Sorting through delimiting characters

Hi

I am trying to write a C++ program that will determine whether a string from a file; from the set of characters "[]" form a group. If so, I want to print the
text that is within the out most bracket group into another file. Some examples are:

xxxx[xxxx]xxxx = [xxxx]
[xxxx[xxxx]xxxx = [xxxx[xxxx]
[xxxx[xxxx]xxxx] = [xxxx[xxxx]xxxx]
x[x[x[xxx]xx]x]x]xx = [x[x[xxx]xx]x]

At the moment I can do the "xxxx[xxxx]xxxx = [xxxx]" example and the reading to and from files I am ok with. I am lost on the thought process I should
take in order to see if I have multiple brackets and whether they form a valid group.

Any help would be great.

Currently I have:

Code:

#include<iostream>
#include<fstream>
#include<cstdio>
#include<cstring>

using namespace std;

#define OPEN	'['		
#define CLOSE   ']'	

int main()
{
	ifstream inData;
	ofstream outData;				
	char filenameIn[100];
	char filenameOut[100];
	char lineOfText[200];
	char chBuffer[BUFSIZ];
        char *Start;							
        char *End;							
	int  nChar;

	cout << "Enter filename to read DATA from: ";
	cin >> filenameIn;
	cout << endl;
	inData.open(filenameIn);
	cout << "Enter filename to write DATA to: ";
	cin >> filenameOut;
	cout << endl;
	outData.open(filenameOut);
	
	if(!inData)
		cout << "ERROR in opening " << filenameIn << endl;

	else
		if(inData)
		{
			cout << "Opening file: " << filenameIn << endl << endl;
			inData.get(lineOfText, 200);
			cout << lineOfText << endl << endl;
			
			if (((Start = strchr(lineOfText, OPEN)) != '\0') && (((End = strchr(Start, CLOSE))) != '\0'))	
			{
				nChar = End - Start + 1;											
				strncpy(chBuffer, Start, nChar);									
				chBuffer[nChar] = '\0';												
				cout << "The characters found are: " << chBuffer << endl << endl;
				outData << chBuffer << endl;
			}
			else
				if(!(Start && End))
					cout << "None found" << endl << endl; 
		}

	inData.close();
	outData.close();
	cout << endl << endl << endl;
	return 0;
}

That right, and thats is what is confusing me. I can't even think of an algorithm let alone the code for that. Finding one true set of "[]" was pretty simple but finding multiple complete sets is driving me nuts.

So taking the last example:

x[x[x[xxx]xx]x]x]xx = [x[x[xxx]xx]x]

I some how have to do something like:

Find start bracket [
See if there is another start [
See if there is another start [
See if there is another start [

If not another start bracket see if there is an ending bracket ]
Else if no ending bracket found, FAIL, no valid word group

If ending bracket found ], verify matching start bracket
See if there is another ending bracket found ]
If another ending bracket found ] ,verify matching start bracket
See if there is another ending bracket found ]
If another ending bracket found ] ,verify matching start bracket
See if there is another ending bracket found ]
If another ending bracket found ] ,verify matching start bracket

Else if no matching start bracket found
print text in the outmost '[]' set.

I could do it this way but only if I know how many bracket sets I am dealing with. My problem is how would you loop this in a way that you could have 10, 20 or however many bracket sets and then only print the character within the out most valid set of brackets.

I though I could do this but I may just give up on this and try something simpiler.

declare an two int variables to act as counters and initialize them to zero. Use a loop to read each char from string one at a time. if char is [ then increase Lcounter by one and send char to storage. if char is ] increase Rcounter by one and send char to storage. send all other char to storage only if Lcounter > 0 and Rcounter <= Lcounter..

Code:

char data[80] = "x[x[x[xxx]xx]x]x]xx ";
char newData[80];
int i, j;
int Rcounter = 0;
int Lcounter = 0;
j = 0;
for(i = 0; i < strlen(data); i++)
{
   if(data[i] == '[')
  { 
     ++Lcounter;
     newData[j++] = data[i];
  }
  else if(data[i] == ']')
  {
     newData[j++] = data[i];
     ++Rcounter;
  }
  else
  {
     if(Lcounter > 0 && Rcounter < Lcounter)
       newData[j++] = data[i];
  }
}
newData[j] = '\0';

cout << newData;

Code:

void findBraces(char *pc)
{
	int open=0, close=0, ocount=0, ccount=0, i;
	for (i=0;pc[i] != '\0';i++)
	{
		if (pc[i] == '[')
		{
			if (ocount == -1)
				open = i+1;
			++ocount;
		}
		else if (pc[i] == ']')
		{
			++ccount;
			if (ccount == ocount && ocount != 0)
				close = i-1;
		}
			
	}
	if (close == 0)
	{
		close = i;
	}
	for (i=open;i<=close;i++)
	{
		if (pc[i] != '[' && pc[i] != ']')
			cout << pc[i];
	}
}

That's my take on it. Although it's late and I haven't tested that, I *think* it *should* work.

So just when I thought it couldn't be done with just loops I get proven wrong. I started looking around and found the vector library and with some of those functions I started to get things going again and then I check back here today and what do I find. That simple loops can indeed get it done. Your examples have taught me a lot about loop usage.

Thanks for the great help