify, what's stopping you from callingdelete?
– Praneeth Peiris
4 hours ago
@PraneethPeiris -deleteon what?
– StoryTeller
4 hours ago
2
An alternate was of destructing an arbitrary type is having ashared_ptr<void>which you can assign anynewd type to and it will remember how to destruct it whilst keeping your code from having to remember the type.
– Mike Vine
3 hours ago
@PraneethPeiris delete on a void* will not call the destructor
– const_iterator
3 hours ago
 |Â
show 1 more comment
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I have class Data which can hold a pointer to an object. I want to be able to call its destructor manually later on, for which I need its address stored in a variable but it seems that taking the address of constructor/destructor is forbidden. Is there any way around this ?
struct Data {
union
long i;
float f;
void* data_ptr;
_data;
std::type_index _typeIndex;
void (*_destructor_ptr)();
template<typename T>
void Init()
if constexpr (std::is_integral<T>::value)
//
else if constexpr (std::is_floating_point<T>::value)
//
else
_data.data_ptr = new T;
_typeIndex = std::type_index(typeid(T));
_destructor_ptr = &T::~T; // << -- can't do this
c++ explicit-destructor-call
asked 4 hours ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have class Data which can hold a pointer to an object. I want to be able to call its destructor manually later on, for which I need its address stored in a variable but it seems that taking the address of constructor/destructor is forbidden. Is there any way around this ?
struct Data {
union
long i;
float f;
void* data_ptr;
_data;
std::type_index _typeIndex;
void (*_destructor_ptr)();
template<typename T>
void Init()
if constexpr (std::is_integral<T>::value)
//
else if constexpr (std::is_floating_point<T>::value)
//
else
_data.data_ptr = new T;
_typeIndex = std::type_index(typeid(T));
_destructor_ptr = &T::~T; // << -- can't do this
c++ explicit-destructor-call
c++ explicit-destructor-call
asked 4 hours ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
asked 4 hours ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
const_iterator
837
asked 4 hours ago
const_iterator
837
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
stackoverflow.com/questions/7905272/… ?
– Kamil Cuk
4 hours ago
5
Just to clarify, what's stopping you from callingdelete?
– Praneeth Peiris
4 hours ago
@PraneethPeiris -deleteon what?
– StoryTeller
4 hours ago
2
An alternate was of destructing an arbitrary type is having ashared_ptr<void>which you can assign anynewd type to and it will remember how to destruct it whilst keeping your code from having to remember the type.
– Mike Vine
3 hours ago
@PraneethPeiris delete on a void* will not call the destructor
– const_iterator
3 hours ago
 |Â
show 1 more comment
stackoverflow.com/questions/7905272/… ?
– Kamil Cuk
4 hours ago
5
Just to clarify, what's stopping you from callingdelete?
– Praneeth Peiris
4 hours ago
@PraneethPeiris -deleteon what?
– StoryTeller
4 hours ago
2
An alternate was of destructing an arbitrary type is having ashared_ptr<void>which you can assign anynewd type to and it will remember how to destruct it whilst keeping your code from having to remember the type.
– Mike Vine
3 hours ago
@PraneethPeiris delete on a void* will not call the destructor
– const_iterator
3 hours ago
stackoverflow.com/questions/7905272/… ?
– Kamil Cuk
4 hours ago
stackoverflow.com/questions/7905272/… ?
– Kamil Cuk
4 hours ago
5
5
Just to clarify, what's stopping you from calling
delete ?– Praneeth Peiris
4 hours ago
Just to clarify, what's stopping you from calling
delete ?– Praneeth Peiris
4 hours ago
@PraneethPeiris -
delete on what?– StoryTeller
4 hours ago
@PraneethPeiris -
delete on what?– StoryTeller
4 hours ago
2
2
An alternate was of destructing an arbitrary type is having a
shared_ptr<void> which you can assign any newd type to and it will remember how to destruct it whilst keeping your code from having to remember the type.– Mike Vine
3 hours ago
An alternate was of destructing an arbitrary type is having a
shared_ptr<void> which you can assign any newd type to and it will remember how to destruct it whilst keeping your code from having to remember the type.– Mike Vine
3 hours ago
@PraneethPeiris delete on a void* will not call the destructor
– const_iterator
3 hours ago
@PraneethPeiris delete on a void* will not call the destructor
– const_iterator
3 hours ago
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Store a lambda, suitably converted:
void (*_destructor_ptr)(void *v);
// ...
