I have specified a few features inside Cargo.toml
:
[features]
complex = []
simple = []
When I build my project I use cargo build --features="complex"
or simple
.
In some functions, I want to return a value based on which feature is used:
fn test() -> u32 {
let x: u32 = 3;
if cfg!(feature = "complex") {
let y: u32 = 2;
x + y
}
if cfg!(feature = "simple") {
let y: u32 = 1;
x + y
}
}
But this doesn't work as it tries to evaluate both expressions. What is the proper way to use the cfg!
macro in my case?
The documentation for cfg!
states:
Boolean evaluation of configuration flags.
That means that cfg!(...)
is replaced with a Boolean (true
/ false
). Your code would look something like this, after it's expanded:
fn test() -> u32 {
let x = 3;
if true {
let y = 2;
x + y
}
if true {
let y = 1;
x + y
}
}
The easiest solution is to add an else
:
fn test() -> u32 {
let x = 3;
if cfg!(feature = "complex") {
let y = 2;
x + y
} else {
let y = 1;
x + y
}
}
You can also use the attribute form of cfg
. In this case, the attribute can prevent the entire next expression from being compiled:
fn test() -> u32 {
let x: u32 = 3;
#[cfg(feature = "complex")]
{
let y: u32 = 2;
x + y
}
#[cfg(feature = "simple")]
{
let y: u32 = 1;
x + y
}
}
as it tries to evaluate both expressions.
No, it doesn't. Evaluation occurs at run-time, and this code cannot even be compiled.
See also:
- Is it possible to conditionally compile a code block inside a function?
- Example of how to use Conditional Compilation Macros in Rust
- How many lines are covered by the Rust conditional compilation attribute?