How to immediately invoke a C++ lambda? - c++

Question

A constructor from a class I'm inheriting requires a non-trivial object to be passed in. Similar to this:
MyFoo::MyFoo() : SomeBase( complexstuff )
{
return;
}
The complexstuff has little to do with MyFoo, so I didn't want to have to pass it in.
Instead of writing some kind of 1-off temporary function that returns complexstuff I used a lambda. What took me a few minutes to figure out is I have to invoke the lambda. So my code now looks like this:
MyFoo::MyFoo() : SomeBase(
[]()
{
/* blah blah do stuff with complexstuff */
return complexstuff;
} () )
{
return;
}
If you didn't catch it, it is subtle. But after the lambda body, I had to put () to tell the compiler to immediately "run" the lambda. Which made sense after I figured out what I had done wrong. Otherwise, without the () to invoke the lambda, gcc says something similar to this:
error: no matching function for call to 'SomeBase(<lambda()>)'
But now that has me thinking -- did I do this correctly? Is there a better way in C++11 or C++14 to tell the compiler that I want it to immediately invoke a lambda I've written? Or is appending an empty () like I did the usual way to do this?

Answer 1

But now that has me thinking -- did I do this correctly?
Yes you did.
Is there a better way in C++11 or C++14 to tell the compiler that I want it to immediately invoke a lambda I've written?
Not that I know of. A lambda is also just a function object, so you need to have a () to call it, there is no way around it (except of course some function that invokes the lambda like std::invoke).
If you want you can drop the () after the capture list, because your lambda doesn't take any parameters.
Or is appending an empty () like I did the usual way to do this?
Yes, it is the shortest way. As said before, std::invoke would also work instead, but it requires more typing. I would say a direct call with () is the usual way it is done.

Answer 2

In C++17 you can use std::invoke. This does the exact same thing as you did, but perhaps you will find this clearer.
#include <iostream>
#include <functional>
void foo(int i)
{
std::cout << i << '\n';
}
int main()
{
foo( std::invoke( []() { return 1; } ) );
}

Answer 3

There's no way to tell the compiler to invoke the lambda immediately. The simplest course (both in terms of complexity and number of typed characters) is what you did already. It's also very idiomatic for anyone who has worked with languages that have closures (I'm thinking JavaScript here).
If you want to avoid the syntax, then either modify SomeBase or complexstuff to execute the callable.
If all you want is syntactic sugar for invoking the lambda, you can always do what something like Alexandrescu's SCOPE_GUARD does, and abuse operator overloading:
Live example
#include <iostream>
namespace detail {
enum class invoke_t{};
template<class Callable>
auto operator+(invoke_t, Callable c) -> decltype(c()) {
return c();
}
}
constexpr detail::invoke_t invoke{};
int main() {
invoke + []() {
std::cout << "called";
};
}
But I wouldn't. Inventing your own DSL will just make your code worse to maintain. Stick to the idioms that utilize plain language constructs.

Answer 4

Is there a better way
You could also consider having a private static member function building the complexstuff, something like
class MyFoo : public Base {
private:
static SomeComplexType compute_complex_stuff() {
SomeComplexType complexstuff;
/*compute the complexstuff */
return complexstuff;
};
public:
MyFoo() : Base(compute_complex_stuff()) {};
};
I don't know if it is better than defining a lambda expression and applying it immediately; that is IMHO a matter of taste; for a short lambda body I would prefer a lambda expression immediately applied (but perhaps some compiler would create the temporary closure in that case, so it might be slower without optimizations; I expect most C++11 compilers to be able to make that optimization).
BTW, GCC provides the statement expression language extension (also understood by Clang) for your purposes. With it you could write
MyFoo::MyFoo : Base (({
SomeComplexType complexstuff;
/*compute the complexstuff */
return complexstuff;
}) {};