std::is_member_function_pointer does not compile if false
What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:
template<class T> do_something() {
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value) {
this->_t->x(); // _t is type of T*
}
}
What happens: The compiler does not compile if T does not bring the function. Small example:
#include <type_traits>
#include <iostream>
class Foo {
public:
void x() { }
};
class Bar { };
int main() {
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;
}
Compiler says:
is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.
c++ typetraits if-constexpr
asked 18 hours ago
jagemuejagemue
15919
add a comment |
What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:
template<class T> do_something() {
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value) {
this->_t->x(); // _t is type of T*
}
}
What happens: The compiler does not compile if T does not bring the function. Small example:
#include <type_traits>
#include <iostream>
class Foo {
public:
void x() { }
};
class Bar { };
int main() {
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;
}
Compiler says:
is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.
c++ typetraits if-constexpr
asked 18 hours ago
jagemuejagemue
15919
you don't check if the type is a function pointer, you try to see a function pointer which did not exist
– Klaus
18 hours ago
Thanks for your comment! But how can I check that?
– jagemue
18 hours ago
stackoverflow.com/questions/257288/…
– Klaus
18 hours ago
Possible duplicate of Is it possible to write a template to check for a function's existence?
– Klaus
18 hours ago
add a comment |
What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:
template<class T> do_something() {
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value) {
this->_t->x(); // _t is type of T*
}
}
What happens: The compiler does not compile if T does not bring the function. Small example:
#include <type_traits>
#include <iostream>
class Foo {
public:
void x() { }
};
class Bar { };
int main() {
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;
}
Compiler says:
is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.
c++ typetraits if-constexpr
asked 18 hours ago
jagemuejagemue
15919
What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:
template<class T> do_something() {
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value) {
this->_t->x(); // _t is type of T*
}
}
What happens: The compiler does not compile if T does not bring the function. Small example:
#include <type_traits>
#include <iostream>
class Foo {
public:
void x() { }
};
class Bar { };
int main() {
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;
}
Compiler says:
is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.
c++ typetraits if-constexpr
c++ typetraits if-constexpr
asked 18 hours ago
jagemuejagemue
15919
asked 18 hours ago
jagemuejagemue
15919
asked 18 hours ago
jagemuejagemue
15919
asked 18 hours ago
jagemuejagemue
15919
asked 18 hours ago
jagemuejagemue
15919
15919
you don't check if the type is a function pointer, you try to see a function pointer which did not exist
– Klaus
18 hours ago
Thanks for your comment! But how can I check that?
– jagemue
18 hours ago
stackoverflow.com/questions/257288/…
– Klaus
18 hours ago
Possible duplicate of Is it possible to write a template to check for a function's existence?
– Klaus
18 hours ago
add a comment |
you don't check if the type is a function pointer, you try to see a function pointer which did not exist
– Klaus
18 hours ago
Thanks for your comment! But how can I check that?
– jagemue
18 hours ago
stackoverflow.com/questions/257288/…
– Klaus
18 hours ago
Possible duplicate of Is it possible to write a template to check for a function's existence?
– Klaus
18 hours ago
you don't check if the type is a function pointer, you try to see a function pointer which did not exist
– Klaus
18 hours ago
you don't check if the type is a function pointer, you try to see a function pointer which did not exist
– Klaus
18 hours ago
Thanks for your comment! But how can I check that?
– jagemue
18 hours ago
Thanks for your comment! But how can I check that?
– jagemue
18 hours ago
stackoverflow.com/questions/257288/…
– Klaus
18 hours ago
stackoverflow.com/questions/257288/…
– Klaus
18 hours ago
Possible duplicate of Is it possible to write a template to check for a function's existence?
– Klaus
18 hours ago
Possible duplicate of Is it possible to write a template to check for a function's existence?
– Klaus
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.
If you want to detect whether it exists or not, you can use the detection idiom. Example:
#include <experimental/type_traits>
template<class T>
using has_x = decltype(&T::x);
template<class T> void do_something(T t) {
if constexpr (std::experimental::is_detected<has_x, T>::value) {
t.x();
}
}
struct Foo {
void x() { }
};
struct Bar { };
int main() {
do_something(Foo{});
do_something(Bar{});
}
live example on godbolt.org
I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:
"checking expression validity in-place with C++17"
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.
If you want to detect whether it exists or not, you can use the detection idiom. Example:
#include <experimental/type_traits>
template<class T>
using has_x = decltype(&T::x);
template<class T> void do_something(T t) {
if constexpr (std::experimental::is_detected<has_x, T>::value) {
t.x();
}
}
struct Foo {
void x() { }
};
struct Bar { };
int main() {
do_something(Foo{});
do_something(Bar{});
}
live example on godbolt.org
I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:
"checking expression validity in-place with C++17"
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
add a comment |
is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.
If you want to detect whether it exists or not, you can use the detection idiom. Example:
#include <experimental/type_traits>
template<class T>
using has_x = decltype(&T::x);
template<class T> void do_something(T t) {
if constexpr (std::experimental::is_detected<has_x, T>::value) {
t.x();
}
}
struct Foo {
void x() { }
};
struct Bar { };
int main() {
do_something(Foo{});
do_something(Bar{});
}
live example on godbolt.org
I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:
"checking expression validity in-place with C++17"
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
add a comment |
is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.
If you want to detect whether it exists or not, you can use the detection idiom. Example:
#include <experimental/type_traits>
template<class T>
using has_x = decltype(&T::x);
template<class T> void do_something(T t) {
if constexpr (std::experimental::is_detected<has_x, T>::value) {
t.x();
}
}
struct Foo {
void x() { }
};
struct Bar { };
int main() {
do_something(Foo{});
do_something(Bar{});
}
live example on godbolt.org
I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:
"checking expression validity in-place with C++17"
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.
If you want to detect whether it exists or not, you can use the detection idiom. Example:
#include <experimental/type_traits>
template<class T>
using has_x = decltype(&T::x);
template<class T> void do_something(T t) {
if constexpr (std::experimental::is_detected<has_x, T>::value) {
t.x();
}
}
struct Foo {
void x() { }
};
struct Bar { };
int main() {
do_something(Foo{});
do_something(Bar{});
}
live example on godbolt.org
I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:
"checking expression validity in-place with C++17"
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
answered 18 hours ago
Vittorio RomeoVittorio Romeo
58.9k17161304
58.9k17161304
add a comment |
add a comment |
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you don't check if the type is a function pointer, you try to see a function pointer which did not exist
– Klaus
18 hours ago
Thanks for your comment! But how can I check that?
– jagemue
18 hours ago
stackoverflow.com/questions/257288/…
– Klaus
18 hours ago
Possible duplicate of Is it possible to write a template to check for a function's existence?
– Klaus
18 hours ago