Figuring out the constness of an object within its destructor
up vote
9
down vote
favorite
I have a class Stuff, with two functions foo (const and non-const):
class Stuff
{
public:
~Stuff() { foo(); }
void foo() const { cout << "const foo" << endl; }
void foo() { cout << "non-const foo" << endl; }
};
Here's what I am trying to do:
If the stuff was const, call const foo in the destructor of Stuff.
if the stuff wasn't const, call non-const foo in the destructor of Stuff.
I was hoping that just defining the destructor as shown above would work, but it turns out that the constness is stripped away right before executing the destructor (it is enforced right after the constructor completes, so I cannot set any flag there either). To make it clearer, here's an example:
{ Stuff stuff; }
{ const Stuff cstuff; }
This code prints "non-const foo" twice. I want it to print "non-const foo", followed by "const foo". Is that possible with C++?
EDIT: A few people are asking for more context. In the real code, stuff is basically a handle to some data. If stuff is accessed in a non-const manner, I assume that the data has been modified, so I need to communicate that to other processes (MPI) using the foo function (after I am done modifying it -> in the destructor, when I release the handle). If it was accessed in a const-manner, I know I don't need to transfer anything, so I am calling non-const foo, that does nothing.
c++ const destructor
asked 6 hours ago
Touloudou
1009
|
show 3 more comments
up vote
9
down vote
favorite
I have a class Stuff, with two functions foo (const and non-const):
class Stuff
{
public:
~Stuff() { foo(); }
void foo() const { cout << "const foo" << endl; }
void foo() { cout << "non-const foo" << endl; }
};
Here's what I am trying to do:
If the stuff was const, call const foo in the destructor of Stuff.
if the stuff wasn't const, call non-const foo in the destructor of Stuff.
I was hoping that just defining the destructor as shown above would work, but it turns out that the constness is stripped away right before executing the destructor (it is enforced right after the constructor completes, so I cannot set any flag there either). To make it clearer, here's an example:
{ Stuff stuff; }
{ const Stuff cstuff; }
This code prints "non-const foo" twice. I want it to print "non-const foo", followed by "const foo". Is that possible with C++?
EDIT: A few people are asking for more context. In the real code, stuff is basically a handle to some data. If stuff is accessed in a non-const manner, I assume that the data has been modified, so I need to communicate that to other processes (MPI) using the foo function (after I am done modifying it -> in the destructor, when I release the handle). If it was accessed in a const-manner, I know I don't need to transfer anything, so I am calling non-const foo, that does nothing.
c++ const destructor
asked 6 hours ago
Touloudou
1009
The destructor destroys the object, by definition it's not const. Const only applies if you are callingstuff.foo()orcstuff.foo().
– Matthieu Brucher
6 hours ago
6
It's not possible to detect whether the object isconstin the constructor or destructor. This might be an XY problem.
– Brian
6 hours ago
1
It is not const both in the constructor and the destructor. You want to know in the constructor if it will be, and in the destructor if it was. Strange
– bruno
6 hours ago
2
As per your edit: It sounds as though you are trying to implement transactional semantics of some kind. Would it perhaps make sense to keep track of whether state was modified using a flag (which only gets set on the mutable code-path), and implementing the conditional behavior in the destructor based on whether this flag was toggled?
– Bitwize
6 hours ago
5
It sounds to me like you might want two different types,MutableStuffandImmutableStuff.
– molbdnilo
6 hours ago
|
show 3 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I have a class Stuff, with two functions foo (const and non-const):
class Stuff
{
public:
~Stuff() { foo(); }
void foo() const { cout << "const foo" << endl; }
void foo() { cout << "non-const foo" << endl; }
};
Here's what I am trying to do:
If the stuff was const, call const foo in the destructor of Stuff.
if the stuff wasn't const, call non-const foo in the destructor of Stuff.
I was hoping that just defining the destructor as shown above would work, but it turns out that the constness is stripped away right before executing the destructor (it is enforced right after the constructor completes, so I cannot set any flag there either). To make it clearer, here's an example:
{ Stuff stuff; }
{ const Stuff cstuff; }
This code prints "non-const foo" twice. I want it to print "non-const foo", followed by "const foo". Is that possible with C++?
