How does the Swift compiler determine which method or operator to call given a
set of overloads at a given call site? This question may conflate two issues:
method overloads and operator overloads, but it seems at least superficially
that they are related. Somehow they end up using different rules as I will
demonstrate shortly.
<code>
class A{
}
class B: A {
}
func +(a:A, b:A){
print("+1")
}
func +(a:A, b:B){
print("+2")
}
func +(a:B, b:A){
print("+3")
}
/*
func +(a:B, b:B){
print("+4")
}
*/
A() + A() // prints +1
A() + B() // prints +1
B() + A() // prints +1
// B() + B() // will come back to this
func f(a: A, b: A){
print("f1")
}
func f(a: B, b: A){
print("f2")
}
func f(a: A, b: B){
print("f3")
}
/*
func f(a: B, b: B){
print("f4")
}
*/
f(A(), b: A()) // prints f1
f(B(), b: A()) // prints f2
f(A(), b: B()) // prints f3
// f(B(), b: B()) // will come back to this
</code>
The output from the various invocations of f() make sense. It seems that Swift
uses similar overload resolution rules as Java where the 'most specific method'
is chosen given a set of methods whose parameters are part of the same class
hierarchy.
Given the code as-is, the outputs from the operator overloads all seem to call
the first operator defined. The order of definition doesn't matter, if +(A, A)
is put last it will still be called by all three invocations of +. Already this
doesn't make sense to me. Why isn't +(A, B) called for the expression A() + B()?
Something strange happens when the function +(B, B) is uncommented out. All of
a sudden the overloads work just like they did in the regular function
definition case.
The full code is:
<code>
func +(a:A, b:A){
print("+1")
}
func +(a:A, b:B){
print("+2")
}
func +(a:B, b:A){
print("+3")
}
func +(a:B, b:B){
print("+4")
}
A() + A() // prints +1
A() + B() // prints +2
B() + A() // prints +3
B() + B() // prints +4
</code>
Without the definition of +(B, B) the expression B() + B() is an error because
it is ambiguous whether to choose +(A, B) or +(B, A). I am fine with that
behavior. But why do A()+B() and B()+A() suddenly refer to the +2 and +3
operators respectively?
The regular function case does not behave this way. Given the 3 uncommented
defintions of f as above the output is as expected, f1, f2, f3 in order. When
the definition of f(B, B) is uncommented the expression f(B(), B()) works as
expected and prints f4. The other expressions are unaffected.
--
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