On Friday, 5 January 2018 at 04:10:54 UTC, Steven Schveighoffer
wrote:
The compiler assumes x is going to be 5 forever, so instead of
loading the value at that address, it just loads 5 into a
register (or maybe it just folds x == 5 into true).
I was curious what dmd did, and the disassembly indeed shows it
just loads 5 into the register and leaves it there - assuming
since it is immutable, it will never change through any pointer
and thus never reloads it from memory at any time.
Interestingly, dmd -O just stubs out the whole function. I guess
it assumes all the defined behavior actually accomplishes nothing
and it is free to optimize out undefined behavior... thus the
function needs no code. Similarly, if the last assert is changed
to x != 5, dmd -O doesn't even actually do a comparison (the
value 5 never appears in the generated code!), it just outputs
the direct call to assertion failure.
