looped loops

Marjolein · #1 21-Jun-2004, 08:39

Hi everybody,

Does anybody know some way to loop loops?
I mean, I can loop 2 times by using

for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
statement;

How can I loop say m times without using m integer variables?

Thanks for any replies!

Regards, Marjolein

dabigmooish · #2 21-Jun-2004, 11:35

I'm not sure what your asking. Are you trying to say is there a way to do a 5 nested loops with only 1 variable? If so then I think the answer is no. Are you trying to say how can you run the same loop X amount of times? In that case you'd only need the two loops like you have. Try and be a bit more specific

machinated · #3 21-Jun-2004, 12:24

Quote:

Originally Posted by Marjolein

Hi everybody,

Does anybody know some way to loop loops?
I mean, I can loop 2 times by using

for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
statement;

How can I loop say m times without using m integer variables?

Thanks for any replies!

Regards, Marjolein

you don't loop 2 times, you loop n*n times in this case.

mithunjacob · #4 21-Jun-2004, 16:42

What is a nested FOR loop? It's basically a series of nos! Check this out:

CPP / C++ / C Code:

for(int i=0;i<10;++i)
 for(int j=0;j<10;++j)
   for(int k=0;k<10;++k)
      cout<<"\n"<<i<<j<<k;

The output would be:
000
001
002
...

Well, what is this actually? Isn't it more like a counter of sorts? So, if I could make a counter of specifiable digits with constraints, if you like, you'd really be having a nested FOR loop..I've used this concept and put it in a tutorial awaiting approval in the Tutorials section...I've added the header file which defines the counter class, anyway...

CPP / C++ / C Code:

//Counter.h
#include <iostream.h>
#include <conio.h>
#include <stdlib.h>


class dig
{
public:
int val;
int min;
int max;

dig( int mi=0, int ma=9)
{
val=mi,max=ma,min=mi;
}

int hitMax()
{
return (val==max);
}

};

class counter
{
dig *digs;

public:
int upd;
int digNo;
int beg;

int continuous;

void init(int n,int c=0)
{
digNo=n;
digs=new dig[digNo];
continuous=c;
upd=1;
beg=1;
}

counter(){}

counter(int n, int c=0) {init(n,c);}

void set()
{
upd=1;
}

void reset()
{
for(int i=0;i<digNo;++i) digs[i].val=digs[i].min;
if(continuous) upd=1;
else upd=0;
}

int val(int no)
{
if(no>=0 && no<digNo) return digs[no].val;
else return -1;
}

void digSet(int no, int ma, int mi, int va)
{
digs[no].max=ma,digs[no].min=mi,digs[no].val=va;
}

~counter()
{
delete digs;
}

int update()
{

if(upd)
{
	for(int i=digNo-1;i>=0;--i)
		{
		if(digs[i].hitMax())
			{
			if(i==0)
				{
				reset();
				break;
				}
			else digs[i].val=digs[i].min;
			}
		else
			{
			++digs[i].val;
			break;
			}
		}
if(beg==1)
	{
	for(int i=0;i<digNo;++i) digs[i].val=digs[i].min;
	beg=0;
	}
}

return upd;
}
};

A small example:

CPP / C++ / C Code:

#include <fstream.h>
#include <conio.h>
#include "counter.h"

void main()
{
counter number(2,1);                                      

/*First argument:no.of digits
->Suppose you wanted all nos. 
   less than 50, then all of   
   you have to do is set the 
   constraints with number.digSet(0,4,0,0);
                       1)0 => First digit
                       2)4 => Max val
                       3)0 => Min val
                       4)0 => Initial val*/


for(;;number.update())
{
        for(int i=0;i<2;++i) cout<<number.val(i);
        if(getch()==27) break;         //27 is ASCII for the Escape key
        clrscr();                                                  
}

}

Marjolein · #5 22-Jun-2004, 00:56

Quote:

Originally Posted by machinated

you don't loop 2 times, you loop n*n times in this case.

Yes, but I wonder if there is some way to loop pow(n,m) times...

Marjolein · #6 22-Jun-2004, 01:07

I am afraid that my programming level is not that high that I can understand what exactly your counter class is doing and how it is related to my question?

sho · #7 22-Jun-2004, 03:57

Quote:

Originally Posted by Marjolein

Hi everybody,

Does anybody know some way to loop loops?
I mean, I can loop 2 times by using

for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
statement;

How can I loop say m times without using m integer variables?

Thanks for any replies!

