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Idiomatic way to prevent slicing?
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Sometimes it can be an annoyance that c++ defaults to allow slicing. For example
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!
This compiles and runs as expected! Though, what if I dont want to enable slicing?
What is the idiomatic way to write foo
such that one cannot slice instances of any derived class?
c++ inheritance object-slicing
add a comment |
Sometimes it can be an annoyance that c++ defaults to allow slicing. For example
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!
This compiles and runs as expected! Though, what if I dont want to enable slicing?
What is the idiomatic way to write foo
such that one cannot slice instances of any derived class?
c++ inheritance object-slicing
The easiest way is to not use inheritance. You have havefoo
be a member variable ofbar
.
– KevinZ
2 days ago
@KevinZ stackoverflow.com/a/55600208/4117728
– user463035818
2 days ago
add a comment |
Sometimes it can be an annoyance that c++ defaults to allow slicing. For example
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!
This compiles and runs as expected! Though, what if I dont want to enable slicing?
What is the idiomatic way to write foo
such that one cannot slice instances of any derived class?
c++ inheritance object-slicing
Sometimes it can be an annoyance that c++ defaults to allow slicing. For example
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
foo y = x; // <- I dont want this to compile!
This compiles and runs as expected! Though, what if I dont want to enable slicing?
What is the idiomatic way to write foo
such that one cannot slice instances of any derived class?
c++ inheritance object-slicing
c++ inheritance object-slicing
edited 2 days ago
user463035818
asked Apr 9 at 19:32
user463035818user463035818
19.1k42971
19.1k42971
The easiest way is to not use inheritance. You have havefoo
be a member variable ofbar
.
– KevinZ
2 days ago
@KevinZ stackoverflow.com/a/55600208/4117728
– user463035818
2 days ago
add a comment |
The easiest way is to not use inheritance. You have havefoo
be a member variable ofbar
.
– KevinZ
2 days ago
@KevinZ stackoverflow.com/a/55600208/4117728
– user463035818
2 days ago
The easiest way is to not use inheritance. You have have
foo
be a member variable of bar
.– KevinZ
2 days ago
The easiest way is to not use inheritance. You have have
foo
be a member variable of bar
.– KevinZ
2 days ago
@KevinZ stackoverflow.com/a/55600208/4117728
– user463035818
2 days ago
@KevinZ stackoverflow.com/a/55600208/4117728
– user463035818
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo
to
struct foo
int a;
foo() = default; // you have to add this because of the template constructor
template<typename T>
foo(const T&) = delete; // error trying to copy anything but a foo
template<typename T>
foo& operator=(const T&) = delete; // error assigning anything else but a foo
;
then you can only ever copy construct or copy assign a foo
to foo
. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.
Note that this doesn't prevent explicit slicing like this:foo y = static_cast<foo&>(x);
. That said, perhaps it's not a problem to OP.
– eerorika
Apr 9 at 20:02
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
Apr 9 at 20:05
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
Apr 9 at 20:06
4
I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.
– NathanOliver
Apr 9 at 20:08
actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)
– user463035818
Apr 10 at 8:23
|
show 1 more comment
Since 2011, the idiomatic way has been to use auto
:
#include <iostream>
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
auto y = x; // <- y is a bar
If you wish to actively prevent slicing, there are a number of ways:
Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:
#include <iostream>
struct foo int a; ;
struct bar
bar(int a, int b)
: foo_(a)
, b(b)
int b;
int get_a() const return foo_.a;
private:
foo foo_;
;
int main()
bar x1,2;
// foo y = x; // <- does not compile
Another more specialised way might be to alter the permissions around copy operators:
#include <iostream>
struct foo
int a;
protected:
foo(foo const&) = default;
foo(foo&&) = default;
foo& operator=(foo const&) = default;
foo& operator=(foo&&) = default;
;
struct bar : foo
bar(int a, int b)
: fooa, bb
int b;
;
int main()
auto x = bar (1,2);
// foo y = x; // <- does not compile
add a comment |
You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:
struct foo
// ...
protected:
foo(foo&) = default;
;
6
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
Apr 9 at 19:42
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'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo
to
struct foo
int a;
foo() = default; // you have to add this because of the template constructor
template<typename T>
foo(const T&) = delete; // error trying to copy anything but a foo
template<typename T>
foo& operator=(const T&) = delete; // error assigning anything else but a foo
;
then you can only ever copy construct or copy assign a foo
to foo
. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.
Note that this doesn't prevent explicit slicing like this:foo y = static_cast<foo&>(x);
. That said, perhaps it's not a problem to OP.
