Can anyone tell me a valid way to get a pointer to a templated lamda (without caputes) operator () ? Already tried two alternatives:
int main()
{
static
auto l = []<bool a, bool b>( unsigned c ) -> unsigned
{
return (unsigned)a b c;
};
using fn_t = unsigned (*)( unsigned );
fn_t fnA = &l.operator ()<true, false>; // doesn't work
fn_t fnB = &decltype(l)::operator ()<true, false>; // also doesn't work
}
clang(-cl) 12:
x.cpp(9,13): error: cannot create a non-constant pointer to member function
fn_t fnA = &l.operator ()<true, false> ; // doesn't work
^~~~~~~~~~~~~~~~~~~~~~~~~~~
x.cpp(10,14): error: address of overloaded function 'operator()' does not match required type
'unsigned int (unsigned int)'
fn_t fnB = &decltype(l)::operator ()<true, false> ; // also doesn't work
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
x.cpp(4,11): note: candidate function template has different qualifiers (expected unqualified but found 'const')
auto l = []<bool a, bool b>(unsigned c) -> unsigned
^
MSVC 2019 latest update:
x.cpp(9): error C2276: '&': illegal operation on bound member function expression
x.cpp(10): error C2440: 'initializing': cannot convert from 'overloaded-function' to 'fn_t'
x.cpp(10): note: None of the functions with this name in scope match the target type
CodePudding user response:
The type of the address of lambda's operator()
is a member function pointer, however, the definition of your fn_t
is just a free function pointer. You should define your fn_t
as:
using fn_t = unsigned int (decltype(l)::*)(unsigned int) const;
Then the following should work:
fn_t fnA = &decltype(l)::operator ()<true, false>;
Or, why not?
auto fnB = &decltype(l)::operator ()<true, false>;
CodePudding user response:
It is the lambda which can be converted to function pointer in some conditions (which are not there here), not its member operator()
. And you cannot convert member pointer to function pointer.
One workaround is to use another lambda.
fn_t fn = [](unsigned c) { return decltype(l){}.operator()<true, false>(c); };