Occurences of a string in another string.

Okay, so i've written the function and it seems to work. The outline was to write a function with this signature: char* findx(char*s, char*x), that finds multiple occurences of string s in x.

My code is:

Code:

char* findx(char* s, char* x)
{
	
	int lens = strlen(s);
	int lenx = strlen(x);
	int count = 0; 
	int found = 0;

	for(int i = 0; i < lenx; ++i) {
		for(int j = count; j < lens; ++j) {
			if(s[j] == x[i]) {
				++ count;
				break;
			}
			count = 0;
		}
		if(count == lens){
			++found;
			count = 0;
		}
	}

		
cout << found;

	return s;
}

As I said, this seems to work fine. Can anybody see anything wrong with it?

Thanks.

How does a single char* return result result in multiple values?

> Can anybody see anything wrong with it?
How many tests have you done?
Can you show us?

That's where I cheated slightly. I wasn't sure from the exercise description what should be returned, so I just call the function as such:

findx("dog", "jklkjfdogjkjkldogjkjkldog");

That then prints out that dog has been found three times.

Well, do you need to count overlapping occurrences? For example, should findx("xx", "xxxx") print 2 or 3?

Originally Posted by laserlight

Well, do you need to count overlapping occurrences? For example, should findx("xx", "xxxx") print 2 or 3?

Ha.....Well that looks a little bit more difficult.

Remember.....babysteps!!

Out of curiosity, how would I go about implementing something like that in Pseudocode, so that I can have a go proper.

Thanks

How would you do it yourself if this were a task in real life?

Originally Posted by Elysia

How would you do it yourself if this were a task in real life?

I would probably ask on here....

Seriously, if i need to find stringa in stringb, would I simply start from stringb[0] and see if string a begins there and check along stringb for strlen(stringa), and then increment stringb index each time and check again from there?

It sounds like you have a strategy for doing it already. Can't say if it's meant to check for overlapping or not.

Originally Posted by Elysia

It sounds like you have a strategy for doing it already. Can't say if it's meant to check for overlapping or not.

I think it will work. Will try it this evening and post the solution.

Originally Posted by darren78

I think it will work. Will try it this evening and post the solution.

Isn't it great when you figure something out!!!

It was a very simple change in the end:

Code:

char* findx(char* s, char* x)
{
	
	int lens = strlen(s);
	int lenx = strlen(x);
	int count = 0; 
	int found = 0;

	for(int i = 0; i < lenx; ++i) {
		for(int j = count; j < lens; ++j) {
			if(s[j] == x[i]) {
				++ count;
				break;
			}
			count = 0;
		}
		if(count == lens){
			++found;
			count = 0;
			i = i-(lens-1); // only part I changed!
		}
	}

		
cout << found;

	return s;
}

It seems to work fine with the test data I used. I inputted various lengths of 'x' strings and it was correct on each occasion!

Anybody any suggestions to make it more efficient or is this fine as it is?

Thanks.

Hmmm? Are you sure that's right? You are subtracting the length of the string s - 1 to i. That means i will < 0.

Originally Posted by Elysia

Hmmm? Are you sure that's right? You are subtracting the length of the string s - 1 to i. That means i will < 0.

Damn. You're right. I am getting the expected output though. Let me have another play around with it!

Code:

char* findx(char* s, char* x)
{
	
	int lens = strlen(s);
	int lenx = strlen(x);
	int count = 0; 
	int found = 0;

	for(int i = 0; i < lenx; ++i) {
		for(int j = count; j < lens; ++j) {
			if(s[j] == x[i]) {
				++ count;
				break;
			}
			count = 0;
		}
		if(count == lens){
			++found;
			count = 0;
			i = i-lens+1; // only part I changed!
		}
	}

		
cout << found;

	return s;
}

Okay, changed slightly. Maintains that i is positive, and it works okay I think. No idea why the previous one still worked though?

What is i = i-lens+1 supposed to do, though? Subtracting the length of a string from your current position in the other string makes no sense.

Originally Posted by Elysia

What is i = i-lens+1 supposed to do, though? Subtracting the length of a string from your current position in the other string makes no sense.

Say for example i want to find how many occurences of 'xxx' there are in 'xxxxx', including overlapping.

So on the first run through:

01234
xxxxx

When found, ie when count == lens..i== 2. I then want it to search again from when [i] was one ahead from the beginning of the previous found 'xxx'. So on this occasion [i] = 2. Using my code i = i-lens+1, i will equal 0, however because the for loop will increment [i] also, the search will start again from [i] = 1.

Hope this makes sense, I don't seem to be very good at explaining myself

Edited: first line should say i==2, not x==2 as previously typed.