Is it possible to pass variable template to function through lambda?

Question

As I know how to pass template function as template argument, I'm now struggling to pass variable template in a similar manner.

Here is minimal example of what I tried:

#define PASS_VARIABLE_TEMPLATE(name) [dummy=nullptr](auto&&...args) \
                                                    {return name<decltype(args)...>;}

//testing
template <typename T>
bool value = std::is_fundamental<T>::value;

template <typename Hax>
void print_bool(Hax h)
{
    std::cout << h(int{}) << std::endl; // no error, wrong output
    //std::cout << h(int{}, float{}) << std::endl; // error, good
}

int main()
{
    print_bool(PASS_VARIABLE_TEMPLATE(value)); //prints 0 instead of 1
}

Demo

If it compiles, then why the output is wrong?

@lorro Oh silly me, forgot to do the decaying! But on other hand that would destroy reference/pointer checking... It should be possible to check this case too: int b; int& ref = a; h(ref); //type is int& — xinaiz
– xinaiz, Commented Oct 9, 2016 at 15:05
You will get the expected results with #define PASS_VARIABLE_TEMPLATE(name) [dummy=nullptr](auto&&args) { return name<typename std::remove_reference<decltype(args)>::type>; } — Sam Varshavchik
– Sam Varshavchik, Commented Oct 9, 2016 at 15:12
@SamVarshavchik Yes, but on the other hand if std::is_rvalue_reference was applied instead, wouldn't it break? — xinaiz
– xinaiz, Commented Oct 9, 2016 at 15:15

WhiZTiM · Accepted Answer · 2016-10-09 18:00:01Z

3

The main problem with your code is that decltype deduces the arguments as an rvalue reference (int&&) because your lambda uses forwarding references to accept the arguments. std::is_fundamental will work well with a bare type.

For your specific snippet, the correct solution is to remove the reference.

#define PASS_VARIABLE_TEMPLATE(name) \
    [dummy=nullptr](auto&&...args){return name<std::remove_reference_t<decltype(args)>...>;}

Now it works. :-) See it Live On Coliru

A slightly more or better generic way will be to additionally remove cv qualifiers. In the end, you may want to use std::decay

#define PASS_VARIABLE_TEMPLATE(name) [dummy=nullptr](auto&&...args) \
{return name<std::decay_t<decltype(args)>...>;}

edited Oct 9, 2016 at 18:00

answered Oct 9, 2016 at 15:58

WhiZTiM

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Yakk - Adam Nevraumont · Accepted Answer · 2016-10-09 15:22:17Z

1

template<class T>struct tag_t{using type=T; constexpr tag_t(){}};
template<class Tag>using tagged_type=typename Tag::type;
template<class T>constexpr tag_t<T> tag{};

These help pass types as values and unpack them.

#define PASS_VARIABLE_TEMPLATE(name) [](auto...args) \
                                                {return name<tagged_type<decltype(args)>...>;}

Inside print_bool you do:

std::cout << h(tag<int>) << std::endl;

Not sure why you do the dummy=nullptr thing.

tag as a template can carry types unmolested.

edited Oct 9, 2016 at 15:22

answered Oct 9, 2016 at 15:21

Yakk - Adam Nevraumont

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1 Comment

xinaiz Over a year ago

dummy is due to GCC bug with non-capturing lambda (check out the accepted answer from link I posted in the question). Btw, great solution.

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