EDIT: A few people are asking for more context. In the real code, stuff is basically a handle to some data. If stuff is accessed in a non-const manner, I assume that the data has been modified, so I need to communicate that to other processes (MPI) using the foo function (after I am done modifying it -> in the destructor, when I release the handle). If it was accessed in a const-manner, I know I don't need to transfer anything, so I am calling non-const foo, that does nothing.
c++ const destructor
asked 6 hours ago
Touloudou
1009
I have a class Stuff, with two functions foo (const and non-const):
class Stuff
{
public:
~Stuff() { foo(); }
void foo() const { cout << "const foo" << endl; }
void foo() { cout << "non-const foo" << endl; }
};
Here's what I am trying to do:
If the stuff was const, call const foo in the destructor of Stuff.
if the stuff wasn't const, call non-const foo in the destructor of Stuff.
I was hoping that just defining the destructor as shown above would work, but it turns out that the constness is stripped away right before executing the destructor (it is enforced right after the constructor completes, so I cannot set any flag there either). To make it clearer, here's an example:
{ Stuff stuff; }
{ const Stuff cstuff; }
This code prints "non-const foo" twice. I want it to print "non-const foo", followed by "const foo". Is that possible with C++?
EDIT: A few people are asking for more context. In the real code, stuff is basically a handle to some data. If stuff is accessed in a non-const manner, I assume that the data has been modified, so I need to communicate that to other processes (MPI) using the foo function (after I am done modifying it -> in the destructor, when I release the handle). If it was accessed in a const-manner, I know I don't need to transfer anything, so I am calling non-const foo, that does nothing.
c++ const destructor
c++ const destructor
asked 6 hours ago
Touloudou
1009
asked 6 hours ago
Touloudou
1009
edited 6 hours ago
asked 6 hours ago
Touloudou
1009
asked 6 hours ago
Touloudou
1009
asked 6 hours ago
Touloudou
1009
1009
The destructor destroys the object, by definition it's not const. Const only applies if you are callingstuff.foo()orcstuff.foo().
– Matthieu Brucher
6 hours ago
6
It's not possible to detect whether the object isconstin the constructor or destructor. This might be an XY problem.
– Brian
6 hours ago
1
It is not const both in the constructor and the destructor. You want to know in the constructor if it will be, and in the destructor if it was. Strange
– bruno
6 hours ago
2
As per your edit: It sounds as though you are trying to implement transactional semantics of some kind. Would it perhaps make sense to keep track of whether state was modified using a flag (which only gets set on the mutable code-path), and implementing the conditional behavior in the destructor based on whether this flag was toggled?
– Bitwize
6 hours ago
5
It sounds to me like you might want two different types,MutableStuffandImmutableStuff.
– molbdnilo
6 hours ago
|
show 3 more comments
The destructor destroys the object, by definition it's not const. Const only applies if you are callingstuff.foo()orcstuff.foo().
– Matthieu Brucher
6 hours ago
6
It's not possible to detect whether the object isconstin the constructor or destructor. This might be an XY problem.
– Brian
6 hours ago
1
It is not const both in the constructor and the destructor. You want to know in the constructor if it will be, and in the destructor if it was. Strange
– bruno
6 hours ago
2
As per your edit: It sounds as though you are trying to implement transactional semantics of some kind. Would it perhaps make sense to keep track of whether state was modified using a flag (which only gets set on the mutable code-path), and implementing the conditional behavior in the destructor based on whether this flag was toggled?
– Bitwize
6 hours ago
5
It sounds to me like you might want two different types,MutableStuffandImmutableStuff.
– molbdnilo
6 hours ago
The destructor destroys the object, by definition it's not const. Const only applies if you are calling
stuff.foo() or cstuff.foo().– Matthieu Brucher
6 hours ago
The destructor destroys the object, by definition it's not const. Const only applies if you are calling
stuff.foo() or cstuff.foo().– Matthieu Brucher
6 hours ago
6
6
It's not possible to detect whether the object is
const in the constructor or destructor. This might be an XY problem.– Brian
6 hours ago
It's not possible to detect whether the object is
const in the constructor or destructor. This might be an XY problem.– Brian
6 hours ago
1
1
It is not const both in the constructor and the destructor. You want to know in the constructor if it will be, and in the destructor if it was. Strange
– bruno
6 hours ago
It is not const both in the constructor and the destructor. You want to know in the constructor if it will be, and in the destructor if it was. Strange
– bruno
6 hours ago
2
2
As per your edit: It sounds as though you are trying to implement transactional semantics of some kind. Would it perhaps make sense to keep track of whether state was modified using a flag (which only gets set on the mutable code-path), and implementing the conditional behavior in the destructor based on whether this flag was toggled?
