Why is everything after the space not printed?

Question

I created a simple program using the acm library that takes a user inputted string and prints it out in morse. This is the program:

import acm.program.*;

public class morse extends Program{
    public void run(){

        String[] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
        String[] morse = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        String word = readLine("Give the word(s) to transcribe: ").toUpperCase();
        println("Morse: ");
        int j = 0;
        for(int i = 0; i < word.length(); i++){
            int k = 0;
            boolean flag = true;
            if(Character.toString(word.charAt(j)).equals(" ")){
                println();
            }else {
                while(!Character.toString(word.charAt(j)).equals(alphabet[k])){
                    if(k < 25){
                        k += 1;
                    }else{
                        println("Letter was not found.");
                        flag = false;
                        break;
                    }
                }
                if(flag){
                    println(morse[k] + " ");
                }
            j += 1;
            }   
        }
    }
}

However, every time the string contains a space, everything after the space is not printed. I seriously cant find the reason behind this. Can anyone help me or even point me somewhere ? Thanks

(The letters after the space are all printed as spaces)

What are `j` and `k` supposed to represent? – khelwood Jan 04 '18 at 15:10 — khelwood, Jan 04 '18 at 15:10
What do you think they are counting? – khelwood Jan 04 '18 at 15:12 — khelwood, Jan 04 '18 at 15:12

luk2302 · Accepted Answer · 2018-01-04T15:13:42.893

2

I do not know why you define i in the for loop but never use it. Your main problem is that when you encounter an space you do not increment j. I think you have two options:

increment j after you call println(); inside the if
drop j completely and simply use i wherever j previously was used (probably the better idea)

General recommendation for your code: You are performing too much weird Character and String logic. You could do

drop the alphabet
get the char from the String the same way you currently do
subtract 'A' from it
use the resulting char as index to access the morse array
drop k and the entire while loop.

edited Jan 04 '18 at 15:13

answered Jan 04 '18 at 15:09

luk2302

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but i increment j after the the if, which means no matter what it will be increased – GravityWell Jan 04 '18 at 15:12
@Gus no, you increment it only inside the `else`. – luk2302 Jan 04 '18 at 15:13
@Gus No, you increment `j` inside your `else` block. It's just misleadingly indented. – khelwood Jan 04 '18 at 15:13
i just realised j was inside the else and just badly indented. That did it. – GravityWell Jan 04 '18 at 15:16
@luk2302 I didnt quite get what you said – GravityWell Jan 04 '18 at 15:16
@Gus you did not get *what*? – luk2302 Jan 04 '18 at 15:19
@luk2302 Why substract 'A' and how the rest will work. Im new to it. – GravityWell Jan 04 '18 at 15:20
@Gus literally subtract `'A'`: `int index = word.charAt(i) - 'A';`. I could show you how the code looks but I feel it is better if you write the code yourself, the entire for loop (including braces) will only be 8 lines of code. – luk2302 Jan 04 '18 at 15:21
@luk2302 yea i dont get why dropping the A would help me. – GravityWell Jan 04 '18 at 15:24
@Gus because the result is the correct index, the same thing you previously called `k`, just a lot quicker and shorter, without `while`, without a lot of comparing, etc. – luk2302 Jan 04 '18 at 15:25
@luk2302 yea but what if the string contained an 'A' that needed to be transcribed ? – GravityWell Jan 04 '18 at 15:28
@Gus: `word.charAt(i) - 'A'` doesnt remove the A from anything. It performs a calculation where the characters are treated as int values and by subtracting the value of 'A' you make sure that you get 0 for 'A', 1 for 'B' etc pp. – OH GOD SPIDERS Jan 04 '18 at 15:30
@Gus `'A' - 'A'` is 0, first element of the array gets accessed, all good :) `'B' - 'A'` is 1, etc. – luk2302 Jan 04 '18 at 15:31

Why is everything after the space not printed?

1 Answers1