Underlying type of std::optional in template

Question

I have a generic code that executes the passsed lambda and checks the outputs. I would like the helper method to be fully deduced, but cannot come up with the solution since the method returns std::optional and I want the processing method to return the underlying type of it. Code:

#include <optional> #include <type_traits> std::optional<int> dummy() { return {42}; }; template<typename F, typename Ret = std::result_of_t<F(void)>> static Ret process(const F& f) { Ret val{}; if(auto res = f(); res) { val = *res;} return val; } int main() { // call auto result = process(&dummy); }

The result is of course type of std::optional<int>. What should I change in my template definition to deduce the return type of process to underlying type of std::optional - int?

Please try to copy/paste the code you actually showed in your question, and attempt to compile it. It won't work. Obvious typos. Please show real code in your question. — Sam Varshavchik
– Sam Varshavchik, Commented Apr 13, 2020 at 11:35
No, it doesn't compile. That's just one of an amazing number of typos and errors, above. You've been on stackoverflow.com for seven years, and you still are not familiar with the requirements for a minimal reproducible example? — Sam Varshavchik
– Sam Varshavchik, Commented Apr 13, 2020 at 11:40
@SamVarshavchik What are you talking about, after the Ret typo fix, it was perfectly fine. — Barry
– Barry, Commented Apr 13, 2020 at 13:31

rafix07 · Accepted Answer · 2020-04-13 11:45:42Z

All standard containers keep information about underlying data type on value_type typedef.

Why not write:

template<typename F, typename Ret = typename std::result_of_t<F(void)>::value_type >

then for optional<int>, Ret is int.

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