– Bitwize
6 hours ago
As per your edit: It sounds as though you are trying to implement transactional semantics of some kind. Would it perhaps make sense to keep track of whether state was modified using a flag (which only gets set on the mutable code-path), and implementing the conditional behavior in the destructor based on whether this flag was toggled?
– Bitwize
6 hours ago
5
5
It sounds to me like you might want two different types,
MutableStuff and ImmutableStuff.– molbdnilo
6 hours ago
It sounds to me like you might want two different types,
MutableStuff and ImmutableStuff.– molbdnilo
6 hours ago
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
14
down vote
accepted
[...]
constandvolatile
semantics (7.1.6.1) are not applied on an object under destruction. They stop being in effect when the
destructor for the most derived object (1.8) starts.
12.4/2 [class.dtor] in N4141.
So no, this is not possible.
answered 6 hours ago
Baum mit Augen
40.1k12114147
But note that you still can callstatic_cast<const Stuff&>(*this).foo()to call the const versions of functions.
– Artyer
6 hours ago
@Artyer But you'd have to know whether or not you want to, which appears to be the core of the question.
– Baum mit Augen
6 hours ago
And in other methods? Is it possible to determine the constancy of the created object from inside the class in principle?
– Dmytro Dadyka
6 hours ago
@DmytroDadyka In normal member functions, you can overload based on constness, as is demonstrated in OP.
– Baum mit Augen
6 hours ago
1
@DmytroDadyka No, constructors have basically the same clause I quoted here, see 12.1/3.
– Baum mit Augen
6 hours ago
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53840945%2ffiguring-out-the-constness-of-an-object-within-its-destructor%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
[...]
constandvolatile
semantics (7.1.6.1) are not applied on an object under destruction. They stop being in effect when the
destructor for the most derived object (1.8) starts.
12.4/2 [class.dtor] in N4141.
So no, this is not possible.
answered 6 hours ago
Baum mit Augen
40.1k12114147
But note that you still can callstatic_cast<const Stuff&>(*this).foo()to call the const versions of functions.
– Artyer
6 hours ago
@Artyer But you'd have to know whether or not you want to, which appears to be the core of the question.
– Baum mit Augen
6 hours ago
And in other methods? Is it possible to determine the constancy of the created object from inside the class in principle?
– Dmytro Dadyka
6 hours ago
@DmytroDadyka In normal member functions, you can overload based on constness, as is demonstrated in OP.
– Baum mit Augen
6 hours ago
1
@DmytroDadyka No, constructors have basically the same clause I quoted here, see 12.1/3.
– Baum mit Augen
6 hours ago
|
show 1 more comment
up vote
14
down vote
accepted
[...]
constandvolatile
semantics (7.1.6.1) are not applied on an object under destruction. They stop being in effect when the
destructor for the most derived object (1.8) starts.
12.4/2 [class.dtor] in N4141.
So no, this is not possible.
answered 6 hours ago
Baum mit Augen
40.1k12114147
But note that you still can callstatic_cast<const Stuff&>(*this).foo()to call the const versions of functions.
– Artyer
6 hours ago
@Artyer But you'd have to know whether or not you want to, which appears to be the core of the question.
– Baum mit Augen
6 hours ago
And in other methods? Is it possible to determine the constancy of the created object from inside the class in principle?
– Dmytro Dadyka
6 hours ago
@DmytroDadyka In normal member functions, you can overload based on constness, as is demonstrated in OP.
– Baum mit Augen
6 hours ago
1
@DmytroDadyka No, constructors have basically the same clause I quoted here, see 12.1/3.
– Baum mit Augen
6 hours ago
|
show 1 more comment
up vote
14
down vote
accepted
up vote
14
down vote
accepted
[...]
constandvolatile
semantics (7.1.6.1) are not applied on an object under destruction. They stop being in effect when the
destructor for the most derived object (1.8) starts.
