Printing Syntax Tree with different indentation at different levels?

Hey guys, I'm trying to print out my syntax tree using different amounts of indentation for each level in the tree. I'm not not quite sure how to achieve this in C. I've created a for loop and a char* that has " " assigned to it... but when I try to add the char* to itself for each increment in the level it just crashes the program. Anyways, here's what I got, let me know what you guys think I should do:

Code:

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int synTreePreOrderWalk( TREENODE * ptree, int (*func)(), int level)
{
    char *c = "  ";
        int i;
        int iRet;
        if( ptree) {        // if current node not null
                // perform action - quit walking if non-zero return value
                for(i = 0; i < level; i++)
                // *c += *c;
                if( iRet = (*func)(ptree, level)) printf("%s" "%s\n", c, ptree->label);
                // visit children
                level++;  // going to a new indentation level
               
                if( iRet = synTreePreOrderWalk( ptree->left, func, level))  printf("%s" "%s\n", c, ptree->label);
                if( iRet = synTreePreOrderWalk( ptree->right, func, level)) printf("%s" "%s\n", c, ptree->label);
                // set level back to my level
                level--;  // visit siblings
               
                //if( iRet = synTreePreOrderWalk( ptree->sibling, func, level)) return iRet;
        }
        return 0;
}

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Oh dear.

*c is pointer with no memory assigned to it...

[edit] quzah correctly corrects me on this, but as s/he says you still cannot add to *c as is

Not exactly:

Code:

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char *c = "  ";

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It's actually pointing at a string literal. And you're not allowed to modify literals.


Quzah.

This is a rough program that shows one way to do it. The indentation level from left to right, depends on the depth that the solution was found at.

If you use some of the other rem'd out blocks of code, you can see it quite clearly.

Code:

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/* Solves using brute force, the 8 tile 2D "rubik's cube" puzzle.
Adak, Nov. 2008

Status: Rough, but works. Has a good, clear,  iterative depth first
search (DFS)!

To Improve: Fix showB, when starting depth is 4 or less.
*/

#include <stdio.h>

#define MaxDepth 30

//function prototypes
void copyB(int);
int isSolved(int);
int move1(int);
int move2(int);
int move3(int);
void showB(int);

int b[2][4][MaxDepth + 1];    //b is shorthand for the board
int t[1][4];                  //t is a temporary holding row
int line[MaxDepth + 1];      //the line of moves that are played

int main(void)  {
  int i, move, deep, depth, done, solved, reset, gar;
// test right shift   
/*  b[0][0][0] = 2; b[0][1][0] = 3; b[0][2][0] = 4; b[0][3][0] = 1;
  b[1][0][0] = 7; b[1][1][0] = 6; b[1][2][0] = 5; b[1][3][0] = 8;
*/
// test row swap
/*  b[0][0][0] = 8; b[0][1][0] = 7; b[0][2][0] = 6; b[0][3][0] = 5;
  b[1][0][0] = 1; b[1][1][0] = 2; b[1][2][0] = 3; b[1][3][0] = 4;
*/
// test Rotate 4 inner tiles
/*  b[0][0][0] = 1; b[0][1][0] = 3; b[0][2][0] = 6; b[0][3][0] = 4;
  b[1][0][0] = 8; b[1][1][0] = 2; b[1][2][0] = 7; b[1][3][0] = 5;
*/
// test depth of 2, solution 1 2
/*  b[0][0][0] = 7; b[0][1][0] = 6; b[0][2][0] = 5; b[0][3][0] = 8;
  b[1][0][0] = 2; b[1][1][0] = 3; b[1][2][0] = 4; b[1][3][0] = 1;
*/
//test depth of 2, solution 2 3  no depth=2 solution found for 1+3
/*  b[0][0][0] = 8; b[0][1][0] = 2; b[0][2][0] = 7; b[0][3][0] = 5;
  b[1][0][0] = 1; b[1][1][0] = 3; b[1][2][0] = 6; b[1][3][0] = 4;
*/
//test depth of 3, solution 2 3 3
/*  b[0][0][0] = 8; b[0][1][0] = 3; b[0][2][0] = 2; b[0][3][0] = 5;
  b[1][0][0] = 1; b[1][1][0] = 6; b[1][2][0] = 7; b[1][3][0] = 4;
*/
//test depth of 7, (2684/1375) solution is 3 3 1 3 1 1 1
  b[0][0][0] = 2; b[0][1][0] = 6; b[0][2][0] = 8; b[0][3][0] = 4;
  b[1][0][0] = 1; b[1][1][0] = 3; b[1][2][0] = 7; b[1][3][0] = 5;

