What output will be produced by the code below?
import Foundation let number = 16 print("\(number) squared is \(number * number), and its square root is \(sqrt(number))")
Answer: This code will not compile.
sqrt() function can be called using two different types of parameter, neither of which are integers. In this example, Swift's type inference will consider
number to be an
Integer, which cannot be used by
sqrt()unless you typecast it.