3 ms·
Not really? When you reduce with function f(T1, T2) -> T3 the result of previous iteration becomes first argument for the next iteration, so they must be of th
by ajuc 12y ago
Not really?
When you reduce with function f(T1, T2) -> T3 the result of previous iteration becomes first argument for the next iteration, so they must be of the same "type".
Or am I missing something?
- dkersten 12y agoAh, no, you're right. I was thinking that mapping can be f(T1)->T2 and how reducing can change types too, but I guess I wasn't paying enough attention because of course the reduce signature is f(accumulator, input) and the output is accumulator, so yea, you're absolutely right: f(T1,T2)->T1