CRTP applied on a template class
Let's consider a CRTP template class Print which is meant to print the derived class:
template <typename T>
struct Print {
auto print() const -> void;
auto self() const -> T const & {
return static_cast<T const &>(*this);
}
private:
Print() {}
~Print() {}
friend T;
};
Because I want to specialize print based on the derived class like we could do this with an override, I don't implement the method yet.
We can wrap an Integer and do so for example:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend Print<Integer>;
};
template <>
auto Print<Integer>::print() const -> void {
std::cout << self().m_i << std::endl;
}
This works so far, now let's say I want to Print a generic version of a wrapper:
template <typename T>
class Wrapper :
public Print<Wrapper<T>>
{
public:
Wrapper(T value) : m_value(std::move(value)) {}
private:
T m_value;
friend Print<Wrapper<T>>;
};
If I specialize my print method with a specialization of the Wrapper it compile and works:
template <>
auto Print<Wrapper<int>>::print() const -> void
{
cout << self().m_value << endl;
}
But if I want to say "for all specializations of Wrapper, do that", it doesn't work:
template <typename T>
auto Print<Wrapper<T>>::print() const -> void
{
cout << self().m_value << endl;
}
If I run this over the following main function:
auto main(int, char**) -> int {
auto i = Integer{5};
i.print();
auto wrapper = Wrapper<int>{5};
wrapper.print();
return 0;
}
The compiler print:
50:42: error: invalid use of incomplete type 'struct Print<Wrapper<T> >'
6:8: error: declaration of 'struct Print<Wrapper<T> >'
Why ? How can I do that ? Is it even possible or do I have to make a complete specialization of my CRTP class ?
c++ c++11 crtp
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
add a comment |
Let's consider a CRTP template class Print which is meant to print the derived class:
template <typename T>
struct Print {
auto print() const -> void;
auto self() const -> T const & {
return static_cast<T const &>(*this);
}
private:
Print() {}
~Print() {}
friend T;
};
Because I want to specialize print based on the derived class like we could do this with an override, I don't implement the method yet.
We can wrap an Integer and do so for example:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend Print<Integer>;
};
template <>
auto Print<Integer>::print() const -> void {
std::cout << self().m_i << std::endl;
}
This works so far, now let's say I want to Print a generic version of a wrapper:
template <typename T>
class Wrapper :
public Print<Wrapper<T>>
{
public:
Wrapper(T value) : m_value(std::move(value)) {}
private:
T m_value;
friend Print<Wrapper<T>>;
};
If I specialize my print method with a specialization of the Wrapper it compile and works:
template <>
auto Print<Wrapper<int>>::print() const -> void
{
cout << self().m_value << endl;
}
But if I want to say "for all specializations of Wrapper, do that", it doesn't work:
template <typename T>
auto Print<Wrapper<T>>::print() const -> void
{
cout << self().m_value << endl;
}
If I run this over the following main function:
auto main(int, char**) -> int {
auto i = Integer{5};
i.print();
auto wrapper = Wrapper<int>{5};
wrapper.print();
return 0;
}
The compiler print:
50:42: error: invalid use of incomplete type 'struct Print<Wrapper<T> >'
6:8: error: declaration of 'struct Print<Wrapper<T> >'
Why ? How can I do that ? Is it even possible or do I have to make a complete specialization of my CRTP class ?
c++ c++11 crtp
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
1
Can you post a complete example? My initial hunch is thatauto self() const -> Tis the culprit, since you actually return whole object. Even though the body casts to a reference return would copy the object. I am to lazy to compile a live example to test.
– luk32
Nov 15 '18 at 15:06
1
I don't use trailing return types much, but doesselfreturn a copy ofthis?
– François Andrieux
Nov 15 '18 at 15:07
Yes you're right I did a copy, I didn't pay attention, but on an int this doesn't solve anything anyway. I corrected it though and add a main function.
