C++ standard has a below statement:
The closure type for a non-generic lambda-expression with no lambda-capture whose constraints (if any) are satisfied has a conversion function to pointer to function with C++ language linkage (10.5) having the same parameter and return types as the closure type's function call operator.
To understand the statement much better, I used the cppinsights to see what the clang compiler says for the below function.
#include <iostream>
using test = void (*)(int);
int main()
{
test t = [](int arg) { std::cout << arg << std::endl; };
}
cppinsights translates the function as:
#include <iostream>
using test = void (*)(int);
int main()
{
class __lambda_7_11
{
public: inline void operator()(int arg) const
{
std::cout.operator<<(arg).operator<<(std::endl);
}
public: using retType_7_11 = void (*)(int);
inline operator retType_7_11 () const
{
return __invoke;
}
private: static inline void __invoke(int arg)
{
std::cout.operator<<(arg).operator<<(std::endl);
}
} __lambda_7_11{};
using FuncPtr_7 = test;
FuncPtr_7 t = static_cast<void (*)(int)>(__lambda_7_11.operator __lambda_7_11::retType_7_11());
}
As usual, the compiler generates an anonymous class with operator() overloaded along with "conversion function to pointer to function" as specified by the standard.
What I don't understand is that why a "static __invoke" function is generated and the "conversion function" is internally calling "__invoke" (directly as a function pointer) without any parameter which is expected by "__invoke"?
Cppinsights is correct in what it does. It returns the function pointer iso calling the function.
It would have been more readable if it would been writen the following way:
inline operator retType_7_11 () const
{
return &__invoke;
}
However, the & ain't required in this case.
What I don't understand is that why a
static __invoke
function is generated and the "conversion function" is internally calling__invoke
(directly as a function pointer) without any parameter which is expected by__invoke
?
The conversion function does not "internally call" __invoke
, it just returns __invoke
, ie, a pointer to a static
member function. Remember that the difference between static
and non- static
member functions is that the former is not bound to a specific instance of that class, and can hence be treated as ordinary function pointers, as opposed to pointers to members. Have a look at the following example:
struct Test {
void f();
static void g();
};
void (Test::*f)() = &Test::f; // pointer to instance-specific member fct.
void (*g)() = &Test::g; // ordinary function pointer
void (*h)() = Test::g; // no ampersand, this is implicitly a function pointer
The latter is what you intend the result of the conversion to be, and what cppinsights shows you is nothing but a technique to implement this conversion: as the lambda expression has an empty closure, the function object generated by the compiler has no state, and a member function of an object with no state can be a static
member function that can bind to an ordinary function pointer.
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