Member Avatar for k22mac

I am taking an intro level java programming class and i am a little stuck at the moment. I'm trying to use a the replace method. My objecti with this program is to gather a phrase from the user and then have it print the same phrase with some characters replaced. The Specific characters are 'a' to '@', 'e' to '3', 'i' to '!', 'o' to '0', 'b' to '8', and ' ', to '_'. I am keeping this as basic as possible so any comments I would appreciate if you keep in mind that anything too much more complex than what i have here will probably not be what i'm looking for. But please anything you can think of, whether it helps or not, will be much appreciated.

import java.util.Scanner;
public class Mutator
{
    public static void main (String[] args)
    {
        Scanner keyboard = new Scanner(System.in);
        System.out.println("Which of our presidents freed the slaves?");
        String phrase = keyboard.nextLine();


        phrase.replace('a','@');
        phrase.replace('e','3');
        phrase.replace('i','!');
        phrase.replace('o','0');
        phrase.replace('b','8');
        phrase.replace(' ','_');


        System.out.println(phrase);
    }
}
  • 4 Contributors
  • 4 Replies
  • 127 Views
  • 2 Hours Discussion Span
  • Latest Post Latest Post by dimasalang

Dani AI

Generated

Short version: Strings are immutable in Java, so each call to replace returns a new String. As pointed out, the result needs to be used (reassigned or printed). 's suggestion to reuse the same variable is the simplest — no need for a fresh variable per character and was right that trying different approaches will show the behavior quickly.

A single-pass build is more efficient and clearer for multiple character mappings than chaining many replace calls (which creates temporary Strings). The example below shows a compact helper that walks the input once and applies the specified substitutions. It also makes it easy to add uppercase handling if needed.

static String mutate(String s) {
    StringBuilder out = new StringBuilder(s.length());
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        switch (c) {
            case 'a': out.append('@'); break;
            case 'e': out.append('3'); break;
            case 'i': out.append('!'); break;
            case 'o': out.append('0'); break;
            case 'b': out.append('8'); break;
            case ' ': out.append('_'); break;
            default: out.append(c);
        }
    }
    return out.toString();
}

Notes and quick troubleshooting:

  • The replace(char,char) method is case-sensitive. To handle A as well, either add case 'A': entries or normalize (with caution) before mapping.
  • For short inputs, reassigning with repeated replace calls is perfectly fine for clarity. For very large strings or many mappings, prefer the single-pass approach above.
  • Close the Scanner when done (e.g., keyboard.close()).
  • Official behavior is documented in the Java API: String.replace(char, char) documentation.

Recommended Answers

All 4 Replies

Member Avatar for dmanw100

Remember that replace() returns a string. It doesn't modify the string that called it directly, but rather it makes a new String object. You're really close, but you need to keep that in mind.

Member Avatar for k22mac

So do I need to create a new variable for each character I'm replacing?

Member Avatar for NormR1

do I need to create a new variable

Try it and see what happens. You'll need a print statement for each variable.

Edited by NormR1
Member Avatar for dimasalang

I suggest to put it to the same variable.

phrase = phrase.replace('a', '@');
commented: Yes! that was exactly what i needed. That got me right where i wanted to be. Thanks for your help!! +0
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