You don't use std::forward()
correctly: you need to give the deduced type as an argument to std::forward()
:
foo (std::forward<Args>(args)...);
Question
Consider the code
template <typename... Args>
void foo (Args&& ...)
{
}
template <typename... Args>
void bar (Args&& ... args)
{
foo (std::forward (args)...);
}
int main ()
{
bar (true);
}
~
gcc 4.7.2 gives the error
error: no matching function for call to ‘forward(bool&)’
note: candidates are:
template<class _Tp> constexpr _Tp&& std::forward(typename std::remove_reference<_Tp>::type&)
note: template argument deduction/substitution failed:
note: template<class _Tp> constexpr _Tp&& std::forward(typename std::remove_reference<_Tp>::type&&)
note: template argument deduction/substitution failed:
Why isn't a literal deduced as an rvalue?
Solution
You don't use std::forward()
correctly: you need to give the deduced type as an argument to std::forward()
:
foo (std::forward<Args>(args)...);