Hello, I have a problem which I'm not sure if there is a nice solution.
I need a sequence of loops imbedded one inside the other, with variable depth.
Well, here is an example.  The following program will loop around all the
3-bit strings.
                        /****** loop3.c ********/
const int K=3;
main()
{
  int m[K],n[K],i[K], j;  for (j=0;j<K;j++) { m[j]=0; n[j]=2; }
  for (i[0]=m[0]; i[0]<n[0]; i[0]++)
        for (i[1]=m[1]; i[1]<n[1]; i[1]++)
          for (i[2]=m[2]; i[2]<n[2]; i[2]++)
                printf("Do something with %2i %2i %2i\n",i[0],i[1],i[2]);

Now what if I want all the K-bit strings, where K is any integer?
Suppose I want to experiment with K=1...20.  Do I have to supply 20
versions of source code with different numbers of 'for' statements?
Obviously, the following naive 'solution' does not work:
                        /****** looploop.c *******/
const int K=3;
main()
{
  int m[K],n[K],i[K],j;  for (j=0;j<K;j++) { m[j]=0; n[j]=2; }
  for (j=0;j<K;j++)
        for (i[j]=m[j]; i[j]<n[j]; i[j]++)
          printf("Do something with %2i %2i %2i\n",i[0],i[1],i[2]);

The reason this does not work can be seen easily from the following program
which is equivalent to loop3.c.
                        /******* goto3.c ********/
const int K=3;
main()
{
  int m[K],n[K],i[K],j;  for (j=0;j<K;j++) { m[j]=0; n[j]=2; }
  i[0]=m[0]; l0:
  i[1]=m[1]; l1:
  i[2]=m[2]; l2:
  printf("Do something with %2i %2i %2i\n",i[0],i[1],i[2]);
  i[2]++;if(i[2]<n[2])goto l2;
  i[1]++;if(i[1]<n[1])goto l1;
  i[0]++;if(i[0]<n[0])goto l0;

That is, the labels are not variable.  However, I've found a solution
                        /****** gotoloop.c ******/
const int K=3;
main()
{
  int m[K],n[K],i[K],j, l;
  for (j=0;j<K;j++) { m[j]=0; n[j]=2; }
  l=0;
 start:
  for (j=l+1;j<K;j++) i[j]=m[j];
  { for (j=0;j<K;j++) printf("%2i ",i[j]);  printf("\n"); }
  for (l=K-1;l>=0;l--)
        {(i[l])++; if(i[l]<n[l]) goto start;}

Apart from using 'goto', this program is not very illuminating, either.
(It helps the understanding a little bit by noting that l denote the current
layer of loop.)  So is there a better solution?

---

Dept of Computer Sciences and Applied Mathematics
Aston University, Birmingham B4 7ET, UK

<fortran code deleted>

<more fortran code deleted>

How about:

#define K 4
main()
{
  int m [K],
       n [K],
       i [K] ;

  int j ;

  for (j=0 ; j<K ; j++)
     { i [j] = m[j]=0 ;
       n[j]=2 ;
     }

  for ( ; ; )
     { for (j = K - 1 ; j >= 0 && i[j] ++ >= n[j] ? i[j] = m[j], 1 : 0 ; j --) ;
       if (j < 0 && i[0] == m[0])
         { break ;
         }
       printf("Do something with ") ;
       for (j = 0 ; j < K ; j ++)
            { printf ("%2d ", i[j]) ;
            }
       printf("\n") ;
     }

William.

  This is a mistake.  I know I will get flamed, and I intend no offense
but that has got to be the oddest formatting style I have ever seen...

--
                  Dave Moore - Student at Large - Reed College

                                  --=---=--

: Hello, I have a problem which I'm not sure if there is a nice solution.
: I need a sequence of loops imbedded one inside the other, with variable depth.
: Well, here is an example.  The following program will loop around all the
: 3-bit strings.
:                       /****** loop3.c ********/
: const int K=3;
: main()
: {
:   int m[K],n[K],i[K], j;  for (j=0;j<K;j++) { m[j]=0; n[j]=2; }
:   for (i[0]=m[0]; i[0]<n[0]; i[0]++)
:       for (i[1]=m[1]; i[1]<n[1]; i[1]++)
:         for (i[2]=m[2]; i[2]<n[2]; i[2]++)
:               printf("Do something with %2i %2i %2i\n",i[0],i[1],i[2]);
: }

You can try recursion:

int K=3;
int m[K],n[K],i[K];

void Loop(int cur)
  {
    if (cur==K)
      printf("Do something with %d %d %d\n",i[0],i[1],i[2]);
    else
      for (i[cur]=m[cur]; i[cur]<n[cur]; i[cur]++)
        Loop(cur+1);
  }

void main(void)
  {
    int j;

    for (j=0; j<K; j++) { m[j]=0; n[j]=2; }
    Loop(0);
  }

  YUAN, Feng
  HP Singapore

Yes. It is easy to see that this code is wrong if one has ever tested it.
(Set K=2 and n[j]=1. It is wrong in two accounts).  
BTW, I just can't see why a code is more "C" than "fortran" simply because
it is both wrong and hard to understand.  It seems that C style is simply
loads of "++--&&||;;?:". :-)

However, thanks to a hint in his example, I found the following humble
"fortran code" which does the job.

  do    { for (l=0; l<K; l++) printf("%2i ", i[l]); printf("\n");
          for (l=K-1; l>=0; l--)
                { i[l]++; if(i[l]<n[l]) break; else i[l]=m[l];}
        } while (l>=0);

Other people have suggested to me of using recursion, which is neat for this
example, but is not suitable for the actual problem I intended to solve for
some other reasons.  

Thanks to every one.  I've learned a lot.

---

Dept of Computer Sciences and Applied Mathematics
Aston University, Birmingham B4 7ET, UK