I recently watched Nico Josuttis’s Keynote Meeting C++ 2022, and just finished reading the Barry Revzin’s post about that same topic.

The code in question is this

// increment even elements:
for (int& i : coll | std::views::filter(isEven)) {
    ++i; // UB: but works
}

My question is (maybe for compiler writers), how far is my imagination going, if I think that the UB mandated by [range.filter.iterator]/1 could make the following code,

#include <iostream>
#include <ranges>
#include <vector>

constexpr auto isEven = [](int i){
    return i % 2 == 0;
};

bool bar();

bool foo() {
    bool b = bar();
    std::vector<int> coll{1,2,3,4,5,6,7};
    if (b)
        for (int& i : coll | std::views::filter(isEven))
            ++i; // UB: but works
    return b;
}

be compiled down to just return false;, i.e. to this?

foo():
        xor     eax, eax
        ret

I mean, the compiler does see that there’s at least one element in coll that satisfies isEven, which means that if b is true and the for loop therefore runs, ++i would rund, which would lead to UB. Hence the compiler could happily conclude that bar() must return false.

Is such “reasoning” entirely impossible for a compiler?