lambda not working with template function

[JUAN DENT]

Hi,

I have the following template function:

template<typename F, typename ...T>
void call( F&& f, T&&... t) {

f(std::forward<T>(t)...);
}

It works fine with:

int i = 99;
const char p[] {"Trying"};

call(g2<int,const char [7]>, i, p);


But not with:


call( [](int i){ cout << "lambda2(int): " << endl; });

I get:

no matching function for call to object of type '(lambda at ....)'

What is going on??

Thanks for the help

Juan

[cire]

Your parameterized call function requires (at least) two arguments and isn't getting them.

[JUAN DENT]

I think that's no it because the first argument is a function (in this case a lambda) and the second would be the integer parameter to the lambda...


Makes sense?

Thanks

Juan

[JUAN DENT]

To be specific, the following call DOES NOT produce errors (and it only takes 1 parameter, the lambda):

call( [](){ cout << "lambda1()\n"; });

[JLBorges]

#include <iostream>

template< typename F, typename ...T > void call( F&& f, T&&... t) {

    std::forward<F>(f)( std::forward<T>(t)... );
}

int main() {
    
    call( [] { std::cout << "nullary closure()\n" ; } ) ; // fine: empty argument pack
    // call( [](){ std::cout << "nullary closure()\n" ; }, 100 ) ; // *** error: too many arguments

    call( [] (int) { std::cout << "unary closure(int)\n" ; }, 100 ) ; // fine: argument pack of size one
    // call( [] (int) { std::cout << "unary closure(int)\n" ; } ) ; // *** error: too few arguments
    // call( [] (int) { std::cout << "unary closure(int)\n" ; }, 100, 200 ) ; // *** error: too many arguments

    call( [] (int,int) { std::cout << "binary closure(int,int)\n" ; }, 100, 200 ) ; // fine: argument pack of size two
    // call( [] (int,int) { std::cout << "binary closure(int,int)\n" ; }, 100 ) ; // *** error: too few arguments
    // call( [] (int,int) { std::cout << "binary closure(int,int)\n" ; }, 100, 200, 300 ) ; // *** error: too many arguments
}

http://coliru.stacked-crooked.com/a/c8055c99158203e2