Shadowing

Reuse a name by shadowing the previous binding.

Shadowing

Writing a new let with the same name shadows the previous variable. This is different from mut: a new variable is created and even its type can change:

let girdi = "42";          // &str
let girdi: i32 = girdi.parse().unwrap();  // i32
let girdi = girdi * 2;      // 84
println!("{girdi}");

It lets you keep the same name through conversion steps without inventing names like girdi_str, girdi_sayi.

Scope and shadowing

Shadowing done in an inner block is valid only inside that block; when the block ends, the outer value returns:

let x = 5;
{
    let x = x * 2;
    println!("iç: {x}");   // 10
}
println!("dış: {x}");      // 5

Exercise Shadowing chain

Start with let deger = " 21 ";. First write let deger = deger.trim();, then let deger: i32 = deger.parse().unwrap();, finally let deger = deger * 2;, and print the result.

Show example solution

fn main() {
    let deger = " 21 ";
    let deger = deger.trim();
    let deger: i32 = deger.parse().unwrap();
    let deger = deger * 2;
    println!("{}", deger);
}

Test yourself

What is the key difference between shadowing and <code>mut</code>?
- Shadowing only works with constants
- Shadowing creates a new variable and can change the type
- mut can change the type
- They are the same
Show answer

Shadowing creates a new variable and can change the type — mut changes the value of the same variable; shadowing with let creates a brand-new binding and the old value becomes unreachable.

Shadow it

let ad = "  Ada  ";
____ ad = ad.trim();

Show answer

let ad = "  Ada  ";
let ad = ad.trim();

Code and result

let x = 5; let x = x + 1;

x = 6 (new variable)
let x = "1"; let x: i32 = ...

Type changed
let x in an inner block

Only in the block
mut x = "a" (x i32)

Compile error

Next lesson: Summary - Variables

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