问题
Consider the following code using the ranges library (from c++20)
#include <iostream>
#include <ranges>
#include <vector>
int main() {
std::vector<int> inputs{1, 2, 3, 4, 5, 6};
auto square_it = [](auto i) {
std::cout << i << std::endl;
return i * 2; };
auto results = inputs | std::views::transform(square_it) | std::views::filter([](auto i){ return i % 3 == 0; });
for(auto r : results) {
// std::cout << r << std::endl;
}
}
The cout in the square function is to log when the square function is called by the ranges library. This code prints
1
2
3
3
4
5
6
6
The question is, why are values that match the filter's predicated are printed twice?
I have seem this code in a presentation in CppCon 2020, where the presenter explains why this happens. According to him, filter iterates until its predicate is satisfied (and of course if needs to call transform each time). Then filter stops and it is ready to be iterated on. After that the actual iteration is started and a value is read from filter, which then calls transform again a second time for the same input.
It is not clear to me why this is necessary. Since ranges compute values lazily and every element pulls data from the one before it, why can't filter just pass the value to whoever is after it in the pipeline as soon as it finds a match?
回答1:
why can't filter just pass the value to whoever is after it in the pipeline as soon as it finds a match?
Because in the iterator model, positioning and accessing are distinct operations. You position an iterator with ++; you access an iterator with *. These are two distinct expressions, which are evaluated at two distinct times, resulting in two distinct function calls that yield two distinct values (++ yields an iterator, * yields a reference).
A filtering iterator, in order to perform its iteration operation, must access the values of its underlying iterator. But that access cannot be communicated to the caller of ++ because that caller only asked to position the iterator, not to get its value. The result of positioning an iterator is a new iterator value, not the value stored in that iterated position.
So there's nobody to return it to.
You can't really delay positioning until after accessing because a user might reposition the iterator multiple times. I mean, you could implement it that way in theory by storing the number of such increments/decrements. But this increases the complexity of the iterator's implementation. Especially since resolving such delayed positioning can happen through something as simple as testing against another iterator or sentinel, which is supposed to be an O(1) operation.
This is simply a limitation of the model of iterators as having both position and value. Alternative models exist where iteration and access are bundled together exist, but they're not how standard library iteration works.
来源：https://stackoverflow.com/questions/64199664/why-c-ranges-transform-filter-calls-transform-twice-for-values-that-match