_destructor_ptr = +(void* v) delete static_cast<T*>(v); ;
Note that you must pass _data.data_ptr for v. If you intend to store a plain function pointer, the lambda may not capture or implicitly refer to _data.data_ptr.
answered 4 hours ago
StoryTeller
84.7k12171237
1
Why do you need the + sign before the lambda ?
– const_iterator
2 hours ago
1
@const_iterator for the plus sign see this one here stackoverflow.com/questions/17822131/…
– Myz
2 hours ago
1
@const_iterator - Sorcery to force coercion into a pointer. It's not really needed here, the assignment itself would do it. But at times you want to ensure it happens, and it's a bit out of force of habit for me.
– StoryTeller
2 hours ago
1
@const_iterator - You get a unique lambda type for each type, not for each object. Converting it to a function pointer also produces one function per type. You can go the more verbose route, but it doesn't gain you anything here. If you like the more descriptive name of the custom struct, then by all means.
– StoryTeller
1 hour ago
2
@const_iterator - Also, if you produced an alternative answer to your question, it belongs in the answer section, not as an edit to your post. You can even accept it instead if you wish. It's how SO is meant to work.
– StoryTeller
1 hour ago
 |Â
show 1 more comment
up vote
5
down vote
There's also this solution if your compiler doesn't support lambdas:
template<typename T>
struct DestructorHelper
static void Destroy(void * v)
delete static_cast<T*>(v);
;
and use it as:
_destructor_ptr = &DestructorHelper<T>::Destroy;
answered 1 hour ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Store a lambda, suitably converted:
void (*_destructor_ptr)(void *v);
// ...
_destructor_ptr = +(void* v) delete static_cast<T*>(v); ;
Note that you must pass _data.data_ptr for v. If you intend to store a plain function pointer, the lambda may not capture or implicitly refer to _data.data_ptr.
answered 4 hours ago
StoryTeller
84.7k12171237
1
Why do you need the + sign before the lambda ?
– const_iterator
2 hours ago
1
@const_iterator for the plus sign see this one here stackoverflow.com/questions/17822131/…
– Myz
2 hours ago
1
@const_iterator - Sorcery to force coercion into a pointer. It's not really needed here, the assignment itself would do it. But at times you want to ensure it happens, and it's a bit out of force of habit for me.
– StoryTeller
2 hours ago
1
@const_iterator - You get a unique lambda type for each type, not for each object. Converting it to a function pointer also produces one function per type. You can go the more verbose route, but it doesn't gain you anything here. If you like the more descriptive name of the custom struct, then by all means.
– StoryTeller
1 hour ago
2
@const_iterator - Also, if you produced an alternative answer to your question, it belongs in the answer section, not as an edit to your post. You can even accept it instead if you wish. It's how SO is meant to work.
– StoryTeller
1 hour ago
 |Â
show 1 more comment
up vote
5
down vote
accepted
Store a lambda, suitably converted:
void (*_destructor_ptr)(void *v);
// ...
_destructor_ptr = +(void* v) delete static_cast<T*>(v); ;
Note that you must pass _data.data_ptr for v. If you intend to store a plain function pointer, the lambda may not capture or implicitly refer to _data.data_ptr.
answered 4 hours ago
StoryTeller
84.7k12171237
1
Why do you need the + sign before the lambda ?
