I\'ve asked this question asking about the guarantees of the lifetime of a reference in a local variable and was referred to this thread in which the exact semantic
Let's analize how Swift works is respect to memory management:
Given the above two above facts, your Something
instance is kept alive by the something
variable which holds a strong reference to it until the end of scope, i.e. until the main
function returns.
Try replacing let something = Something()
by _ = Something()
and you see that the expected behaviour that you want:
Something.init()
Something.deinit
in main()
Same behaviour occurs when holding a weak reference to the instance: weak var something = Something()
Note that there is a slight difference here between the something
variable and the Something
instance. The variable holds a reference to the instance, not the instance itself. Thus, the reference lifetime is the same as the variable lifetime, and this is what affect the instance's lifetime. Why the variable is not destroyed sooner, e.g. when the compiler detects it's no longer needed, I don't know.
If you had made your variable an optional and set it to nil, you would have explicitly told the compiler where you want the reference to be released.
e.g.
func main() {
var something:Something? = Something()
something = nil
print("in main()")
}
Otherwise you are letting the compiler decide execution order on its own. In the same way that it could decide to execute any line of code out of order, it can also decide when to deallocate local variables.
In your example "something" is an non-optional local variable, so it is likely that the compiler will decide to allocate it on the stack. This will mean that the variable's value (on the stack) will hold the reference until the function returns.