how to indicate white space in java

Question

public class solution {
    public static void main(String[] args) {
        String str = "The horse jumped over the sheep";
        String str1 = "";
        for(int i = 0; i!='\0'; i++); {
            if(str.charAt(i) != ' ')
                str1=str1+str.charAt(i);
        }
        System.out.println(str1);
    }
}

The program shows error at variable i. Is there any other means to print without white spaces without any in-built function?

"The program shows error at variable i" -> what does that mean ? which line ? which error ? — vincrichaud, Jul 04 '18 at 14:51
You have a number of bugs in your code. try `for (int i = 0; i < str.length(); i++)` note there is no `;` at the end. — Peter Lawrey, Jul 04 '18 at 14:51

Shanu Gupta · Answer 1 · 2018-07-04T15:32:30.967

2

You have a ; after for loop here:

for (int i = 0; i != '\0'; i++)
    ;

Remove it and variable i would then be recognized.

Also your comparision of i with '\0' is wrong. I think you want to compare it till the end of the string str.

Which should be done like:

String str = "The horse jumped over the sheep";
String str1 = "";
for (int i = 0; i < str.length(); i++) {
    if (str.charAt(i) != ' ')
        str1 = str1 + str.charAt(i);
}
System.out.println(str1);

Outputs:

Thehorsejumpedoverthesheep

PS: I suggest you to read this post : String, StringBuffer, and StringBuilder as well.

edited Jul 04 '18 at 15:32

answered Jul 04 '18 at 14:52

Shanu Gupta

3,699
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and `'\0' == 0` – Peter Lawrey Jul 04 '18 at 14:52
1

Can u add chars to strings like that? I think you need Character.toString() – orwinmc Jul 04 '18 at 15:00
bt '\0' also indicates the null which is the end character of all the strings,right? – priya priya Jul 04 '18 at 15:05
2

@priyapriya Not in `Java`. The length of the `String` is determined by its `length` property. In `C`, `strings` need to be `null` terminated since they are just a plain array of chars – StaticBeagle Jul 04 '18 at 15:17
@Jetstream5500 You won't need `Character.toString()`. `char` would be promoted to `String`. – Shanu Gupta Jul 04 '18 at 15:24
@Jetstream5500 you still shouldn't add chars that way, though. A StringBuilder must be used so that it smooths the cost of resizing the resulting String. – kumesana Jul 04 '18 at 15:30
@ShanuGupta I tried it just now and realized I was wrong. I have updated my code to reflect that. – orwinmc Jul 04 '18 at 15:43

orwinmc · Answer 2 · 2018-07-04T15:40:03.847

(1) You need to remove the ; after the for loop to get rid of the error.

(2) I changed your for loop to a for-each loop as there was an error in your comparison (and it simplifies the code)

(3) I modified the code to include StringBuilder which will be more efficient when str becomes extremely long

String str = "a b c d";
StringBuilder str2 = new StringBuilder();
for (char c : str.toCharArray()) {
    if (c != ' ') {
        str2.append(c);
    }
}
System.out.println(str2);

how to indicate white space in java

2 Answers2