On Friday, 17 March 2017 at 13:10:23 UTC, Jonathan M Davis wrote:
it looks like break and continue _are_ used at compile time, since it prints
They are working exactly the same way as any other loop. The fact that it is unrolled and the dead code removed from the binary is an implementation detail that doesn't change the semantics of the loop, just like any other loop unroll optimization.
`continue` is more like `goto` than it is like `static if`, even when looping over AliasSeq.
whereas if the break were just compiled in to be run at runtime, you never would have seen any of the strings past "hello" in the outer loop get printed.
No, they did a normal break at runtime. But they are breaking the INNER loop, so the outer loop continues. The generated code simply removed the dead portions since the optimizer realized there was no point carrying it around.
You are totally overthinking this, loops still work the same way regardless of what they are iterating over.
The only special thing about an AliasSeq thing is that the current value is available at compile time, so it is legal to use in more places (then the compiler implementation generates the code differently to make it happen, but that's the same idea as passing -funroll-loops or -O3 and getting different code, they don't change how break works.)
Remember that this is basically doing loop unrolling, which is not necessarily the same thing as you would expect from a "static foreach." It's not like we're building a string mixin here.
Yes.
