Closure with explicit vs implicit capturing

[mkbrwr (Mykhailo Tymchyshyn)]

Encountered the next code

var thing = "cars"

let closure1 = { [thing] in
print("I love (thing)")
}

let closure2 = {
print("I love (thing)")
}

thing = "airplanes"

closure1() // I love cars
closure2() // I love airplanes

Why is the behaviour different for explicit and implicit capturing ?

[tera (Mike)]

Expressions — The Swift Programming Language (Swift 5.5)
Capture Lists
...
The entries in the capture list are initialized when the closure is created. For each entry in the capture list, a constant is initialized to the value of the constant or variable that has the same name in the surrounding scope. For example in the code below, a is included in the capture list but b is not, which gives them different behavior.

var a = 0
var b = 0
let closure = { [a] in
    print(a, b)
}

a = 10
b = 10
closure()
// Prints "0 10"

There are two different things named a , the variable in the surrounding scope and the constant in the closure’s scope, but only one variable named b . The a in the inner scope is initialized with the value of the a in the outer scope when the closure is created, but their values aren’t connected in any special way. This means that a change to the value of a in the outer scope doesn’t affect the value of a in the inner scope, nor does a change to a inside the closure affect the value of a outside the closure. In contrast, there’s only one variable named b —the b in the outer scope—so changes from inside or outside the closure are visible in both places.

[mkbrwr (Mykhailo Tymchyshyn)]

Turns out there is document that answers my question directly, thanks for pointing it out. Now it's clear!