C++11 lambda function definition in if-else

Question

How do I do something like this, but in a way that will compile, and hopefully without insane typedefs?

auto b;
auto g;
if (vertical)
{
    b = [=, &b_](int x, int y) -> bool { return b_[x + y*w]; };
    g = [=, &g_](int x, int y) -> int& { return g_[x + y*w]; };
}
else
{
    b = [=, &b_](int x, int y) -> bool { return b_[y + x*w]; };
    g = [=, &g_](int x, int y) -> int& { return g_[y + x*w]; };
}

Answer 1

The reason this does not compile is the improper use of auto. You can avoid conditional creation of lambdas altogether by changing their definition a little:

int mx = vertical ? 1 : w;
int my = vertical ? w : 1;
auto b = [=, &b_](int x, int y) -> bool { return b_[mx*x + my*y]; };
auto g = [=, &g_](int x, int y) -> int& { return g_[mx*x + my*y]; };

The idea is to set multipliers mx and my conditionally. This lets you construct your lambdas unconditionally, so you could use auto in your declaration.