initialize string dynamically in java

Question

I want to initialize a string as follows:

public int function (  int count )  { 
String s_new = "88888...  'count' number of 8's "    <-- How to do this
//Other code 
}

Currently I'm not sure how to do this, so I've declared an int array ( int[] s_new ) instead and Im using for loops to initialize this int array.

EDIT: I meant that I need to initialize a string containing only 8's ... the number of times the digit 8 occurs is 'count' times.

Is the `88888...` part of the string supposed to represent the value of `count`? ie the user passes in 5, you put 5 eights there? — Durandal, Jan 18 '14 at 20:28

score 4 · Answer 1 · answered Jan 18 '14 at 20:30

4

You can use Guava's Strings.repeat() method:

String str = Strings.repeat("8", count);

answered Jan 18 '14 at 20:30

Rohit Jain

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score 3 · Answer 2 · edited May 23 '17 at 11:56

3

Try:

String s_new = "";
for (int i = 0; i < count; i++) {
    s_new += "8";
}
return s_new;

Now, this is a naive solution. A better solution (as is posted in other answers here) will use a StringBuffer or StringBuilder to accomplish this in a more efficient manner.

Also, further reading on the difference between those two options: Difference between StringBuilder and StringBuffer

edited May 23 '17 at 11:56

Community

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answered Jan 18 '14 at 20:28

BlackVegetable

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It compiles fine but is creating a new string object each iteration – zapl Jan 18 '14 at 20:32
@zapl Clarified that this is a naive solution, and not optimal. – BlackVegetable Jan 18 '14 at 20:33

score 3 · Accepted Answer · edited May 23 '17 at 12:29

3

In these cases, it is recommended to use a StringBuilder:

StringBuilder sb = new StringBuilder();
String s = "";
int count = 8;

for (int i = 0; i < count; i++) {
    sb.append('8');
}

s = sb.toString();

System.out.println(s);

Output:

88888888

edited May 23 '17 at 12:29

Community

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answered Jan 18 '14 at 20:30

Christian Tapia

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This is a more efficient way to accomplish this than mine (memory/time). +1 – BlackVegetable Jan 18 '14 at 20:31
Java has a pretty smart compiler, all string `+` operations are actually converted to `StringBuilder.append's` so I doubt it, but when dealing with 8 characters output you need to be sure to optimize everything – Philipp Gayret Jan 18 '14 at 20:32
@Philipp I was unaware of how the Java compiler handled such. But, yeah. Those 8 character outputs could take a mighty long time. ;) – BlackVegetable Jan 18 '14 at 20:34
1

@Philipp `+` is internally implemented via `StringBuilder` but the compiler will do that in a loop here. `new StringBuilder(s).append('8').toString()` each time. You have move the builder out of the loop manually – zapl Jan 18 '14 at 20:36
@Philipp not in loops I believe and this isn't part of the Java spec so cannot be relied upon in non-Oracle compilers. – Boris the Spider Jan 18 '14 at 20:36

score 2 · Answer 4 · answered Jan 18 '14 at 20:30

2

You can build strings using the StringBuilder class.

StringBuilder sb = new StringBuilder();
for (int i = 0; i < count; i++)
    sb.append('8')
String s_new = sb.toString();

s_new would then have as much 8 as you have count.

answered Jan 18 '14 at 20:30

zapl

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score 1 · Answer 5 · answered Jan 18 '14 at 21:02

1

Solution on pure Java using arrays:

public String repeat(char ch, int count) {
    char[] chars = new char[count];
    Arrays.fill(chars, ch);
    return new String(chars);
}

answered Jan 18 '14 at 21:02

user and

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initialize string dynamically in java

5 Answers5