– Steranoid
Nov 15 '18 at 15:15
Yoda notation why like you?auto print() const -> void;
– SergeyA
Nov 15 '18 at 15:24
I like rust ... Well, that's pretty much the only reason :P
– Steranoid
Nov 15 '18 at 15:25
add a comment |
Let's consider a CRTP template class Print which is meant to print the derived class:
template <typename T>
struct Print {
auto print() const -> void;
auto self() const -> T const & {
return static_cast<T const &>(*this);
}
private:
Print() {}
~Print() {}
friend T;
};
Because I want to specialize print based on the derived class like we could do this with an override, I don't implement the method yet.
We can wrap an Integer and do so for example:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend Print<Integer>;
};
template <>
auto Print<Integer>::print() const -> void {
std::cout << self().m_i << std::endl;
}
This works so far, now let's say I want to Print a generic version of a wrapper:
template <typename T>
class Wrapper :
public Print<Wrapper<T>>
{
public:
Wrapper(T value) : m_value(std::move(value)) {}
private:
T m_value;
friend Print<Wrapper<T>>;
};
If I specialize my print method with a specialization of the Wrapper it compile and works:
template <>
auto Print<Wrapper<int>>::print() const -> void
{
cout << self().m_value << endl;
}
But if I want to say "for all specializations of Wrapper, do that", it doesn't work:
template <typename T>
auto Print<Wrapper<T>>::print() const -> void
{
cout << self().m_value << endl;
}
If I run this over the following main function:
auto main(int, char**) -> int {
auto i = Integer{5};
i.print();
auto wrapper = Wrapper<int>{5};
wrapper.print();
return 0;
}
The compiler print:
50:42: error: invalid use of incomplete type 'struct Print<Wrapper<T> >'
6:8: error: declaration of 'struct Print<Wrapper<T> >'
Why ? How can I do that ? Is it even possible or do I have to make a complete specialization of my CRTP class ?
c++ c++11 crtp
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
Let's consider a CRTP template class Print which is meant to print the derived class:
template <typename T>
struct Print {
auto print() const -> void;
auto self() const -> T const & {
return static_cast<T const &>(*this);
}
private:
Print() {}
~Print() {}
friend T;
};
Because I want to specialize print based on the derived class like we could do this with an override, I don't implement the method yet.
We can wrap an Integer and do so for example:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend Print<Integer>;
};
template <>
auto Print<Integer>::print() const -> void {
std::cout << self().m_i << std::endl;
}
This works so far, now let's say I want to Print a generic version of a wrapper:
template <typename T>
class Wrapper :
public Print<Wrapper<T>>
{
public:
Wrapper(T value) : m_value(std::move(value)) {}
private:
T m_value;
friend Print<Wrapper<T>>;
};
If I specialize my print method with a specialization of the Wrapper it compile and works:
template <>
auto Print<Wrapper<int>>::print() const -> void
{
cout << self().m_value << endl;
}
But if I want to say "for all specializations of Wrapper, do that", it doesn't work:
template <typename T>
auto Print<Wrapper<T>>::print() const -> void
{
cout << self().m_value << endl;
}
If I run this over the following main function:
auto main(int, char**) -> int {
auto i = Integer{5};
i.print();
auto wrapper = Wrapper<int>{5};
wrapper.print();
return 0;
}
The compiler print:
50:42: error: invalid use of incomplete type 'struct Print<Wrapper<T> >'
6:8: error: declaration of 'struct Print<Wrapper<T> >'
Why ? How can I do that ? Is it even possible or do I have to make a complete specialization of my CRTP class ?
c++ c++11 crtp
c++ c++11 crtp
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
edited Nov 15 '18 at 15:14
Steranoid
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
asked Nov 15 '18 at 14:59
SteranoidSteranoid
485
485
1
Can you post a complete example? My initial hunch is thatauto self() const -> Tis the culprit, since you actually return whole object. Even though the body casts to a reference return would copy the object. I am to lazy to compile a live example to test.
– luk32
Nov 15 '18 at 15:06
1
I don't use trailing return types much, but doesselfreturn a copy ofthis?