– eerorika
Apr 9 at 20:02
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
Apr 9 at 20:05
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
Apr 9 at 20:06
4
I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.
– NathanOliver
Apr 9 at 20:08
actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)
– user463035818
Apr 10 at 8:23
|
show 1 more comment
I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo
to
struct foo
int a;
foo() = default; // you have to add this because of the template constructor
template<typename T>
foo(const T&) = delete; // error trying to copy anything but a foo
template<typename T>
foo& operator=(const T&) = delete; // error assigning anything else but a foo
;
then you can only ever copy construct or copy assign a foo
to foo
. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.
Note that this doesn't prevent explicit slicing like this:foo y = static_cast<foo&>(x);
. That said, perhaps it's not a problem to OP.
– eerorika
Apr 9 at 20:02
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
Apr 9 at 20:05
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
Apr 9 at 20:06
4
I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.
– NathanOliver
Apr 9 at 20:08
actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)
– user463035818
Apr 10 at 8:23
|
show 1 more comment
I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo
to
struct foo
int a;
foo() = default; // you have to add this because of the template constructor
template<typename T>
foo(const T&) = delete; // error trying to copy anything but a foo
template<typename T>
foo& operator=(const T&) = delete; // error assigning anything else but a foo
;
then you can only ever copy construct or copy assign a foo
to foo
. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.
I'm not sure if there is a named idiom for it but you can add a deleted function to the overload set that is a better match then the base classes slicing operations. If you change foo
to
struct foo
int a;
foo() = default; // you have to add this because of the template constructor
template<typename T>
foo(const T&) = delete; // error trying to copy anything but a foo
template<typename T>
foo& operator=(const T&) = delete; // error assigning anything else but a foo
;
then you can only ever copy construct or copy assign a foo
to foo
. Any other type will pick the function template and you'll get an error about using a deleted function. This does mean that your class, and the classes that use it can no longer be an aggregate though. Since the members that are added are templates, they are not considered copy constructors or copy assignment operators so you'll get the default copy and move constructors and assignment operators.
edited Apr 9 at 19:56
answered Apr 9 at 19:43
NathanOliverNathanOliver
98.8k16138218
98.8k16138218
Note that this doesn't prevent explicit slicing like this:foo y = static_cast<foo&>(x);
. That said, perhaps it's not a problem to OP.
– eerorika
Apr 9 at 20:02
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
Apr 9 at 20:05
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
Apr 9 at 20:06
4
I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.
– NathanOliver
Apr 9 at 20:08
actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)
– user463035818
Apr 10 at 8:23
|
show 1 more comment
Note that this doesn't prevent explicit slicing like this:foo y = static_cast<foo&>(x);
. That said, perhaps it's not a problem to OP.
– eerorika
Apr 9 at 20:02
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
Apr 9 at 20:05
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
Apr 9 at 20:06
4
I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.
– NathanOliver
Apr 9 at 20:08
actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)
– user463035818
Apr 10 at 8:23
Note that this doesn't prevent explicit slicing like this:
foo y = static_cast<foo&>(x);
. That said, perhaps it's not a problem to OP.– eerorika
Apr 9 at 20:02
Note that this doesn't prevent explicit slicing like this:
foo y = static_cast<foo&>(x);
. That said, perhaps it's not a problem to OP.– eerorika
Apr 9 at 20:02
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
Apr 9 at 20:05
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
Apr 9 at 20:05
1
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
Apr 9 at 20:06
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
Apr 9 at 20:06
4
4
I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.
– NathanOliver
Apr 9 at 20:08
I look at it as reverse SFINAE. You make the overloads you want to compile, and then add a deleted template stopping everything else.
