select code based on cfg attribute not true [rust]

Question

I want rust code compiled when the compilation attribute debug_assertions is false (or not enabled), i.e. a "debug build".
Is this possible? What is the syntax?

For example, I can compile function build_type when compiling for debug build (i.e. option --release is not passed to command cargo build).

#[cfg(debug_assertions)]
pub fn build_type() -> String {
    String::from("debug")
}

In this case, I want a "release version" of the function,

#[cfg(debug_assertions)]
pub fn build_type() -> String {
    String::from("debug")
}

#[!cfg(debug_assertions)]
pub fn build_type() -> String {
    String::from("release")
}

However, the syntax #[!cfg(debug_assertions)] results in cargo build error expected identifier, found '!'.

Other failed syntax variations were:

• #[cfg(!debug_assertions)]
• #[cfg(debug_assertions = false)]
• #[cfg(debug_assertions = "false")]

Answer 1

In book Rust By Example, the section on cfg has this snipped code sample:

// And this function only gets compiled if the target OS is *not* linux
#[cfg(not(target_os = "linux"))]
fn are_you_on_linux() {
    println!("You are *not* running linux!");
}

Use syntax #[cfg(not(debug_assertions))] to conditionally compile for cargo build --release.

The book The Rust Reference has the full list of conditional compilation predicates.

Answer 2

You can do this, based on rust book

if cfg!(not(target_os = "linux")) {
    // Your code
} else {
    // Your else code
}

And to check if it is a debug binary, you can do the same, but with the debug_assertions cfg, like this

if cfg!(debug_assertions) {
    // Your code if is debug
}