string operations: looking for a specified letter

[hirak1984]

hey thnx buddy that is really tricky.

:D

Nope, sorry; there's an overloaded version available. Here's the spoiler:

int count= 0;
for (int i= 0; (i= indexOf(yourString, i)) != -1); i++, count++);
return count;

kind regards,

Jos

[ria3]

Thank you so much for your help Hirak, BSCode, and JosAH!
I love the tricks on this forum, lol :)

[nmadct]

very clever. I hadn't thought of that. Maybe I need to get more creative with my loops...:)

I would argue that you should NOT think of that... "creative" code is harder to read!

int count = 0;
int i = 0;
while (true) {
   i = yourString.indexOf(i);
   if (i == -1)
      break;
   i++;
   count++;
}
return count;

This does exactly the same thing but at a glance it's more obvious what it's doing.

[nmadct]

Not that I'm trying to spoil your cleverness, hehe!

Also, I meant to write: i = yourString.inde xOf(charToCount , i);

So it works instead of looping infinitely.

[hirak1984]

ohk... ohk....
cool down..

no more arguments,plz. :D

I would argue that you should NOT think of that... "creative" code is harder to read!

int count = 0;
int i = 0;
while (true) {
   i = yourString.indexOf(i);
   if (i == -1)
      break;
   i++;
   count++;
}
return count;

This does exactly the same thing but at a glance it's more obvious what it's doing.

[sandyw]

What did I start the battle of "Counters"! !!
You got to love it,

sandy

[JosAH]

I would argue that you should NOT think of that... "creative" code is harder to read!

1: int count = 0;
2: int i = 0;
3: while (true) {
4:    i = yourString.indexOf(yourChar, i);
5:    if (i == -1)
6:       break;
7:    i++;
8:    count++;
9: }

This does exactly the same thing but at a glance it's more obvious what it's doing.

That's all in the eye of the beholder (I numbered your code lines and fixed the
little typo). The way I think goes like this:

1: and 2: ok initialization stuff
3: ok, an infinite loop
4: huh? ah, got it.
5: 6: double huh? So it isn't in infinite loop after all; so he wanted to say:

while ((i= yourString.indexOf(yourChar, i)) != -1) ...

7: 8: See? that's the increment part of the loop, so the following would've done
the job:

for( ; (i= yourString.indexOf(yourChar, i)) != -1; i++, count++);

(staring at the top of the page again: aha! that 'i' variable just has a loop scope
so this is what he meant:

for(int i= 0 ; (i= yourString.indexOf(yourChar, i)) != -1; i++, count++);

kind regards,

Jos ;-)

[prometheuzz]

Ha, you people with your multy line solutions!

/**
 * @param s     the string to look for
 * @param t     the text to look through for 's'
 * @return      the number of times 's' is in 't'
 */
public static int count(String s, String t) {
  return (t.length()-t.replaceAll(s, "").length())/s.length();
}

; )

[JosAH]

Ha, you people with your multy line solutions!

/**
 * @param s     the string to look for
 * @param t     the text to look through for 's'
 * @return      the number of times 's' is in 't'
 */
public static int count(String s, String t) {
  return (t.length()-t.replaceAll(s, "").length())/s.length();
}

; )

I did expect something to be casted to double though given a gory solution
like that; and then some ...

kind regards,

Jos ;-)