Something that might help you.

[avansc]

I was going to post this as a challenge, but i thought its probably be a little easy, and maybe to boring to do.
But none the less its a pretty neat function. I'm sure C/C++ has similar functions, but i like writing my own.
if any have ideas on how to make it better id like to hear them.


oh yeah, the function returns a int telling you the position of a string n in a string h, if its not a substring it return -1

Code: Select all

// findString returs the possition of a substring n "needle" if it exsists is h "haystack", returns -1 if non found.

int findSTR(char *n, char *h)
{
	int p1 = 0;
	int p2 = 0;
	
	while(h[p1] != NULL)
	{
		p2 = 0;
		while(n[p2] == h[p1+p2])
		{
			p2++;
			if(n[p2] == NULL)
				return p1;

			if(h[p1+p2] == NULL)
				return  -1;
		}
		p1++;
	}
	return  -1;
}

[/b]

[MarauderIIC]

For the curious, the standard fns are:
C++:

Code: Select all

#include <string>
using namespace std;
...
string a;
//you can use an unsigned int for string::size_type, usually.
string::size_type location = a.find("findme", 0);
if (location == string::npos)
    cout << "nomatch" << endl;
else 
    cout << "match @ " << location << endl;

http://www.cppreference.com/wiki/string/find

If you really want to do it in C, though, use the standard

Code: Select all

#include <string.h>
//...
    char* str1 = "this is a string of characters";
    char* str2 = "a string";
    char* result = strstr( str1, str2 );
    if( result == NULL ) printf( "Could not find '%s' in '%s'\n", str2, str1 );
    else printf( "Found a substring: '%s'\n", result );

When run, the above code displays this output:
Found a substring: 'a string of characters'

http://www.cppreference.com/wiki/c/string/strstr /// http://www.cplusplus.com/reference/clib ... trstr.html
I would guess that to get the character number it was found at -- if you really want to -- you could probably do (int)(result - str1)

[avansc]

yeah, the reason i put this down was because it would help on my next challange, but i think it was a bit optimistic.
i wanted to make a challenge where you could do a search like this but use regular expressions.

int find(char *n, char *h) then having you n be something like '[0-9][*][0-9]' so return if a string started with a number and ended with a n umber, but i dont thing that will be the challenge. its a little boring, however, if you wanna have a leg up on someone in the industry teach yourself how to use regular expressions in your programs.

[Amarant]

Here's an adaption of your code with basically the same functionality, except that it doesn't include the 2nd if statement and returns a char * instead of an index.

Code: Select all

char * findSTR(char * needle, char * haystack)
{
   for(; *haystack; haystack++)
   {
      int i = 0;
      while(needle[i] == haystack[i])
      {
         if(needle[++i] == '\0')
            return haystack;
      }
   }
   return 0;
}

[avansc]

Here's an adaption of your code with basically the same functionality, except that it doesn't include the 2nd if statement and returns a char * instead of an index.

Code: Select all

char * findSTR(char * needle, char * haystack)
{
   for(; *haystack; haystack++)
   {
      int i = 0;
      while(needle[i] == haystack[i])
      {
         if(needle[++i] == '\0')
            return haystack;
      }
   }
   return 0;
}

its nice, except that the functionality of the function was to find where in the string it stats if it is there.

[Amarant]

True, and that's why I said it had 'basically' the same functionality and not 'exactly' the same functionality.

[avansc]

True, and that's why I said it had 'basically' the same functionality and not 'exactly' the same functionality.

yeah. but i do like your code. its elegant.
hey, have you looked at the new challenge, its number 4. its a very nice challenge. its hard. but ive you oppertunity to wrote beautiful code.
im almost finished.

[Arce]

This probably would have been useful to know about 2 days ago when i had to write a program to solve random complex infix, postfix, and prefix math problems...XD

Meh, you learn new shit every day.

Edit: not imaginary numbers. XD multi-operational.