Etydhv

I am in middle of writing some generic code for future library. I came across following problem inside template function. Imagine following code:

template<class F>
auto foo(F &&f) 
 auto result = std::forward<F>(f)(/*some args*/);
 //do some generic stuff
 return result;

It will work fine unless I pass to it function that returns void like that

foo([]());

Now of course I could use some std::enable_if magic to check return type and make specialization for function returning void look like this:

template<class F, class = /*enable if stuff*/>
void foo(F &&f) 
 std::forward<F>(f)(/*some args*/);
 //do some generic stuff

But that would awfully duplicate code for actually logically equivalent function. Can this be done easily in generic way for both void-returning and non-void-returning functions in elegant way?

EDIT:
there is data dependency between function f() and generic stuff I want to do, so I do not take code like this into account:

template<class F>
auto foo(F &&f) 
 //do some generic stuff
 return std::forward<F>(f)(/*some args*/);

if you can place the "some generic stuff" in the destructor of a bar class (inside a security try/catch block, if you're not sure that doesn't throw exceptions, as pointed by Drax), you can simply write

template <typename F>
auto foo (F &&f)
 
 bar b;

 return std::forward<F>(f)(/*some args*/);
 

So the compiler compute f(/*some args*/), exec the destructor of b and return the computed value (or nothing).

Observe that return func();, where func() is a function returning void, is perfectly legal.

Some specialization, somewhere, is necessary. But the goal here is to avoid specializing the function itself. However, you can specialize a helper class.

Tested with gcc 9.1 with -std=c++17.

#include <type_traits>
#include <iostream>

template<typename T>
struct return_value 


 T val;

 template<typename F, typename ...Args>
 return_value(F &&f, Args && ...args)
 : valf(std::forward<Args>(args)...)
 
 

 T value() const
 
 return val;
 
;

template<>
struct return_value<void> 

 template<typename F, typename ...Args>
 return_value(F &&f, Args && ...args)
 
 f(std::forward<Args>(args)...);
 

 void value() const
 
 
;

template<class F>
auto foo(F &&f)

 return_value<decltype(std::declval<F &&>()(2, 4))> rf, 2, 4;

 // Something

 return r.value();


int main()

 foo( [](int a, int b) return; );

 std::cout << foo( [](int a, int b) return a+b; ) << std::endl;

The best way to do this, in my opinion, is to actually change the way you call your possibly-void-returning functions. Basically, we change the ones that return void to instead return some class type Void that is, for all intents and purposes, the same thing and no users really are going to care.

struct Void ;

All we need to do is to wrap the invocation. The following uses C++17 names (std::invoke and std::invoke_result_t) but they're all implementable in C++14 without too much fuss:

// normal case: R isn't void
template <typename F, typename... Args, 
 typename R = std::invoke_result_t<F, Args...>,
 std::enable_if_t<!std::is_void<R>::value, int> = 0>
R invoke_void(F&& f, Args&&... args) 
 return std::invoke(std::forward<F>(f), std::forward<Args>(args)...);


// special case: R is void
template <typename F, typename... Args, 
 typename R = std::invoke_result_t<F, Args...>,
 std::enable_if_t<std::is_void<R>::value, int> = 0>
Void invoke_void(F&& f, Args&&... args) 
 // just call it, since it doesn't return anything
 std::invoke(std::forward<F>(f), std::forward<Args>(args)...);

 // and return Void
 return Void;

The advantage of doing it this way is that you can just directly write the code you wanted to write to begin with, in the way you wanted to write it:

template<class F>
auto foo(F &&f) 
 auto result = invoke_void(std::forward<F>(f), /*some args*/);
 //do some generic stuff
 return result;

And you don't have to either shove all your logic in a destructor or duplicate all of your logic by doing specialization. At the cost of foo([]) returning Void instead of void, which isn't much of a cost.

And then if Regular Void is ever adopted, all you have to do is swap out invoke_void for std::invoke.

In case you need to use result (in non-void cases) in "some generic stuff", I propose a if constexpr based solution (so, unfortunately, not before C++17).

Not really elegant, to be honest.

First of all, detect the "true return type" of f (given the arguments)

using TR_t = std::invoke_result_t<F, As...>;

Next a constexpr variable to see if the returned type is void (just to simplify a little the following code)

constexpr bool isVoidTR std::is_same_v<TR_t, void> ;

Now we define a (potentially) "fake return type": int when the true return type is void, the TR_t otherwise

using FR_t = std::conditional_t<isVoidTR, int, TR_t>;

Then we define the result value as "fake return type" (so int in void case)

FR_t result ;

Then a (void) instruction to avoid a possible "set but not used" warning

(void)result;

Now, using if constexpr, the two case to exec f (this, IMHO, is the ugliest part because we have to write two times the same f invocation)

if constexpr ( isVoidTR )
 std::forward<F>(f)(std::forward<As>(args)...);
else
 result = std::forward<F>(f)(std::forward<As>(args)...);

After this, the "some generic stuff" that can use result (in non-void cases) and also `isVoidTR).

To conclude, another if constexpr

if constexpr ( isVoidTR )
 return;
else
 return result;

The following is a full compiling C++17 example

#include <memory>
#include <type_traits>

template <typename F, typename ... As>
auto foo (F && f, As && ... args)
 
 // true return type
 using TR_t = std::invoke_result_t<F, As...>;

 constexpr bool isVoidTR std::is_same_v<TR_t, void> ;

 // (possibly) fake return type
 using FR_t = std::conditional_t<isVoidTR, int, TR_t>;

 FR_t result ; // is int in case of void

 (void)result; // to avoid "set but not used" warning, in void case

 if constexpr ( isVoidTR )
 std::forward<F>(f)(std::forward<As>(args)...);
 else
 result = std::forward<F>(f)(std::forward<As>(args)...);

 // some generic stuff (potentially depending from result,
 // in non-void cases)

 if constexpr ( isVoidTR )
 return;
 else
 return result;
 

int main ()
 
 foo([]());

 //auto a foo([]()) ; // compilation error: foo() is void

 auto b foo([](auto a0, auto...) return a0; , 1, 2l, 3ll) ;

 static_assert( std::is_same_v<decltype(b), int> );