How to conditionally compile in Rust [duplicate] - rust

Question

I've followed quite a bit of the documentation and tried to reuse an example, but I can't get my code to work.
My Cargo.toml looks like this:
[package]
name = "Blahblah"
version = "0.3.0"
authors = ["ergh <derngummit#ahwell.com"]
[dependencies]
[[bin]]
name = "target"
path = "src/main.rs"
[features]
default=["mmap_enabled"]
no_mmap=[]
mmap_enabled=[]
I'd like to test my code locally with a different buffer origin than mmap based on what feature configuration I pass to the cargo build command. I have this in my code:
if cfg!(mmap_enabled) {
println!("mmap_enabled bro!");
...
}
if cfg!(no_mmap) {
println!("now it's not");
...
}
The compiler doesn't see the code in either of the if statement bodies, so I know that both of the cfg! statements are evaluating to false. Why?
I've read Conditional compilation in Rust 0.10? and I know it's not an exact duplicate because I'm looking for a functioning example.

Answer 1

The correct way to test for a feature is feature = "name", as you can see in the documentation you linked if you scroll a bit:
As for how to enable or disable these switches, if you’re using Cargo,
they get set in the [features] section of your Cargo.toml:
[features]
# no features by default
default = []
# Add feature "foo" here, then you can use it.
# Our "foo" feature depends on nothing else.
foo = []
When you do this, Cargo passes along a flag to rustc:
--cfg feature="${feature_name}"
The sum of these cfg flags will determine which ones get activated,
and therefore, which code gets compiled. Let’s take this code:
#[cfg(feature = "foo")]
mod foo {
}
In your case using the cfg! macro, this would map to cfg!(feature = "foo").