Why does vectorization behave differently for almost the same code?

Question

Here are free functions that do the same but in the first case the loop is not vectorized but in the other cases it is. Why is that?

#include <vector>

typedef std::vector<double> Vec;

void update(Vec& a, const Vec& b, double gamma) {
    const size_t K = a.size();
    for (size_t i = 0; i < K; ++i) { // not vectorized
        a[i] = b[i] * gamma - a[i];
    }
}

void update2(Vec& a, const Vec& b, double gamma) {
    for (size_t i = 0; i < a.size(); ++i) { // vectorized
        a[i] = b[i] * gamma - a[i];
    }
}

void update3(Vec& a, size_t K, const Vec& b, double gamma) {
    for (size_t i = 0; i < K; ++i) { // vectorized
        a[i] = b[i] * gamma - a[i];
    }
}

int main(int argc, const char* argv[]) {
    Vec a(argc), b;
    update(a, b, 0.5);
    update2(a, b, 0.5);
    update3(a, a.size(), b, 0.5);
    return 0;
}

Relevant messages from the compiler (VS2013):

1>  c:\home\dima\trws\trw_s-v1.3\trws\test\vector.cpp(7) : info C5002: loop not vectorized due to reason '1200'
1>  c:\home\dima\trws\trw_s-v1.3\trws\test\vector.cpp(13) : info C5001: loop vectorized
1>  c:\home\dima\trws\trw_s-v1.3\trws\test\vector.cpp(19) : info C5001: loop vectorized

From comment by @tony

Reason 1200: "Loop contains loop-carried data dependences that prevent vectorization. Different iterations of the loop interfere with each other such that vectorizing the loop would produce wrong answers, and the auto-vectorizer cannot prove to itself that there are no such data dependences." source

Answer 1

I guess it's some deeply internal compiler implementation issue, like at what stage did auto-vectorizer "kick in" and what's the state of the internal representation of the code at that time. When I tried on MSVC2017, it worked more in line with what one would expect. It auto-vectorized update() and update3(), but not update2(), with the reason given 501 for line 14, which is documented as:

Induction variable is not local; or upper bound is not loop-invariant.