Obtaining function pointer to lambda?
Solution 1
This fails:
auto *b = [](int i) { return i; };
because the lambda is not a pointer. auto
does not allow for conversions. Even though the lambda is convertible to something that is a pointer, that's not going to be done for you - you have to do it yourself. Whether with a cast:
auto *c = static_cast<int(*)(int)>([](int i){return i;});
Or with some sorcery:
auto *d = +[](int i) { return i; };
Solution 2
Especially when it can implicitly convert a unique, lambda type to a function pointer:
But it cannot convert it to "a function pointer". It can only convert it to a pointer to a specific function signature. This will fail:
int (*h)(float) = a;
Why does that fail? Because there is no valid implicit conversion from a
to h
here.
The conversion for lambdas is not compiler magic. The standard simply says that the lambda closure type, for non-capturing, non-generic lambdas, has an implicit conversion operator for function pointers matching the signature of its operator()
overload. The rules for initializing int (*g)(int)
from a
permit using implicit conversions, and thus the compiler will invoke that operator.
auto
doesn't permit using implicit conversion operators; it takes the type as-is (removing references, of course). auto*
doesn't do implicit conversions either. So why would it invoke an implicit conversion for a lambda closure and not for a user-defined type?
Solution 3
The lambda code doesn't work for the same reason this doesn't work:
struct foo {
operator int*() const {
static int x;
return &x;
}
};
int* pint = foo{};
auto* pint2 = foo{}; // does not compile
or even:
template<class T>
void test(T*) {};
test(foo{});
The lambda has an operator that implicitly converts it to a (particular) function pointer, just like foo
.
auto
does not do conversion. Ever. Auto behaves like a class T
parameter to a template function where its type is deduced.
As the type on the right hand side is not a pointer, it cannot be used to initialize an auto*
variable.
Lambdas are not function pointers. Lambdas are not std::function
s. They are auto-written function objects (objects with an operator()
).
Examine this:
void (*ptr)(int) = [](auto x){std::cout << x;};
ptr(7);
it compiles and works in gcc (not certain if it is an extension, now that I think about it). However, what would auto* ptr = [](auto x){std::cout << x;}
supposed to do?
However, unary +
is an operator that works on pointers (and does nearly nothing to them), but not in foo
or a lambda.
So
auto* pauto=+foo{};
And
auto* pfun=+[](int x){};
Both work, magically.
oconnor0
Updated on June 04, 2022Comments
-
oconnor0 almost 2 years
I want to be able to obtain a function pointer to a lambda in C++.
I can do:
int (*c)(int) = [](int i) { return i; };
And, of course, the following works - even if it's not creating a function pointer.
auto a = [](int i) { return i; };
But the following:
auto *b = [](int i) { return i; };
Gives this error in GCC:
main.cpp: In function 'int main()': main.cpp:13:37: error: unable to deduce 'auto*' from '<lambda closure object>main()::<lambda(int)>{}' auto *b = [](int i) { return i; }; ^ main.cpp:13:37: note: mismatched types 'auto*' and 'main()::<lambda(int)>'
It seems arbitrary that a lambda can be converted to a function pointer without issue, but the compiler cannot infer the function type and create a pointer to it using
auto *
. Especially when it can implicitly convert aunique, lambda type
to a function pointer:int (*g)(int) = a;
I've create a little test bed at http://coliru.stacked-crooked.com/a/2cbd62c8179dc61b that contains the above examples. This behavior is the same under C++11 and C++14.
-
Jaa-c almost 8 yearsWhat does this do? Never seen this kind of
+
syntax before. -
Barry almost 8 years@Jaa-c Added a link.
-
Barry almost 8 years"not certain if it is an extension" I'd expect that to work. You can take a specific function pointer to a function template (e.g.
template <class T> void foo(T) { } void(*p)(int) = foo;
-
Jaa-c almost 8 yearsThanks - now I'll have to use it at least once. Reviewer's gonna hate me.
-
Yakk - Adam Nevraumont almost 8 years@Barry Sure I expect that to work. But
auto
lambdas are relatively new, and clang rejects it. Me, I'd even want implicit cast to any function that is convertion-compatible, so evenvoid(*f)(double) = [](int x){std::cout << x;};
would compile (like it does withstd::function<void(double) f = [](int x){std::cout << x;};
) -
Yakk - Adam Nevraumont almost 8 years@Jaa-c just don't
auto *d = ++[](int i){ return i; };
-- a unary plus too many! -
KyleKnoepfel almost 8 yearsTo steal someone else's line from I-don't-know where: this is awesome, yet totally messed up.
-
Yakk - Adam Nevraumont almost 8 yearsplus 1 for +, I forgot about that.
-
Chris Beck almost 8 years@Yakk: I am pleasantly surprised that both
++[] ...
and+++[]
fail to compile, I guess the+
thing is less evil than I thought at first glance -
tomsmeding almost 8 years@Chris Maybe because those make a prefix increment operator instead of a unary plus operator? ;)
-
T.C. almost 8 yearsNote that
+
fails on at least some MSVC. Because their lambdas have four conversion operators to function pointers :/