– NathanOliver
Apr 9 at 20:08
actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)
– user463035818
Apr 10 at 8:23
actually I was a bit hestitant to accept this answer. The technique is great, but in fact it opens the door to specializing all kinds of unwanted assignments, though if I have to choose between the javaish "protect against every possible stupidity at any cost" vs a pythonic "we are all adults" then I know what to pick ;)
– user463035818
Apr 10 at 8:23
|
show 1 more comment
Since 2011, the idiomatic way has been to use auto
:
#include <iostream>
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
auto y = x; // <- y is a bar
If you wish to actively prevent slicing, there are a number of ways:
Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:
#include <iostream>
struct foo int a; ;
struct bar
bar(int a, int b)
: foo_(a)
, b(b)
int b;
int get_a() const return foo_.a;
private:
foo foo_;
;
int main()
bar x1,2;
// foo y = x; // <- does not compile
Another more specialised way might be to alter the permissions around copy operators:
#include <iostream>
struct foo
int a;
protected:
foo(foo const&) = default;
foo(foo&&) = default;
foo& operator=(foo const&) = default;
foo& operator=(foo&&) = default;
;
struct bar : foo
bar(int a, int b)
: fooa, bb
int b;
;
int main()
auto x = bar (1,2);
// foo y = x; // <- does not compile
add a comment |
Since 2011, the idiomatic way has been to use auto
:
#include <iostream>
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
auto y = x; // <- y is a bar
If you wish to actively prevent slicing, there are a number of ways:
Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:
#include <iostream>
struct foo int a; ;
struct bar
bar(int a, int b)
: foo_(a)
, b(b)
int b;
int get_a() const return foo_.a;
private:
foo foo_;
;
int main()
bar x1,2;
// foo y = x; // <- does not compile
Another more specialised way might be to alter the permissions around copy operators:
#include <iostream>
struct foo
int a;
protected:
foo(foo const&) = default;
foo(foo&&) = default;
foo& operator=(foo const&) = default;
foo& operator=(foo&&) = default;
;
struct bar : foo
bar(int a, int b)
: fooa, bb
int b;
;
int main()
auto x = bar (1,2);
// foo y = x; // <- does not compile
add a comment |
Since 2011, the idiomatic way has been to use auto
:
#include <iostream>
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
auto y = x; // <- y is a bar
If you wish to actively prevent slicing, there are a number of ways:
Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:
#include <iostream>
struct foo int a; ;
struct bar
bar(int a, int b)
: foo_(a)
, b(b)
int b;
int get_a() const return foo_.a;
private:
foo foo_;
;
int main()
bar x1,2;
// foo y = x; // <- does not compile
Another more specialised way might be to alter the permissions around copy operators:
#include <iostream>
struct foo
int a;
protected:
foo(foo const&) = default;
foo(foo&&) = default;
foo& operator=(foo const&) = default;
foo& operator=(foo&&) = default;
;
struct bar : foo
bar(int a, int b)
: fooa, bb
int b;
;
int main()
auto x = bar (1,2);
// foo y = x; // <- does not compile
Since 2011, the idiomatic way has been to use auto
:
#include <iostream>
struct foo int a; ;
struct bar : foo int b; ;
int main()
bar x1,2;
auto y = x; // <- y is a bar
If you wish to actively prevent slicing, there are a number of ways:
Usually the most preferable way, unless you specifically need inheritance (you often don't) is to use encapsulation:
#include <iostream>
struct foo int a; ;
struct bar
bar(int a, int b)
: foo_(a)
, b(b)
int b;
int get_a() const return foo_.a;
private:
foo foo_;
;
int main()
bar x1,2;
// foo y = x; // <- does not compile
Another more specialised way might be to alter the permissions around copy operators:
#include <iostream>
struct foo
int a;
protected:
foo(foo const&) = default;
foo(foo&&) = default;
foo& operator=(foo const&) = default;
foo& operator=(foo&&) = default;
;
struct bar : foo
bar(int a, int b)
: fooa, bb
int b;
;
int main()
auto x = bar (1,2);
// foo y = x; // <- does not compile
answered Apr 9 at 19:43
Richard HodgesRichard Hodges
57.1k658105
57.1k658105
add a comment |
add a comment |
You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:
struct foo
// ...
protected:
foo(foo&) = default;
;
6
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
Apr 9 at 19:42
add a comment |
You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:
struct foo
// ...
protected:
foo(foo&) = default;
;
6
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
Apr 9 at 19:42
add a comment |
You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:
struct foo
// ...
protected:
foo(foo&) = default;
;
You can prevent the base from being copied outside of member functions of derived classes and the base itself by declaring the copy constructor protected:
struct foo
// ...
protected:
foo(foo&) = default;
;
answered Apr 9 at 19:41
eerorikaeerorika
90.1k665137
90.1k665137
6
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
Apr 9 at 19:42
add a comment |
6
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
Apr 9 at 19:42
6
6
but then I cannot copy
foo
s anymore :( I'd like to prevent only copying a bar to a foo if possible– user463035818
Apr 9 at 19:42
but then I cannot copy
foo
s anymore :( I'd like to prevent only copying a bar to a foo if possible– user463035818
Apr 9 at 19:42
add a comment |
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The easiest way is to not use inheritance. You have have
foo
be a member variable ofbar
.– KevinZ
2 days ago
@KevinZ stackoverflow.com/a/55600208/4117728
– user463035818
2 days ago