Regards, Marjolein

Well, you could do what you are trying to do just by using a conditional statement like the following:

CPP / C++ / C Code:

int counter = 0;
for (int i = 1; i <= n; i++) {
          if (counter == m) break;
          for (int j = 1; j <= n; j++) {
                    if (i == n) {
                              i = 0;
                              counter++;
                    }
                    statement;
          }
}

that way you could loop n^m times... you can easily modify this to use more variables if you want them as indices for arrays or somethin like that...

BTW what are you going to use it for?? am curious...

Marjolein · #8 22-Jun-2004, 05:10

Thanks! I think I can work with this.
I am trying to use it for going through a search tree of variable length...

machinated · #9 22-Jun-2004, 09:16

Quote:

Originally Posted by sho

Well, you could do what you are trying to do just by using a conditional statement like the following:

CPP / C++ / C Code:

int counter = 0;
for (int i = 1; i <= n; i++) {
          if (counter == m) break;
          for (int j = 1; j <= n; j++) {
                    if (i == n) {
                              i = 0;
                              counter++;
                    }
                    statement;
          }
}

that way you could loop n^m times... you can easily modify this to use more variables if you want them as indices for arrays or somethin like that...

BTW what are you going to use it for?? am curious...

this doesn't run n^m times. it runs n*n*m. If you want it to run n^m, you will have to type it m times.

mithunjacob · #10 22-Jun-2004, 13:00

Quote:

Originally Posted by Marjolein

Hi everybody,

Does anybody know some way to loop loops?
I mean, I can loop 2 times by using

for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
statement;

How can I loop say m times without using m integer variables?

Thanks for any replies!

Regards, Marjolein

I'm sorry you didn't understand my reply, but I'll elaborate..

As I said, by using nested FOR loops, you're really simulating a counter. So, suppose you wanted m loops, then just declare:

counter Cnt(m,1);

How would you use it?

CPP / C++ / C Code:

for(int i=4;i<n;++i)
 for(int j=4;j<n;++j)
  for(int k=4;k<n;++k)
    .....

Well, as you can see the common constraint is n being the maximum & 4 as minimum, so you use:
count Cnt(m,1);
for(int i=0;i<m;++i) Cnt.digSet(i,n,4,4);
where:
i = designated the loop no.
n = maximum no
4 = minimum no
4 = initial no

Now, suppose I wanted to implement this:

for(int i=3;i<4;++i)
for(int j=3;j<4;++j)//Obviously the <4 makes it useless, but you get the point?
for(int k=3;k<4;++k)
for(int l=0;l<10;++l);
cout<<"\n"<<i<<j<<k<<l;

I would do this (let's assume there are 4 loops)

CPP / C++ / C Code:

#include "counter.cpp"
void main()
{
int m=4,n=3;
counter Cnt(m);
for(int i=0;i<3;++i) Cnt.digSet(i,n,n,n);
Cnt.digSet(3,9,0,0);
Cnt.set();

while(Cnt.update())
{
cout<<"\n";
        for(i=0;i<m;++i) cout<<Cnt.val(i);
}
}

Now suppose you wanted some specific constraint for each loop:

CPP / C++ / C Code:

#include <iostream.h>
void main()
{
int matrix[2][2][2]={1,2,
                      3,4,

                      5,6,
                      7,8};

for(int i=0;i<2;++i)
{
cout<<"\nPage "<<i+1<<": ";
        for(int j=0;j<2;++j)
        {
        cout<<"\n";
                for(int k=0;k<2;++k) cout<<"\t"<<matrix[i][j][k];
        }
}
}

This, using the counter class would be implemented like this:

CPP / C++ / C Code:

#include <iostream.h>
#include "counter.h"
void main()
{
int matrix[2][2][2]={1,2,
                      3,4,

                      5,6,
                      7,8};
counter Cnt(3);

for(int i=0;i<3;++i) Cnt.digSet(i,1,0,0);
Cnt.set();
while(Cnt.update())
        {
        if(Cnt.val(2)==0 && Cnt.val(1)==0) cout<<"\nPage "<<Cnt.val(0)+1<<":";
        if(Cnt.val(2)==0)                  cout<<"\n";
        cout<<"\t"<<matrix[ Cnt.val(0) ][ Cnt.val(1) ][ Cnt.val(2) ];
        }
}

Just in case you're wondering I've set the maximum value to be 1 as it never crosses 1 (you know, (0,0,1), (0,1,0), (0,1,1), (1,0,0) ...... )

Well, I hope you understood my class now!

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looped loops

Look at it this way....

Sorry..