– François Andrieux
Nov 15 '18 at 15:07
Yes you're right I did a copy, I didn't pay attention, but on an int this doesn't solve anything anyway. I corrected it though and add a main function.
– Steranoid
Nov 15 '18 at 15:15
Yoda notation why like you?auto print() const -> void;
– SergeyA
Nov 15 '18 at 15:24
I like rust ... Well, that's pretty much the only reason :P
– Steranoid
Nov 15 '18 at 15:25
add a comment |
1
Can you post a complete example? My initial hunch is thatauto self() const -> Tis the culprit, since you actually return whole object. Even though the body casts to a reference return would copy the object. I am to lazy to compile a live example to test.
– luk32
Nov 15 '18 at 15:06
1
I don't use trailing return types much, but doesselfreturn a copy ofthis?
– François Andrieux
Nov 15 '18 at 15:07
Yes you're right I did a copy, I didn't pay attention, but on an int this doesn't solve anything anyway. I corrected it though and add a main function.
– Steranoid
Nov 15 '18 at 15:15
Yoda notation why like you?auto print() const -> void;
– SergeyA
Nov 15 '18 at 15:24
I like rust ... Well, that's pretty much the only reason :P
– Steranoid
Nov 15 '18 at 15:25
1
1
Can you post a complete example? My initial hunch is that
auto self() const -> T is the culprit, since you actually return whole object. Even though the body casts to a reference return would copy the object. I am to lazy to compile a live example to test.– luk32
Nov 15 '18 at 15:06
Can you post a complete example? My initial hunch is that
auto self() const -> T is the culprit, since you actually return whole object. Even though the body casts to a reference return would copy the object. I am to lazy to compile a live example to test.– luk32
Nov 15 '18 at 15:06
1
1
I don't use trailing return types much, but does
self return a copy of this?– François Andrieux
Nov 15 '18 at 15:07
I don't use trailing return types much, but does
self return a copy of this?– François Andrieux
Nov 15 '18 at 15:07
Yes you're right I did a copy, I didn't pay attention, but on an int this doesn't solve anything anyway. I corrected it though and add a main function.
– Steranoid
Nov 15 '18 at 15:15
Yes you're right I did a copy, I didn't pay attention, but on an int this doesn't solve anything anyway. I corrected it though and add a main function.
– Steranoid
Nov 15 '18 at 15:15
Yoda notation why like you?
auto print() const -> void;– SergeyA
Nov 15 '18 at 15:24
Yoda notation why like you?
auto print() const -> void;– SergeyA
Nov 15 '18 at 15:24
I like rust ... Well, that's pretty much the only reason :P
– Steranoid
Nov 15 '18 at 15:25
I like rust ... Well, that's pretty much the only reason :P
– Steranoid
Nov 15 '18 at 15:25
add a comment |
1 Answer
1
active
oldest
votes
You can do this in a bit of a roundabout way so long as you're careful.
Live Demo
Your Print class will rely on yet another class PrintImpl to do the printing.
#include <type_traits>
template<class...>
struct always_false : std::false_type{};
template<class T>
struct PrintImpl
{
void operator()(const T&) const
{
static_assert(always_false<T>::value, "PrintImpl hasn't been specialized for T");
}
};
You'll partially specialize this PrintImpl for your Wrapper class:
template<class T>
struct PrintImpl<Wrapper<T>>
{
void operator()(const Wrapper<T>& _val) const
{
std::cout << _val.m_value;
}
};
And make sure that Wrapper declares this PrintImpl to be a friend:
friend struct PrintImpl<Wrapper<T>>;
The Print class creates an instance of PrintImpl and calls operator():
void print() const
{
PrintImpl<T>{}(self());
}
This works so long as your specializations are declared before you actually instantiate an instance of the Print class.
You can also fully specialize PrintImpl<T>::operator() for your Integer class without writing a class specialization:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend PrintImpl<Integer>;
};
template <>
void PrintImpl<Integer>::operator()(const Integer& wrapper) const {
std::cout << wrapper.m_i << std::endl;
}
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
That's a nice way, isn't creating the PrintImpl each time the function is called a little bit heavy though ?