– const_iterator
2 hours ago
1
@const_iterator for the plus sign see this one here stackoverflow.com/questions/17822131/…
– Myz
2 hours ago
1
@const_iterator - Sorcery to force coercion into a pointer. It's not really needed here, the assignment itself would do it. But at times you want to ensure it happens, and it's a bit out of force of habit for me.
– StoryTeller
2 hours ago
1
@const_iterator - You get a unique lambda type for each type, not for each object. Converting it to a function pointer also produces one function per type. You can go the more verbose route, but it doesn't gain you anything here. If you like the more descriptive name of the custom struct, then by all means.
– StoryTeller
1 hour ago
2
@const_iterator - Also, if you produced an alternative answer to your question, it belongs in the answer section, not as an edit to your post. You can even accept it instead if you wish. It's how SO is meant to work.
– StoryTeller
1 hour ago
 |Â
show 1 more comment
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Store a lambda, suitably converted:
void (*_destructor_ptr)(void *v);
// ...
_destructor_ptr = +(void* v) delete static_cast<T*>(v); ;
Note that you must pass _data.data_ptr for v. If you intend to store a plain function pointer, the lambda may not capture or implicitly refer to _data.data_ptr.
answered 4 hours ago
StoryTeller
84.7k12171237
Store a lambda, suitably converted:
void (*_destructor_ptr)(void *v);
// ...
_destructor_ptr = +(void* v) delete static_cast<T*>(v); ;
Note that you must pass _data.data_ptr for v. If you intend to store a plain function pointer, the lambda may not capture or implicitly refer to _data.data_ptr.
answered 4 hours ago
StoryTeller
84.7k12171237
answered 4 hours ago
StoryTeller
84.7k12171237
answered 4 hours ago
StoryTeller
84.7k12171237
answered 4 hours ago
StoryTeller
84.7k12171237
84.7k12171237
1
Why do you need the + sign before the lambda ?
– const_iterator
2 hours ago
1
@const_iterator for the plus sign see this one here stackoverflow.com/questions/17822131/…
– Myz
2 hours ago
1
@const_iterator - Sorcery to force coercion into a pointer. It's not really needed here, the assignment itself would do it. But at times you want to ensure it happens, and it's a bit out of force of habit for me.
– StoryTeller
2 hours ago
1
@const_iterator - You get a unique lambda type for each type, not for each object. Converting it to a function pointer also produces one function per type. You can go the more verbose route, but it doesn't gain you anything here. If you like the more descriptive name of the custom struct, then by all means.
– StoryTeller
1 hour ago
2
@const_iterator - Also, if you produced an alternative answer to your question, it belongs in the answer section, not as an edit to your post. You can even accept it instead if you wish. It's how SO is meant to work.
– StoryTeller
1 hour ago
 |Â
show 1 more comment
1
Why do you need the + sign before the lambda ?
– const_iterator
2 hours ago
1
@const_iterator for the plus sign see this one here stackoverflow.com/questions/17822131/…
– Myz
2 hours ago
1
@const_iterator - Sorcery to force coercion into a pointer. It's not really needed here, the assignment itself would do it. But at times you want to ensure it happens, and it's a bit out of force of habit for me.
– StoryTeller
2 hours ago
1
@const_iterator - You get a unique lambda type for each type, not for each object. Converting it to a function pointer also produces one function per type. You can go the more verbose route, but it doesn't gain you anything here. If you like the more descriptive name of the custom struct, then by all means.
– StoryTeller
1 hour ago
2
@const_iterator - Also, if you produced an alternative answer to your question, it belongs in the answer section, not as an edit to your post. You can even accept it instead if you wish. It's how SO is meant to work.
– StoryTeller
1 hour ago
1
1
Why do you need the + sign before the lambda ?
– const_iterator
2 hours ago
Why do you need the + sign before the lambda ?
