Concatenate all of the strings in an array

[Kalona Ark]

I am trying to write a method that will concate all the strings in this array called String[] s.

public static String concateLines(String[] s, String separator) {
       String result = "";
    if (s.length > 0) {
        result = s[0];    // start with the first element
        for (int i=1; i<s.length; i++) {
            result = result + separator + s[i];
        }
    }
    return result;

[Olivier Legat]

Your program seems all right to me. What kind of errors are you getting? You could also try to use a while loop. Like this for example:

int i;
while(s[i] != null)
{
    result = result + separator + s[i];
    i++;
}

But this will only work if all the elements of the array contain something

[Kalona Ark]

When I tried to get the output I get this error: <identifier> expected

I used this to get the output.

System.out.println(result);

[Jeanne Boyarsky]

Kalona,
Everything you've posted so far is fine. Can you post the rest of the program?

[Kalona Ark]

I am having a lot of trouble calling the methods for countCharacter() method (which is suppose to count all characters)and concateLines() method (which is to put the array of string on one line). I have searched the internet and continue to search, but I can not understand I have to do in order to get it to work.

import java.io.*;
import java.util.*;
 
public class Linetasks{
 
 
 
public static boolean isEndOfSentence(char c) {
if ((c == '.') || (c == '?') || (c == '!')) {
return true;
}
else {
return false;
}
}
 
public static long countCharacters(String line) {
long numCharacters = 0;
int index = 0;
while (index < line.length()) {
char c = line.charAt(index++);
numCharacters++;
}
return numCharacters;
}
public static String concateLines(String[] s, String separator) {
       String result = "";
    if (s.length > 0) {
        result = s[0];   
   for (int i=1; i<s.length; i++) {
            result = result + separator + s[i];
        }
    }
    return result;
}
public static void main(String args[]){
 
String[] s = {
"This line is first.",
"And this one come next.",
"Lastly, we have this one!",
"Are there any more lines?"};
 
// this is for array 
 
for (int i=0; i<s.length; i++ )
{
char c[] = s[i].substring(s[i].length()-1).toCharArray(); 
 
if (isEndOfSentence(c[0])) {
System.out.println("The end of sentence is found.");
}
else {
System.out.println("The end of the sentence is not found.");
}
}
[B]String m = s[0].length.copyValueOf(s);[/B]
}
}

[ November 24, 2007: Message edited by: Kalona Ark ]

[Kalona Ark]

Thanks to everyone that helped me. I finally got it.

[Panseer Kaur]

Originally posted by Olivier Legat:
Your program seems all right to me. What kind of errors are you getting? You could also try to use a while loop. Like this for example:

int i;
while(s[i] != null)
{
    result = result + separator + s[i];
    i++;
}

But this will only work if all the elements of the array contain something

That code didn't work at all for me.


Exception in thread "main" java.lang.Error: Unresolved compilation problem:
The operator != is undefined for the argument type(s) int, null

at Child.main(cTest.java:6)


I didn't think that method would work just because you might exceed array bounds but this problem stopped that from happening. Should that work?

[Panseer Kaur]

Oppps I tried comparing nulls with int's. That was my mistake sorry.

[Olivier Legat]

The whole point of the "while(s[i] != null)" is to stop whenever there's nothing left in the array. That's why it doesn't work unless all the elements are full, it will stop when ever an element is empty.

E.g.

s[0]= "hi"
s[1]= "dude"
s[2]= null **in other words: empty**
s[3]= "bye"

It will concatenate "hi" and "dude" but will stop at s[2]

[Stan James]

Have you tried that loop with no nulls in the array? I would expect something undesirable to happen.

[Nicholas Jordan]

[Olivier Legat:]   The whole point of the "while(s[i] != null)" is to stop whenever there's nothing left in the array

I did not look to see what Stan is looking at, but I can figure out from his post what he is looking at. That way of looping through an array that you show was used before Java came along and now all arrays carry length information so you can do int end = s.length; I would tell you some more but you can take this and what Stan said and build another way of doing it. What do you think Stan is looking at in your code that would cause him to post ?

Also, I do not think the original poster really has it working yet. String result = ""; sets result and Strings cannot be later set with result = result + separator + s[i]; as Kalona Ark shows. StringBuffer or StringBuilder, along with several other ways of doing it will work. Along with considering what Stan is saying and figuring out an approach from that way.

[Ernest Friedman-Hill]

Originally posted by Nicholas Jordan:
String result = ""; sets result and Strings cannot be later set with result = result + separator + s[i]; as Kalona Ark shows.

Of course it can. String objects are immutable, but references to String objects are not and can be reassigned at will.

[Jesper de Jong]

A note about efficiency...

The "+" operator for Strings in Java is implemented behind the scenes by using StringBuilder (or StringBuffer in Java versions older than 5.0). The compiler converts the use of "+" with strings behind the scenes like this:

// This...
String s = "a" + "b";
 
// ...Is converted to something like this:
StringBuilder sb = new StringBuilder("a");
sb.append(b);
String s = sb.toString();

If you use the "+" operator on Strings in a loop, then what you are actually doing is creating a StringBuilder object at each iteration in the loop, that StringBuilder is initialised with the content of the string, then the other string is concatenated, and a new String object is created with the characters in the StringBuilder. So your program gets converted to something like this:

public static String concateLines(String[] s, String separator) {
    String result = "";
    if (s.length > 0) {
        result = s[0];    // start with the first element
        for (int i = 1; i < s.length; i++) {
            StringBuilder sb = new StringBuilder(result);
            sb.append(separator);
            sb.append(s[i]);
            result = sb.toString();
        }
    }
    return result;
}

That's not very efficient - in each iteration, a new StringBuilder object has to be created, data from 'result' is copied into the StringBuilder, then 'separator' and 's[i]' are concatenated, and a new String object is created in which the contents of the StringBuilder are copied. So, while it's not apparent from your source code, Java is doing a lot of unnecessary creating of objects and copying of data.

Your program will be much more efficient when you manually use StringBuilder yourself:

public static String concateLines(String[] s, String separator) {
    if (s.length > 0) {
        StringBuilder sb = new StringBuilder(s[0]);
        for (int i = 1; i < s.length; i++) {
            sb.append(separator);
            sb.append(s[i]);
        }
    }
    return sb.toString();
}

Using the "+" operator on Strings in a loop is a common efficiency trap in Java programs.