simple question

[Peter Phung]

Can anyone tell me how to count the number of occurences of a character in a string?

[Dave Vick]

Peter
This sounds like a homework problem - while most of the people here love to help people out we generally don't like to just do their homework for them.
Post what you have got so far and let us know where you are stuck and we'll help you out.

[Dirk Schreckmann]

Hello,
A few different methods come to mind - I'm not sure which would be most efficient.
I'd bet you can figure one out by looking at the API Specification for String.
Good Luck,
-Dirk Schreckmann

[Peter Phung]

so far i've got:
char c = '@';
int index = 0;
int count = 0;

while ((index = params.indexOf(c, index)) != -1) {
count++;
}
System.out.println(count);
but i'm getting stuck in the loop and don't know what to do with it.

[Dirk Schreckmann]

Hello,
You have an infinite loop. Your index never changes so (index = params.indexOf(c, index)) always produces the position of the desired character which is never -1 (unless the character isn't in the String, which would produce a -1) - and so the infinite loop.
Also, your code doesn't do what your thread topic is asking about. Your code doesn't count anything. It can be modified to find the position of your desired character by incrementing index and not count - but I'm not sure that is what you are trying to do.
My powers of guessing are dwindling - it's too late... must sleep... I probably don't have the wherewithal to spell my name correctly...
Good Luck,
-Fred Flinstone
...or make up a good joke.

[Peter Phung]

Can anyone suggest a way of counting characters then?

[Kaspar Dahlqvist]

This is one way!

class Test4 {
static final synchronized public void main(String[] a) {
String s = a[0]; 
int i = 0;
Integer c = new Integer("0");
for (s += "Nisse oppnade dorren och steg in.ee"; i < s.length() :wink: {;
if (s.charAt(i) == new Character('e').charValue())
c = new Integer(c.intValue() + 1); i += 2;
{i--;}
if (s.length() == i)
System.out.println(c.intValue());
}
return;
}
}

//Kaspar