Just wondering...

[destin]

Is there a way to count the number of times... lets say the letter 'f' appears in a string? If someone knows can you just tell me the function or whatever and explain how to use it? thanks.. (by the way.. i'd want this in C.. i dont kno C++ )

[redhead]

Actualy it is a very simple implementation.

#include <stdio.h>

int count_char(char* str, char needle);

int main()
{
  char line[] = "this is a test in the number of a char is pressent";
  printf("Count of %c in \"%s\" is found to be: %d\n", 'e', line, count_char(line, 'e'));
  return 0;
}

int count_char(char* str, char needle)
{
  int count = 0;
  while(*str)
  {
    if(*str == needle)
      count++;
    str++;
  }
  return count;
}

Altho this requires the string to be terminated with '\0' or NULL, if your string might not be NULL terminated I havn't got an easy one right now.. perhaps some alteration with using strlen() to determan the length befor the runthrough of it all.. But this one will run just as quick as any system command..

Sorry I'm too lazy to check if there even exist one in string.h or any standard library function.

[destin]

Oh, thanks. I made my own little program out of this..:

#include <stdio.h>
#include <conio.h>

int count_char(char* str, char needle);

int main()
{
  char *line, letter;
  
  printf("Count what letter? ");
  letter=getche();
  printf("\r\nIn what string? ");
  gets(line);
  printf("You typed '%c' %d times.", letter, count_char(line, letter));
  return 0;
}

int count_char(char* str, char needle)
{
  int count = 0;
  while(*str)
  {
    if(*str == needle)
      count++;
    str++;
  }
  return count;
}

But i dont understand the count_char() function .. would you mind explaining it?

[Joel]

Cool.. I also did this one in C:

#include <stdio.h>

int CountChrs(char* s, char c)
{
    // for move through the 's' array
    int pos = 0;
    // for counting the times of 'c'
    int index = 0;
    // Let the loop begin
    while(s[pos] != '\0')
    {
         // Find if the current char is the same as 'c'
         if (s[pos] == c) index++; // if so... post-increment out var
         pos++;  // if not or it is either... just move one char of the 's'     
    }
    s[pos] = 0;
    // return the times of 'c'
	return index;
}

int main(void)
{
    char c = 'a';
    char str[] = "Calafo";
    printf("Total of %c in %s are: %i\n", c, str, CountChrs(str,c));
	system("pause"); // Not portable
	return 0;
}

[redhead]

sure...

int count_char(char* str, char needle)

Since we're counting chars we ned to be able to call the function with a string (char array) and the char to look for

int count = 0;

We need to keep track of how many we've found during the count, thus the integer here, if tehres none, it's allready set at 0, since this is strict ANSI C, we need to declare the variabel befor anything else is done in teh function.

while(*str)

For as long as the string has something besides NULL in it, we should do stuff.

if(*str == needle)
  count++;

if our current location in the string equals the char we're looking for, increment the counter to reflect our current count.

str++;

Shift our current location in teh string one storage size to the right, or left, or up, or down, depending on how you like to see it

return count;

when we exit from teh while loop, we return with our pressent count.

When dealing with strings or char arrays, ther memory usage is actualy consisting of one part beeing the value stored at the possition in teh string aswell as the memory location teh string is occupying.
The *str == needle compares the value stored at current memory location with needle.
The str++ increments the memory value, thus moving us one place further into the memory occupied by teh string, thus possitioning us at the next char in teh string.

Sorry, thats just about the easiest way I can explain it, without draving it up in hand..
Anyway here it is drawn:

Memory representation of:

char str[] = "string"
           .---.---.---.---.---.---.----.
Index:     | 0 | 1 | 3 | 4 | 5 | 6 | 7  |
           |---|---|---|---|---|---|----|
Char:      |'s'|'t'|'r'|'i'|'n'|'g'|'\0'|
           '---'---'---'---'---'---'----'
mem point:   ^
             '

After str++;

           .---.---.---.---.---.---.----.
Index:     | 0 | 1 | 3 | 4 | 5 | 6 | 7  |
           |---|---|---|---|---|---|----|
Char:      |'s'|'t'|'r'|'i'|'n'|'g'|'\0'|
           '---'---'---'---'---'---'----'
mem point:       ^
                 '

So when you use str[i] you're refering to the index line in the memory, when you address *str you're addressing the char line in the memory, when you use str++, instead of changing the index and looking at the char section from that line, you change the mem point, and see it from that side.

Now the limitation of my ascii skills, prevents me from adding the pointer to next item handling in case 's' is located at memory location 0xff2341de and 't' is located at memory location 0xfff2132d but any compiler logic book, like the dragon book, will tell you all about how this is handlet.. Actualy the thing teh str++ changes is infact the mem pointer shift next item pointer but as I said any compiler designs book will elaborate this to the fullest.

[destin]

Ohh.. now i get it.. thanks redhead.

[redhead]

A few things on your program:

#include <conio.h>

this is not needed if you think strict ANSI C.

char *line
...
  gets(line);

This isn't apropriate the line dosn't have any allocated memory to it, and gets() dosnt check for buffer overrun, thus a potential violation and segmentation fault just vaiting to happen.

#define LENGTH 256
char line[LENGTH+1];
fgets(line, LENGTH, stdin);

is a better and more secure way.

letter = getche();

isn't any good, it is not cross OS comliant, and is what needs the #include <conio.h> statement.

letter = (char)fgetc(stdin);
fflush(stdin);

is an ANSI C compliant way, and thus cross OS compliant.

[destin]

fgetc?
fflush?
fgets?..lol im not familiar with any of those

[redhead]

Originally Posted by destin

fgetc?
fflush?
fgets?..lol im not familiar with any of those

Look at what's in stdio.h

[destin]

also.. i came across a problem after replacing my code with yours.. it wouldnt let me input a string after i typed the letter.. replacing

fgets(line, LENGTH, stdout);

with

fgets(line, LENGTH, stdin);

fixed it

[redhead]

yeah I noticed about 1 minut after I posted it, and fixed it.. Just wasn't aware you'd allready snatched my false placed correction.

[destin]

lol yeah.. cuz i just checked and it was fixed.. i thought i had copied wrong. hehe

[destin]

Now what if i wanted it to count the uppercase/lowercase of that letter too.. for example..

Count what letter? a
In what string? A dog ate.
You typed 'a' 2 times.

rather than it counting it once for only the a in ate.

[redhead]

it is case sensitive, try it and you'll see.. It is left as an excersise for the viewer to make it case insensitive.