– Steranoid
Nov 15 '18 at 15:20
@Steranoid: It's a necessary evil because you cannot partially specialize functions. So we have to use a class.
– AndyG
Nov 15 '18 at 15:21
2
1. It is likely optimized away. 2. You can make the specialized function a static function instead--it wouldn't have to beoperator().
– Dark Falcon
Nov 15 '18 at 15:22
What's the purpose ofalways_falsehere?
– bipll
Nov 15 '18 at 16:03
1
@Steranoid: Fair enough, I see your point. I think that we can get the full function specialization you desire as well as the safety of thestatic_assertfrom my code (and no extra inheritance): wandbox.org/permlink/VFPy7rTdXm9I3fYh
– AndyG
Nov 16 '18 at 15:53
|
show 12 more comments
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1 Answer
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1 Answer
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oldest
votes
You can do this in a bit of a roundabout way so long as you're careful.
Live Demo
Your Print class will rely on yet another class PrintImpl to do the printing.
#include <type_traits>
template<class...>
struct always_false : std::false_type{};
template<class T>
struct PrintImpl
{
void operator()(const T&) const
{
static_assert(always_false<T>::value, "PrintImpl hasn't been specialized for T");
}
};
You'll partially specialize this PrintImpl for your Wrapper class:
template<class T>
struct PrintImpl<Wrapper<T>>
{
void operator()(const Wrapper<T>& _val) const
{
std::cout << _val.m_value;
}
};
And make sure that Wrapper declares this PrintImpl to be a friend:
friend struct PrintImpl<Wrapper<T>>;
The Print class creates an instance of PrintImpl and calls operator():
void print() const
{
PrintImpl<T>{}(self());
}
This works so long as your specializations are declared before you actually instantiate an instance of the Print class.
You can also fully specialize PrintImpl<T>::operator() for your Integer class without writing a class specialization:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend PrintImpl<Integer>;
};
template <>
void PrintImpl<Integer>::operator()(const Integer& wrapper) const {
std::cout << wrapper.m_i << std::endl;
}
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
That's a nice way, isn't creating the PrintImpl each time the function is called a little bit heavy though ?
– Steranoid
Nov 15 '18 at 15:20
@Steranoid: It's a necessary evil because you cannot partially specialize functions. So we have to use a class.
– AndyG
Nov 15 '18 at 15:21
2
1. It is likely optimized away. 2. You can make the specialized function a static function instead--it wouldn't have to beoperator().
– Dark Falcon
Nov 15 '18 at 15:22
What's the purpose ofalways_falsehere?
– bipll
Nov 15 '18 at 16:03
1
@Steranoid: Fair enough, I see your point. I think that we can get the full function specialization you desire as well as the safety of thestatic_assertfrom my code (and no extra inheritance): wandbox.org/permlink/VFPy7rTdXm9I3fYh
– AndyG
Nov 16 '18 at 15:53
|
show 12 more comments
You can do this in a bit of a roundabout way so long as you're careful.
Live Demo
Your Print class will rely on yet another class PrintImpl to do the printing.
#include <type_traits>
template<class...>
struct always_false : std::false_type{};
template<class T>
struct PrintImpl
{
void operator()(const T&) const
{
static_assert(always_false<T>::value, "PrintImpl hasn't been specialized for T");
}
};
You'll partially specialize this PrintImpl for your Wrapper class:
template<class T>
struct PrintImpl<Wrapper<T>>
{
void operator()(const Wrapper<T>& _val) const
{
std::cout << _val.m_value;
}
};
And make sure that Wrapper declares this PrintImpl to be a friend:
friend struct PrintImpl<Wrapper<T>>;
The Print class creates an instance of PrintImpl and calls operator():
void print() const
{
PrintImpl<T>{}(self());
}
This works so long as your specializations are declared before you actually instantiate an instance of the Print class.
