How to search a letter in words with POINTER

[Neo Takaredase]

Hi expert..Im newbie and i want to made a program to search a letter using a pointer..
Example there is a word WORLD
then the program will ask what letter u want to looking for..example its R and the program will said R is in index 3 and all R letter in the word is 1..CAN ANYBODY HELP ME

[Cubbi]

Based on the need to use pointers, you're accessing letters stored in an array of characters, so, the program would go somewhat like this:

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    char arr[] = "WORLD";
    cout << "Enter a letter to search for\n";

    char c;
    cin >> c;

    char* beg = arr;
    char* end = arr + sizeof arr;
    char* pos = find(beg, end, c);
    int cnt = count(beg, end, c);

    if(pos == end)
        cout << "Not found\n";
    else
        cout << "The letter " << c << " found " << cnt << " times. "
             << "The first time at position " << 1 + pos - beg << '\n';
}

online demo: http://ideone.com/SmCg3

Note that this is case-sensitive, enter r instead of R and it will say 'not found'
(edit: missed the count requirement, added)

[Neo Takaredase]

Hmm..i don't know that for sure im newbie could you tell me if the word dinamic..??

[vlad from moscow]

@Cubbi

,

you have demonstrated some halved approach. If you used algorithm then it is more natural to use also standard functions sdt::begin and std::end. So either you should write function find yourself or you use all standard functions to make the approach more complete.:)

@Neo Takaredase

In fact you need two functions. One will search a given character in a character array and other will count how many times the character is present in the array.

I will show the first function

size_t find( const char s[], char c )
{
   const char *p = s;
   while ( *p && *p != c ) ++p;

   return ( p - s );
}

In the main you will use the following test to determine whether the character is found.

size_t pos = find( word, 'R' );

if ( pos != std::strlen( word ) ) cout << "The letter " << 'R' << " is found at position " << pos << '\n';

[Neo Takaredase]

i dont even know how to use alogarithm..any help how to use it?

[Neo Takaredase]

@vlad
what is the 2nd function i can't make it @@

[vlad from moscow]

It is almost the same as the first.

size_t count( const char s[], char c )
{
   size_t n = 0;
   for ( const char *p = s; *p; ++p )
   {
      if ( *p == c ) ++n;
   }

   return ( n );
}

[Neo Takaredase]

#include <iostream>
#include <algorithm>
#include <conio.h>
using namespace std;
size_t find( const char s[], char c )
{
   const char *p = s;
   while ( *p && *p != c ) ++p;
   return ( p - s );
}
size_t count( const char s[], char c )
{
   size_t n = 0;
   for ( const char *p = s; *p; ++p )
   {
      if ( *p == c ) ++n;
   }

   return ( n );
}

int main()
{   char word[100];
    cout << "Enter a word : ";
    cin>>word;
    cout << "Enter a letter to search for : ";
    char c;
    cin >> c;
    size_t pos = find ( word , c );
    size_t count = count ( word ,c );
    if ( pos != std::strlen( word ) ) cout << "The letter " << c << " is found at position " << pos << "The number of letter is" << count <<'\n';
    
    getch();
}

is it right?? but it's look like it wrong haha...

[vlad from moscow]

You need no header <algorithm> because you do not use any function from it.
As for the program if you get a correct result then the program is correct. By the way the position of a letter starting from 0.

Also you have to check that the position is not equal to std::strlen( word ) that means (in case of the equality) that the character is not found. So to use function std::strlen you have to include header <cstring>

[Neo Takaredase]

ok thanks vlad...that really help me thanks a lot..