Spring 99, CSE 428: Solution of Assignment 3

Exercise #1: 
1 (call by value) the output is 2
2 (call by reference) the output is 3
3)(call by value-result) the output is 1

Exercise #2
1) (static scope) the output is 2
2) (dynamic scope) the output is 1

Exercise #3
The best solution is the following: 

/* Solution 1 */

void search_exit(int i, int j, int *found){
     if (A[i][j] == '*') 
          *found = 0;
     else if (i==0 || i==m || j==0 || j==n) 
          *found = 1;
     else {
          A[i][j] = '*';
          search_exit(i-1,j,found);
          if (! *found){ 
             search_exit(i,j+1,found);
             if (! *found){ 
                search_exit(i+1,j,found);
                if (! *found)
                   search_exit(i,j-1,found);
             }
          }
     }
     if (*found) A[i][j]='o'; /* new instruction (wrt the LN) */
}


Another possibility is the following

/* Solution 2 */

void search_exit(int i, int j, int *found){
     if (A[i][j] == '*') 
          *found = 0;
     else if (i==0 || i==m || j==0 || j==n) 
          {
          *found = 1; 
          A[i][j] = 'o';  /* new instruction */
          }
     else {
          A[i][j] = '*';
          search_exit(i-1,j,found);
          if (*found)
               A[i][j] = 'o';  /* new instruction */
          else { 
               search_exit(i,j+1,found);
               if (*found)
                    A[i][j] = 'o';  /* new instruction */
               else { 
                    search_exit(i+1,j,found);
                    if (*found)
                         A[i][j] = 'o';  /* new instruction */
                    else {
                         search_exit(i,j-1,found);
                         if (*found) A[i][j] = 'o';  /* new instruction */
                    }
               }
          }
     }
}

This second solution is equivalent to the first (although redundant). 

Finally, there is another possibility: 

/* Solution 3 */

void search_exit(int i, int j, int *found){
     if (A[i][j] == '*') 
          *found = 0;
     else if (i==0 || i==m || j==0 || j==n) 
          *found = 1; 
     else {
          A[i][j] = '*';
          search_exit(i-1,j,found);
          if (*found)
               A[i-1][j] = 'o';  /* new instruction */
          else { 
               search_exit(i,j+1,found);
               if (*found)
                    A[i][j+1] = 'o';   /* new instruction */
               else { 
                    search_exit(i+1,j,found);
                    if (*found)
                         A[i+1][j] = 'o';    /* new instruction */
                    else {
                         search_exit(i,j-1,found);
                         if (*found) A[i][j-1] = 'o';   /* new instruction */
                    }
               }
          }
     }
}

This third solution is "almost" correct: the difference with the 
other two solutions is that it does not put an 'o' in the starting 
place.