Idiomatic way to prevent slicing?
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Sometimes it can be an annoyance that c++ defaults to allow slicing. For example
#include <iostream>
struct foo { int a; };
struct bar : foo { int b; };
int main() {
bar x{1,2};
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 idomatic 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
#include <iostream>
struct foo { int a; };
struct bar : foo { int b; };
int main() {
bar x{1,2};
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 idomatic 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
#include <iostream>
struct foo { int a; };
struct bar : foo { int b; };
int main() {
bar x{1,2};
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 idomatic 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
#include <iostream>
struct foo { int a; };
struct bar : foo { int b; };
int main() {
bar x{1,2};
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 idomatic way to write foo
such that one cannot slice instances of any derived class?
c++ inheritance object-slicing
c++ inheritance object-slicing
edited 5 hours ago
rrauenza
3,55921835
3,55921835
asked 8 hours ago
user463035818user463035818
18.8k42970
18.8k42970
add a comment |
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
8 hours ago
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
8 hours ago
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
8 hours ago
3
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
8 hours ago
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 x{1,2};
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 x{1,2};
// 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)
: foo{a}, b{b}
{}
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;
};
4
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
8 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'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
8 hours ago
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
8 hours ago
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
8 hours ago
3
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
8 hours ago
add a 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
8 hours ago
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
8 hours ago
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
8 hours ago
3
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
8 hours ago
add a 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 8 hours ago
answered 8 hours ago
NathanOliverNathanOliver
98.5k16138218
98.5k16138218
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
8 hours ago
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
8 hours ago
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
8 hours ago
3
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
8 hours ago
add a 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
8 hours ago
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
8 hours ago
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
8 hours ago
3
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
8 hours ago
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
8 hours ago
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
8 hours ago
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
8 hours ago
if I understand correctly this is a nice way to prevent implicit conversions for function parameters in general
– user463035818
8 hours ago
1
1
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
8 hours ago
@user463035818 Yep. I've been using it since I've asked that Q.
– NathanOliver
8 hours ago
3
3
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
8 hours ago
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
8 hours ago
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 x{1,2};
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 x{1,2};
// 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)
: foo{a}, b{b}
{}
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 x{1,2};
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 x{1,2};
// 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)
: foo{a}, b{b}
{}
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 x{1,2};
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 x{1,2};
// 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)
: foo{a}, b{b}
{}
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 x{1,2};
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 x{1,2};
// 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)
: foo{a}, b{b}
{}
int b;
};
int main() {
auto x = bar (1,2);
// foo y = x; // <- does not compile
}
answered 8 hours ago
Richard HodgesRichard Hodges
57k658105
57k658105
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;
};
4
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
8 hours ago
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;
};
4
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
8 hours ago
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 8 hours ago
eerorikaeerorika
89.8k664136
89.8k664136
4
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
8 hours ago
add a comment |
4
but then I cannot copyfoo
s anymore :( I'd like to prevent only copying a bar to a foo if possible
– user463035818
8 hours ago
4
4
but then I cannot copy
foo
s anymore :( I'd like to prevent only copying a bar to a foo if possible– user463035818
8 hours ago
but then I cannot copy
foo
s anymore :( I'd like to prevent only copying a bar to a foo if possible– user463035818
8 hours ago
add a comment |
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