I want to work with the numbers from the array numbers in the function. I just want multiply 2 with 3, 3 with 3, 5 with 3 and 2 with 3. However, I get an error. What am I doing wrong?
import Foundation
let numbers = [2, 3, 5, 2];
func num(number: [Int]...) {
for item in number {
let result = item * 3
print (result)
}
}
num(numbers)
Binary operator '*' cannot be applied to operands of type '[Int]' and 'Int'
You are passing an array of arrays to the function. The function parameter to num() should just be [Int] or Int...
Difference between Int... and [Int]
with Int... you can pass a variable amount of params without explicitly using an array:
func num(mult: Int, arr: Int...){
for n in arr{
println("num = \(n * 3)")
}
}
num(1,2,3,4) // first param (1) == mult(iplier), rest are ints captured by arr
while with an array ([Int]), you have to explicitly pass an array:
func num(mult: Int, arr: [Int]){
for n in arr{
println("num = \(n * 3)")
}
}
num(1, [2,3,4]) // 1 == mult(iplier,), rest are the Ints to operate with
Second solution is clearly cleaner.
I have never personally used Int..., but it definitely has its place...
Needless to say that those (see @Qbyte's comment below on why this statement is wrong)Int... params need to go at the end of the parameter list...
[Int] and Int...?
Int...) can now appear anywhere in the parameter list not only at the end.
Either delete the ... to be able to pass an array :
let numbers = [2, 3, 5, 2]
func num(number: [Int]) {
for item in number {
let result = item * 3
print(result)
}
}
num(numbers)
or you can pass the values directly with a vararg:
func num(number: Int...) {
for item in number {
let result = item * 3
print(result)
}
}
num(2, 3, 5, 2)
You are passing an array of arrays of Int (Array<Array<Int>>).
Array<Int>, which is [Int], which is Int..., but Int... parameter value can't be an array of [Int], it should be sequence of Ints.
E.g.
func foo(bar: Int...) {
// bar is [Int]
}
foo(0, 3, 13, 6) // parameter is sequence of `Int`s
Also, you could use .map() to apply transform to each element of array. And, using Int type as parameter is not efficient way of writing Swift code (you an not pass Array<UInt> as input parameter, etc.), protocol IntegerArithmeticType provides functions for * operator. All default Swift integer types conform to this protocol. So, the only way here is to use generic function with type T, where T: IntegerArithmeticType. Here is the final code:
let numbers = [2, 3, 5, 2]
/// Multilplies each element of array of `IntegerArithmeticTypes` by `multiplier`.
///
/// - Parameter multiplier: `IntegerArithmeticTypes` multiplier for each element of array.
/// - Parameter vals: Array of `IntegerArithmeticTypes` in which each element
/// will be multiplied by `multiplier`
///
/// - Returns: Array with multiplied values from `values`
public func multipliedMap<T: IntegerArithmeticType>(multiplier: T, _ vals [T]) -> Array<T> {
return vals.map { $0 &* multiplier }
}
dump(multipliedMap(3, numbers))
// prints:
//
// ▿ 4 elements
// - [0]: 6
// - [1]: 9
// - [2]: 15
// - [3]: 6
PS: I used &* operator, because IntegerArithmeticType value may potentially overflow.
Try this :
var numbers = [2, 3, 5, 2];
func num(number: [Int])
{
var result = 0
for var j = 0; j < number.count - 1 ; j++
{
result = number[j] * number[j + 1]
print ("\(result)")
}
}
print(num(numbers))
number: [Int]...does not mean what you think it does.func num(number: [Int]...)it...