On 30.09.2016 15:02, Toni Suter via swift-users wrote:
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:
I do believe this *should* be a bug in Swift's parser. It is not obvious
the intention of the user, i.e. which function *user* want to call.
IMO there should be no hidden or complex rules on how Swift selects the
candidate for function calls if parameters suits for a number of
implementations.
IMO The only rule should be - if it is not clear from the parameter list
which func should be called - compilation error must be thrown. This will
be "Swifty" which is "Safe by default". Current situation is not safe by
default.
// 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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