Hi,
I am trying to get a better understanding of Swift's function overload
resolution rules.
As far as I can tell, if there are multiple candidates for a function call,
Swift favors
functions for which the least amount of parameters have been ignored /
defaulted. For example:
// Example 1
func f(x: Int) { print("f1") }
func f(x: Int, y: Int = 0) { print("f2") }
f(x: 0) // f1
// Example 2
func f(x: Int, y: Int = 0) { print("f1") }
func f(x: Int, y: Int = 0, z: Int = 0) { print("f2") }
f(x: 0) // f1
It also looks like Swift favors functions with default-value parameters over
functions with variadic parameters:
func f(x: Int = 0) { print("f1") }
func f(x: Int...) { print("f2") }
f() // f1
f(x: 1) // f1
f(x: 1, 2) // f2 (makes sense because f1 would not work here)
f(x: 1, 2, 3) // f2 (makes sense because f1 would not work here)
But then I tested functions with default-value parameters and variadic
parameters and things start to get weird.
For example, this must be a bug, right?
func f(x: Int..., y: Int = 0) { print(x, y) }
func f(x: Int...) { print(x) }
f() // []
f(x: 1) // [1]
f(x: 1, 2) // [1, 2] 0
f(x: 1, 2, 3) // [1, 2, 3]
I think, in this example, it should always call the second overload, because
no parameter is ignored / defaulted. What do you think?
Thanks and best regards,
Toni_______________________________________________
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