How to replace nth character of a string with another

Question

How could I replace nth character of a String with another one?

func replace(myString:String, index:Int, newCharac:Character) -> String {     // Write correct code here     return modifiedString } 

For example, replace("House", 2, "r") should be equal to "Horse".

Answer 1

Solutions that use NSString methods will fail for any strings with multi-byte Unicode characters. Here are two Swift-native ways to approach the problem:

You can use the fact that a String is a sequence of Character to convert the string to an array, modify it, and convert the array back:

func replace(myString: String, _ index: Int, _ newChar: Character) -> String {     var chars = Array(myString)     // gets an array of characters     chars[index] = newChar     let modifiedString = String(chars)     return modifiedString }  replace("House", 2, "r") // Horse 

Alternately, you can step through the string yourself:

func replace(myString: String, _ index: Int, _ newChar: Character) -> String {     var modifiedString = String()     for (i, char) in myString.characters.enumerate() {         modifiedString += String((i == index) ? newChar : char)     }     return modifiedString } 

Since these stay entirely within Swift, they're both Unicode-safe:

replace("🏠🏡🏠🏡🏠", 2, "🐴") // 🏠🏡🐴🏡🏠 

Answer 2

In Swift 4 it's much easier.

let newString = oldString.prefix(n) + char + oldString.dropFirst(n + 1) 

This is an example:

let oldString = "Hello, playground" let newString = oldString.prefix(4) + "0" + oldString.dropFirst(5) 

where the result is

Hell0, playground 

The type of newString is Substring. Both prefix and dropFirst return Substring. Substring is a slice of a string, in other words, substrings are fast because you don't need to allocate memory for the content of the string, but the same storage space as the original string is used.