I have the following const:
const IS_WSL: bool = is_wsl!();
and I'd like to be able to use this with the cfg attibute to perform conditional compilation. Something like:
#[cfg(const = "IS_WSL")] // what goes here?
const DOWNLOLADS: &'static str = "/mnt/c/Users/foo/Downloads";
#[cfg(not(const = "IS_WSL"))]
const DOWNLOADS: &'static str = "/home/foo/Downloads";
Obviously this syntax doesn't work, but is there any way to achieve what I'm describing?
I'm aware of custom rustc flags, but would like to avoid doing that, since there's a fair amount of logic that I'd rather not try to write in bash
The answer is not. You have to use something like build script to achieve that.
It cannot work because cfg-expansion occurs at an earlier pass in the compiler than constant evaluation.
cfg expansion works at the same time as macro expansion. Both can affect name resolution (macros can create new names, which other macros, or even the same macro, can later refer to) which forces us to use a fixed-point algorithm (resolve names, expand macros, resolve names, expand macros... until no more names can be resolved, i.e. a fixed point was reached). const evaluation takes a place after type checking, sometimes (with generic_const_exprs) even during codegen. If it could affect macro expansion, we would have a giant fixed-point loop resolve names - expand macros - resolve names - expand macros... until a fixed point is reached, then lower to HIR - type-check - evaluate constants (or even lower to MIR - monomorphize and evaluate constants) - and back to name resolution. Besides slowing the compiler down a lot, it'll also make it significantly more complex, not really something the rustc team wants.