– const_iterator
2 hours ago
1
1
@const_iterator for the plus sign see this one here stackoverflow.com/questions/17822131/…
– Myz
2 hours ago
@const_iterator for the plus sign see this one here stackoverflow.com/questions/17822131/…
– Myz
2 hours ago
1
1
@const_iterator - Sorcery to force coercion into a pointer. It's not really needed here, the assignment itself would do it. But at times you want to ensure it happens, and it's a bit out of force of habit for me.
– StoryTeller
2 hours ago
@const_iterator - Sorcery to force coercion into a pointer. It's not really needed here, the assignment itself would do it. But at times you want to ensure it happens, and it's a bit out of force of habit for me.
– StoryTeller
2 hours ago
1
1
@const_iterator - You get a unique lambda type for each type, not for each object. Converting it to a function pointer also produces one function per type. You can go the more verbose route, but it doesn't gain you anything here. If you like the more descriptive name of the custom struct, then by all means.
– StoryTeller
1 hour ago
@const_iterator - You get a unique lambda type for each type, not for each object. Converting it to a function pointer also produces one function per type. You can go the more verbose route, but it doesn't gain you anything here. If you like the more descriptive name of the custom struct, then by all means.
– StoryTeller
1 hour ago
2
2
@const_iterator - Also, if you produced an alternative answer to your question, it belongs in the answer section, not as an edit to your post. You can even accept it instead if you wish. It's how SO is meant to work.
– StoryTeller
1 hour ago
@const_iterator - Also, if you produced an alternative answer to your question, it belongs in the answer section, not as an edit to your post. You can even accept it instead if you wish. It's how SO is meant to work.
– StoryTeller
1 hour ago
 |Â
show 1 more comment
up vote
5
down vote
There's also this solution if your compiler doesn't support lambdas:
template<typename T>
struct DestructorHelper
static void Destroy(void * v)
delete static_cast<T*>(v);
;
and use it as:
_destructor_ptr = &DestructorHelper<T>::Destroy;
answered 1 hour ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
5
down vote
There's also this solution if your compiler doesn't support lambdas:
template<typename T>
struct DestructorHelper
static void Destroy(void * v)
delete static_cast<T*>(v);
;
and use it as:
_destructor_ptr = &DestructorHelper<T>::Destroy;
answered 1 hour ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
5
down vote
up vote
5
down vote
There's also this solution if your compiler doesn't support lambdas:
template<typename T>
struct DestructorHelper
static void Destroy(void * v)
delete static_cast<T*>(v);
;
and use it as:
_destructor_ptr = &DestructorHelper<T>::Destroy;
answered 1 hour ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There's also this solution if your compiler doesn't support lambdas:
template<typename T>
struct DestructorHelper
static void Destroy(void * v)
delete static_cast<T*>(v);
;
and use it as:
_destructor_ptr = &DestructorHelper<T>::Destroy;
answered 1 hour ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
const_iterator
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
const_iterator
837
answered 1 hour ago
const_iterator
837
837
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
const_iterator is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
const_iterator is a new contributor. Be nice, and check out our Code of Conduct.
Â
draft saved
draft discarded
const_iterator is a new contributor. Be nice, and check out our Code of Conduct.
const_iterator is a new contributor. Be nice, and check out our Code of Conduct.
const_iterator is a new contributor. Be nice, and check out our Code of Conduct.
Â
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f52737451%2fcan-i-call-a-virtual-destructor-using-a-function-pointer%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
stackoverflow.com/questions/7905272/… ?
– Kamil Cuk
4 hours ago
5
Just to clarify, what's stopping you from calling
delete?– Praneeth Peiris
4 hours ago
@PraneethPeiris -
deleteon what?– StoryTeller
4 hours ago
2
An alternate was of destructing an arbitrary type is having a
shared_ptr<void>which you can assign anynewd type to and it will remember how to destruct it whilst keeping your code from having to remember the type.– Mike Vine
3 hours ago
@PraneethPeiris delete on a void* will not call the destructor
– const_iterator
3 hours ago