find "\n" in simple string

[hugh7]

I'm trying to find the number of newline characters in a simple string but keep getting 0 for the variable numline.
What can I do to find the two little newline commands?
Here is my code:
(as you can see I'm a beginner)
//stringstep

#include <iostream>

using namespace std;

int endline=0 ;
int numline=0 ;
int counter=0 ;

// this is my string sample
char harb2[ ]=&quot; Find the end&quot; &quot;\n&quot; &quot;Then you will see&quot; &quot;\n&quot; ;
int lsizharb=strlen(harb2) ;

int main( )
{

for (counter=0;counter<=lsizharb; counter++)
if(harb2[counter]==0) // this seems to get end of string marker
endline++;
if(harb2[counter]=='\n') // this ends up with 0 !
numline++;

cout <<numline << endl;

return 0;

}

 

[CodingNovice]

WRONG>>>

for (counter=0;counter<=lsizharb; counter++)
if(harb2[counter]==0) // this seems to get end of string marker
endline++;
if(harb2[counter]=='\n') // this ends up with 0 !
numline++;

RIGHT>>>>>

for (counter=0;counter<=lsizharb; counter++)
{
if(harb2[counter]==0) // this seems to get end of string marker
endline++;
if(harb2[counter]=='\n') // this ends up with 0 !
numline++;
}

 

[ArkM]

Or try:

#include <cstring>
int CountChars(const char* s, char ch = '\n')
{
  int count;
  for (count = 0; (s=strchr(s,ch)) != 0; ++s, ++count)
    ;
  return count;
}

 

[hugh7]

Hi CD
..Ah the braces! Thanks for the help

Hugh

 

[hugh7]

Hi ArkM,

I'll give your suggestion a try.
Thanks

Hugh

 

[blazero]

Heh? I don't understand your method. Personally, I would have used:
...
char harb2[ ]=&quot; Find the end&quot; &quot;\n&quot; &quot;Then you will see &quot;\n&quot;;
int nCount=0;
int harbLen=sizeof harb2 / sizeof harb2[0];

for(int i=0;i<harbLen && harb2[ i ] != 0;i++)
if(harb2[ i ] == '\n')
nCount++;
...

 

[hugh7]

Hi Blazero,

Do you mean my 'method'?
I thought I would step through the string , character by character and find the line breaks. That's it.
Since I'm only a beginner with C++ this is all I could come up with.
I'll study your example. It certainly looks more compact.
I've been reading about the string library and have lots to think about and practice.
Thanks for your reply

Hugh