12.4/2 [class.dtor] in N4141.
So no, this is not possible.
answered 6 hours ago
Baum mit Augen
40.1k12114147
[...]
constandvolatile
semantics (7.1.6.1) are not applied on an object under destruction. They stop being in effect when the
destructor for the most derived object (1.8) starts.
12.4/2 [class.dtor] in N4141.
So no, this is not possible.
answered 6 hours ago
Baum mit Augen
40.1k12114147
answered 6 hours ago
Baum mit Augen
40.1k12114147
answered 6 hours ago
Baum mit Augen
40.1k12114147
answered 6 hours ago
Baum mit Augen
40.1k12114147
40.1k12114147
But note that you still can callstatic_cast<const Stuff&>(*this).foo()to call the const versions of functions.
– Artyer
6 hours ago
@Artyer But you'd have to know whether or not you want to, which appears to be the core of the question.
– Baum mit Augen
6 hours ago
And in other methods? Is it possible to determine the constancy of the created object from inside the class in principle?
– Dmytro Dadyka
6 hours ago
@DmytroDadyka In normal member functions, you can overload based on constness, as is demonstrated in OP.
– Baum mit Augen
6 hours ago
1
@DmytroDadyka No, constructors have basically the same clause I quoted here, see 12.1/3.
– Baum mit Augen
6 hours ago
|
show 1 more comment
But note that you still can callstatic_cast<const Stuff&>(*this).foo()to call the const versions of functions.
– Artyer
6 hours ago
@Artyer But you'd have to know whether or not you want to, which appears to be the core of the question.
– Baum mit Augen
6 hours ago
And in other methods? Is it possible to determine the constancy of the created object from inside the class in principle?
– Dmytro Dadyka
6 hours ago
@DmytroDadyka In normal member functions, you can overload based on constness, as is demonstrated in OP.
– Baum mit Augen
6 hours ago
1
@DmytroDadyka No, constructors have basically the same clause I quoted here, see 12.1/3.
– Baum mit Augen
6 hours ago
But note that you still can call
static_cast<const Stuff&>(*this).foo() to call the const versions of functions.– Artyer
6 hours ago
But note that you still can call
static_cast<const Stuff&>(*this).foo() to call the const versions of functions.– Artyer
6 hours ago
@Artyer But you'd have to know whether or not you want to, which appears to be the core of the question.
– Baum mit Augen
6 hours ago
@Artyer But you'd have to know whether or not you want to, which appears to be the core of the question.
– Baum mit Augen
6 hours ago
And in other methods? Is it possible to determine the constancy of the created object from inside the class in principle?
– Dmytro Dadyka
6 hours ago
And in other methods? Is it possible to determine the constancy of the created object from inside the class in principle?
– Dmytro Dadyka
6 hours ago
@DmytroDadyka In normal member functions, you can overload based on constness, as is demonstrated in OP.
– Baum mit Augen
6 hours ago
@DmytroDadyka In normal member functions, you can overload based on constness, as is demonstrated in OP.
– Baum mit Augen
6 hours ago
1
1
@DmytroDadyka No, constructors have basically the same clause I quoted here, see 12.1/3.
– Baum mit Augen
6 hours ago
@DmytroDadyka No, constructors have basically the same clause I quoted here, see 12.1/3.
– Baum mit Augen
6 hours ago
|
show 1 more comment
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53840945%2ffiguring-out-the-constness-of-an-object-within-its-destructor%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
The destructor destroys the object, by definition it's not const. Const only applies if you are calling
stuff.foo()orcstuff.foo().– Matthieu Brucher
6 hours ago
6
It's not possible to detect whether the object is
constin the constructor or destructor. This might be an XY problem.– Brian
6 hours ago
1
It is not const both in the constructor and the destructor. You want to know in the constructor if it will be, and in the destructor if it was. Strange
– bruno
6 hours ago
2
As per your edit: It sounds as though you are trying to implement transactional semantics of some kind. Would it perhaps make sense to keep track of whether state was modified using a flag (which only gets set on the mutable code-path), and implementing the conditional behavior in the destructor based on whether this flag was toggled?
– Bitwize
6 hours ago
5
It sounds to me like you might want two different types,
MutableStuffandImmutableStuff.– molbdnilo
6 hours ago