Character String Number

[rdove]

VB

This function determines how many occurances of a character are in a string.

Public Function checkString()

   Dim strChar As String
   Dim Char As String
  
   strChar = "Johnny Boy"
   Char = Split(strChar, "y")

  return UBound(Char)
  'returns 2

End Function

[joe_bruin]

now waitaminute,
is this intended as a demostration of the language, or of the api?
your function just calls another function. it doesn't really demonstrate that you are capable of finding the number of times a character appears in a string.
how about we see a reimplementation of instr() in vb instead of just calling it?

for my part, here it is in c:

int char_count(char *s, char c)
{
  int num = 0;

  for( ; *s != '\0'; s++)
  {
    if(*s == c) num++;
  }
  return num;
}

[rdove]

[quote]Originally posted by joe_bruin
now waitaminute,
is this intended as a demostration of the language, or of the api?
your function just calls another function. it doesn't really demonstrate that you are capable of finding the number of times a character appears in a string.
how about we see a reimplementation of instr() in vb instead of just calling it?


Yes it does.....the instr function is a keyword function in vb. Just like the dateadd one I posted earlier. You can use the instr function outside of another function, just take off the function:

dim strChar as String = "Johnny Boy"
dim intNum as Integer

intNum = instr(strChar, "y")

'intNum = 2

This is why I like vb so much because I don't need to write 4 or 5 lines of code to do things like this. Why re-invent the wheel.....

[Belisarius]

I don't know, I think standard libraries are fair game. After all, there's no sense in reinventing the wheel. I know the J2SE implementation of most objects/methods are going to be a lot more efficent than my implementations.

Anyways, Java:

public int getLength(){
  StringTokenizer toks = new StringTokenizer("Johnny Boy", "y");
  return toks.countTokens() - 1;
}

Changed it as I didn't notice that the goal was to count "y's" orignially.

[Valmont]

I am not sure what the goal exactly is so this proggy calculates the amount of characters in a string and counts the occurances of a certain character in a string.

 #include <string>
#include <iostream>
#include <algorithm>
using namespace std;

int main(int argc, char* argv[])
{
	string str = "hello johnny";

	int i=str.size(); //i=12
	int p = count(str.begin(), str.end(), 'h'); //p=2
	
	return 0;
}

[npa]

Originally posted by rdove
VB

This function determines if and how many characters are in a string.

No it doesn't it returns the first index in a given string
of the specified character.

[rdove]

Originally posted by npa
No it doesn't it returns the first index in a given string
of the specified character.

you know what npa..you are right!!! I dunno what I was thinking...
This returns the number of characters in a string

dim strChar as String = "Johnny Boy"
dim intNum as Integer

intNum = LEN(strChar)

'intNum = 10

[Valmont]

And how do you count the number of occurances of a character in a string?

[rdove]

Originally posted by Valmont
And how do you count the number of occurances of a character in a string?

Look at the first post......Thats how you do it in VB....

[Valmont]

instr(strChar, "y")

This returns (a long/variant) the position of the first occurance of "y".
Not the amount of "y"'s.

[rdove]

Originally posted by Valmont
instr(strChar, "y")

This returns (a long/variant) the position of the first occurance of "y".
Not the amount of "y"'s.

Look at the first post.........Happy ????

[Valmont]

erm... you edited the first post?

I didn't notice
Sorry mate.

[Epsilon]

Perl makes this so simple it's almost silly.

sub count_y
{
$string = "Johnny Boy";
$count = ($string =~ tr/y/y/);
return $count;
}

This essentially just replaces every 'y' character with another 'y' character, and returns the total number of replacements to $count. Perl is beautiful.

[Thraxian]

Another VB Example:

Public Function LetterCount(Text As String, Letter As String) As Long
   LetterCount = Len(Text) - Len(Replace$(Text, Letter, vbNullString))
End Function

As to the first post, it had errors in my VB environment. Split returns a string array, not a string. Also, Return is not valid in this context; instead, the result should be assigned to the function variable. Unless, this is VB.NET, in which case, the language was not clearly specified.