Thread: Counting occurances of two letters in a particular order?

Create a bool such as: lastCharWas and set it to false initially. Then when you find the first character you are looking for set it to true. Then when looking for the next character, if 'lastCharWas' is true, and the next character is whatever, increment the count. Then set 'lastCharWas' to false.

Originally Posted by Brokenhope

Create a bool such as: lastCharWas and set it to false initially. Then when you find the first character you are looking for set it to true. Then when looking for the next character, if 'lastCharWas' is true, and the next character is whatever, increment the count. Then set 'lastCharWas' to false.

Thanks for the tip. I couldnt get the bool function to work so i used int set to 0 or 1. Here is my code now but its just counting the "s". Am I barking up the right tree?

Off to find dinner. I'll check back after my belly quits complaining!

Code:

#include <stdio.h>


int main(void)
{
	char ch;
	int is_count = 0;
	int lastchar = 0;

	printf("Key in your word or phrase:  \n");
	
	while ((ch = getchar()) !='#')
	
	{
		if (ch == 'i')
			lastchar = 1;
		
		if (ch == 's' && lastchar == 1)
			is_count++;
			
	}
	lastchar = 0;
	printf("\"is\" count %d\n", is_count);

	getchar();
	getchar();
	return 0;
}

Originally Posted by Adak

Every time you find an 's', you need to reset a variable.

like this:

Code:

			is_s_count = FALSE;
		}

		is_s_count = FALSE;

my god, I never thought this thread would last this long. Probably never broke out
the pen & paper like I originally suggested.

And I still don't see that 3rd if() to test if both are "i" and "s" were set. I have the
working solution for this, but I require some thought and effort by the OP.

Adak, can I IM you my code? I tested it and it seems to work fine. I'm not sure if it meets
the "state machine" implementation, that may/may not be required.