Romeo Sierra Romeo Sierra - 21 days ago 7
Swift Question

Functions that return another function in Swift

Ola,

Let's say that there is a function named

fnOuter(name:String, mks:Double...)
. I want this function to return another function which I could separately write like
fnInner(msg:String)->(String, Double, Double, Character)
. How one could achieve this?
I am a newbie to Swift and tried out the following. But ultimately it ends up with
function 'std' was used as a property; add () to call it
thrown by the Swift compiler. What am I doing wrong here? How could I fix this? Or is this even possible?

func fnRetFn(name:String, mks:Double...) -> (() -> ((String, Double, Double, Character))){
var msg = "Hello " + String(name) + "!"

func calculate() -> (String, Double, Double, Character){
var total:Double = 0.0
var i: Double = 0.0
for mk in mks{
total += mk
i += 1
}

var avg = total / i

var grd : Character
if avg >= 75.0{
grd = "A"
}

else if avg >= 55.0{
grd = "B"
}

else{
grd = "F"
}

return (msg, total, avg, grd)
}

return calculate
}

var outputFn = fnRetFn(name:"Mike", mks:75.3, 87.2)
var std = outputFn
print("\(std.0)")
print("\(std.1)")
print("\(std.2)")
print("\(std.3)")


Edit 1
Please note that the fnInner(msg:String) should return a tuple, not another function.

Answer

With

var outputFn = fnRetFn(name:"Mike", mks:75.3, 87.2)
var std = outputFn

both outputFn and std are references to the same function which was returned by fnRetFn(...):

print(std) // (Function)

To call the function you'll have to append the argument list in parentheses. For a function taking no arguments that is the empty list ():

var outputFn = fnRetFn(name:"Mike", mks:75.3, 87.2)
var std = outputFn()

And now std is the tuple returned from calling the "inner" function:

print(std) // ("Hello Mike!", 162.5, 81.25, "A")
print(std.0, std.1, std.2, std.3) // Hello Mike! 162.5 81.25 A

(Unrelated to your problem, but note that both variables should be constants declared with let.)