You can also fully specialize PrintImpl<T>::operator() for your Integer class without writing a class specialization:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend PrintImpl<Integer>;
};
template <>
void PrintImpl<Integer>::operator()(const Integer& wrapper) const {
std::cout << wrapper.m_i << std::endl;
}
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
That's a nice way, isn't creating the PrintImpl each time the function is called a little bit heavy though ?
– Steranoid
Nov 15 '18 at 15:20
@Steranoid: It's a necessary evil because you cannot partially specialize functions. So we have to use a class.
– AndyG
Nov 15 '18 at 15:21
2
1. It is likely optimized away. 2. You can make the specialized function a static function instead--it wouldn't have to beoperator().
– Dark Falcon
Nov 15 '18 at 15:22
What's the purpose ofalways_falsehere?
– bipll
Nov 15 '18 at 16:03
1
@Steranoid: Fair enough, I see your point. I think that we can get the full function specialization you desire as well as the safety of thestatic_assertfrom my code (and no extra inheritance): wandbox.org/permlink/VFPy7rTdXm9I3fYh
– AndyG
Nov 16 '18 at 15:53
|
show 12 more comments
You can do this in a bit of a roundabout way so long as you're careful.
Live Demo
Your Print class will rely on yet another class PrintImpl to do the printing.
#include <type_traits>
template<class...>
struct always_false : std::false_type{};
template<class T>
struct PrintImpl
{
void operator()(const T&) const
{
static_assert(always_false<T>::value, "PrintImpl hasn't been specialized for T");
}
};
You'll partially specialize this PrintImpl for your Wrapper class:
template<class T>
struct PrintImpl<Wrapper<T>>
{
void operator()(const Wrapper<T>& _val) const
{
std::cout << _val.m_value;
}
};
And make sure that Wrapper declares this PrintImpl to be a friend:
friend struct PrintImpl<Wrapper<T>>;
The Print class creates an instance of PrintImpl and calls operator():
void print() const
{
PrintImpl<T>{}(self());
}
This works so long as your specializations are declared before you actually instantiate an instance of the Print class.
You can also fully specialize PrintImpl<T>::operator() for your Integer class without writing a class specialization:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend PrintImpl<Integer>;
};
template <>
void PrintImpl<Integer>::operator()(const Integer& wrapper) const {
std::cout << wrapper.m_i << std::endl;
}
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
You can do this in a bit of a roundabout way so long as you're careful.
Live Demo
Your Print class will rely on yet another class PrintImpl to do the printing.
#include <type_traits>
template<class...>
struct always_false : std::false_type{};
template<class T>
struct PrintImpl
{
void operator()(const T&) const
{
static_assert(always_false<T>::value, "PrintImpl hasn't been specialized for T");
}
};
You'll partially specialize this PrintImpl for your Wrapper class:
template<class T>
struct PrintImpl<Wrapper<T>>
{
void operator()(const Wrapper<T>& _val) const
{
std::cout << _val.m_value;
}
};
And make sure that Wrapper declares this PrintImpl to be a friend:
friend struct PrintImpl<Wrapper<T>>;
The Print class creates an instance of PrintImpl and calls operator():
void print() const
{
PrintImpl<T>{}(self());
}
This works so long as your specializations are declared before you actually instantiate an instance of the Print class.
You can also fully specialize PrintImpl<T>::operator() for your Integer class without writing a class specialization:
class Integer :
public Print<Integer>
{
public:
Integer(int i) : m_i(i) {}
private:
int m_i;
friend PrintImpl<Integer>;
};
template <>
void PrintImpl<Integer>::operator()(const Integer& wrapper) const {
std::cout << wrapper.m_i << std::endl;
}
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
edited Nov 20 '18 at 15:33
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
answered Nov 15 '18 at 15:15
AndyGAndyG
26.8k77097
26.8k77097
That's a nice way, isn't creating the PrintImpl each time the function is called a little bit heavy though ?
