Consider
template <typename S, typename T, typename F>
vector<T> map(const vector<S> &ss, F f){
vector<T> ts;
ts.reserve(ss.size());
std::transform(ss.begin(), ss.end(), std::back_inserter(ts), f);
return ts;
}
int main(){
vector<int> is = {...};
vector<double> ts = map(is, [](int i){return 1.2*i;});
}
because the compiler 'couldn't deduce template parameter T'. Specifying map
of type
template <typename S, typename T>
vector<T> map(const vector<S> &ss, std::function<T(S)> f)
also doesn't work, because it doesn't match lambdas.
What's the correct way?
CodePudding user response:
Something like this:
template <typename S, typename F>
auto map(const std::vector<S> &ss, F f) -> std::vector<std::decay_t<decltype(f(ss[0]))>>
{
std::vector<std::decay_t<decltype(f(ss[0]))>> ts;
ts.reserve(ss.size());
std::transform(ss.begin(), ss.end(), std::back_inserter(ts), std::move(f));
return ts;
}
I've opted for a trailing return type, to be able to use f
and ss
in decltype
. You could just use auto
return type, but explicitly specifying the type makes the function SFINAE-friendly.
Or, a slightly overengineered version with hottest C 20 features:
template <
template <typename...> typename C = std::vector,
typename S, typename F
>
[[nodiscard]] C<std::decay_t<std::indirect_result_t<F, std::ranges::iterator_t<S>>>>
map(S &&ss, F f)
{
C<std::decay_t<std::indirect_result_t<F, std::ranges::iterator_t<S>>>> ts;
ts.reserve(std::ranges::size(ss));
std::ranges::transform(ss, std::back_inserter(ts), std::move(f));
return ts;
}
CodePudding user response:
this works
std::vector<double> ts = map<int, double>(is, [](int i) {return 1.2 * i; });