How do println! macro captures variable?

[yousuf]

Here is the code:

fn main() 
{
    let mut a: i32 = 57;
    let b: &mut i32 = &mut a;
    let c: &i32 = &b;
    //fun( b, c);     // error
    //fun(&mut b, &c);   //error too

    println!("{} {}", b, c );   // works
}

fn fun( b: &mut i32, c: &i32) { }

It seems to me that println! macro takes the variable(b, c) differently than the function fun. What could be the reasonable explanations behind it in simple English?

Thanks.

[alice]

The macro does indeed capture the variables in a different way. That's because the macro actually expands to something that contains &b and &c rather than just containing b and c.

fn main() {
    let mut a: i32 = 57;
    let b: &mut i32 = &mut a;
    let c: &i32 = &b;

    fun(&b, &c);
    println!("{} {}", b, c);
}

fn fun(b: &&mut i32, c: &&i32) {}

Note that fun takes a reference to a reference in this example.

[erelde]

You can actually read what macros expand to with cargo-expand or in the playground under Tools > Expand Macros.