– Steranoid
Nov 15 '18 at 15:20
@Steranoid: It's a necessary evil because you cannot partially specialize functions. So we have to use a class.
– AndyG
Nov 15 '18 at 15:21
2
1. It is likely optimized away. 2. You can make the specialized function a static function instead--it wouldn't have to beoperator().
– Dark Falcon
Nov 15 '18 at 15:22
What's the purpose ofalways_falsehere?
– bipll
Nov 15 '18 at 16:03
1
@Steranoid: Fair enough, I see your point. I think that we can get the full function specialization you desire as well as the safety of thestatic_assertfrom my code (and no extra inheritance): wandbox.org/permlink/VFPy7rTdXm9I3fYh
– AndyG
Nov 16 '18 at 15:53
|
show 12 more comments
That's a nice way, isn't creating the PrintImpl each time the function is called a little bit heavy though ?
– Steranoid
Nov 15 '18 at 15:20
@Steranoid: It's a necessary evil because you cannot partially specialize functions. So we have to use a class.
– AndyG
Nov 15 '18 at 15:21
2
1. It is likely optimized away. 2. You can make the specialized function a static function instead--it wouldn't have to beoperator().
– Dark Falcon
Nov 15 '18 at 15:22
What's the purpose ofalways_falsehere?
– bipll
Nov 15 '18 at 16:03
1
@Steranoid: Fair enough, I see your point. I think that we can get the full function specialization you desire as well as the safety of thestatic_assertfrom my code (and no extra inheritance): wandbox.org/permlink/VFPy7rTdXm9I3fYh
– AndyG
Nov 16 '18 at 15:53
That's a nice way, isn't creating the PrintImpl each time the function is called a little bit heavy though ?
– Steranoid
Nov 15 '18 at 15:20
That's a nice way, isn't creating the PrintImpl each time the function is called a little bit heavy though ?
– Steranoid
Nov 15 '18 at 15:20
@Steranoid: It's a necessary evil because you cannot partially specialize functions. So we have to use a class.
– AndyG
Nov 15 '18 at 15:21
@Steranoid: It's a necessary evil because you cannot partially specialize functions. So we have to use a class.
– AndyG
Nov 15 '18 at 15:21
2
2
1. It is likely optimized away. 2. You can make the specialized function a static function instead--it wouldn't have to be
operator().– Dark Falcon
Nov 15 '18 at 15:22
1. It is likely optimized away. 2. You can make the specialized function a static function instead--it wouldn't have to be
operator().– Dark Falcon
Nov 15 '18 at 15:22
What's the purpose of
always_false here?– bipll
Nov 15 '18 at 16:03
What's the purpose of
always_false here?– bipll
Nov 15 '18 at 16:03
1
1
@Steranoid: Fair enough, I see your point. I think that we can get the full function specialization you desire as well as the safety of the
static_assert from my code (and no extra inheritance): wandbox.org/permlink/VFPy7rTdXm9I3fYh– AndyG
Nov 16 '18 at 15:53
@Steranoid: Fair enough, I see your point. I think that we can get the full function specialization you desire as well as the safety of the
static_assert from my code (and no extra inheritance): wandbox.org/permlink/VFPy7rTdXm9I3fYh– AndyG
Nov 16 '18 at 15:53
|
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Can you post a complete example? My initial hunch is that
auto self() const -> Tis the culprit, since you actually return whole object. Even though the body casts to a reference return would copy the object. I am to lazy to compile a live example to test.– luk32
Nov 15 '18 at 15:06
1
I don't use trailing return types much, but does
selfreturn a copy ofthis?– François Andrieux
Nov 15 '18 at 15:07
Yes you're right I did a copy, I didn't pay attention, but on an int this doesn't solve anything anyway. I corrected it though and add a main function.
– Steranoid
Nov 15 '18 at 15:15
Yoda notation why like you?
auto print() const -> void;– SergeyA
Nov 15 '18 at 15:24
I like rust ... Well, that's pretty much the only reason :P
– Steranoid
Nov 15 '18 at 15:25