//test? depth of 18 (2134/8765) solution is 231231313131231312
/*  b[0][0][0] = 2; b[0][1][0] = 1; b[0][2][0] = 3; b[0][3][0] = 4;
  b[1][0][0] = 8; b[1][1][0] = 7; b[1][2][0] = 6; b[1][3][0] = 5;
*/
//test 19 ply deep: 2834/1765
/*  b[0][0][0] = 2; b[0][1][0] = 8; b[0][2][0] = 3; b[0][3][0] = 4;
  b[1][0][0] = 1; b[1][1][0] = 7; b[1][2][0] = 6; b[1][3][0] = 5;
*/
  printf("\n\tThe puzzle is solved when it has 1234 in the first row,");
  printf("\n\tand 8765 in the second row.");

  printf("\n\n\tMoves are noted in this program, by number:\n\n");

  printf("\t1 = A right shift of both rows: 1234 becomes 4123\n");
  printf("\t                                8765        5876\n\n");

  printf("\t2 = A swap of the two rows: 1234  becomes  8765\n");
  printf("\t                            8765          1234\n\n");

  printf("\t3 = A clockwise rotation of the four central numbers:\n\n");
  printf("\t    1234 becomes 1724\n");
  printf("\t    8765        8635\n\n");
 
  printf("\tThese are the only three moves allowed in this puzzle.\n\n");
  printf("\tThe boards are displayed according to their depth in the\n");
  printf("\tsearch tree. The starting position (depth 0), is at the far\n");
  printf("\tleft. Every eight characters to the right, shows another\n");
  printf("\tdepth in the search tree, up to 10 ply deep.\n\n");

  printf("\n Our Starting position is:\n\n");
  showB(0);

  printf("\t\t\t    Press Enter to Continue\n\n ");
  gar = getchar();

  deep = solved = reset = 0;
  move = 1; depth = 4;

  while(1)  {  //main DFS loop
      ++deep;
      if(move == 1)
        solved = move1(deep);
      else if(move == 2)
        solved = move2(deep);
      else
        solved = move3(deep);
/*
      if(move == 1)
        printf("\n Right shift:");
      else if(move == 2)
        printf("\n Row swap:");
      else
        printf("\n Rotate 4 #'s:");

      printf("\t Line: ");        //shows the line of moves being searched
      for(i = 0; i < deep; i++) 
        printf(" %d", line[i]);
*/
      if(solved || depth >= MaxDepth)
        break;


      for(i = 0, done = 0; i < depth; i++)  { //are we done? 
        if(line[i] == 3)
            done = 1;
        else  {
            done = 0;
            break;
        }
      }
      if(done && depth <= MaxDepth)  {
        deep = 0;
        depth += 1;
        printf("\n Depth: %d", depth);
        done = 0;
        reset = 1;
      }
      if(reset)  {    //depth was reset, move needs to reset to 1 again
        move = 1;
        reset = 0;
        continue;
      }
      if(deep >= depth)  {
        if(move < 3)
            line[--deep] = 0;
        else
            while(line[deep-1] == 3) {
              line[--deep] = 0;
            }
        if(++move > 3) {
            deep -= 1;
            move = line[deep] + 1;
            if(move > 1)
              reset = 1;
        }
      }
  }

  if(!solved)
      printf("\n\n\t\t\t  No Solution was Found %d", done);

  printf("\n\n\t\t\t      Press Enter to Quit\n");
  gar = getchar(); gar++;
  return 0;
}
void copyB(int d)  {  //copy the old board position, into the new depth
  int r, c;   
 
  for(r = 0; r < 2; r++)
      for(c = 0; c < 4; c++)
        b[r][c][d] = b[r][c][d-1];
}
int move1(int d)  {  //shifts all numbers one column to the right &  wraps
  int r, c, solved, temp;
  solved = 0;
  copyB(d);
     
  for(r = 0; r < 2; r++)  {
      temp = b[r][3][d];
      for(c = 2; c >-1 ; c--) 
        b[r][c+1][d] = b[r][c][d];
      b[r][0][d] = temp;
  }
 
  line[d-1] = 1;
  solved = isSolved(d);
 
  return solved;
}
int move2(int d)  { //swaps rows
  int c, solved;
  solved = 0;
  copyB(d);
 
  for(c = 0; c < 4; c++)  {
      t[0][c] = b[0][c][d];
      b[0][c][d] = b[1][c][d];
      b[1][c][d] = t[0][c];
  }

  line[d-1] = 2;
  solved = isSolved(d);
 
  return solved;
}
int move3(int d) { //Rotates the 4 inner numbers, clockwise
  int solved, temp;
  solved = 0;
  copyB(d);

  temp = b[0][1][d];
  b[0][1][d] = b[1][1][d];
  b[1][1][d] = b[1][2][d];
  b[1][2][d] = b[0][2][d];
  b[0][2][d] = temp;

  line[d-1] = 3;
  solved = isSolved(d);
 
  return solved;
}       
void showB(int d)  {  //shows the board array, if needed
  int r, c, tab;
  printf("\n\n");
  if(d < 11)  {
      for(tab = 0; tab < d; tab++)
        putchar('\t');
      for(r = 0; r < 2; r++) {
        for(c = 0; c < 4; c++) 
            printf("%d ", b[r][c][d]);
        putchar('\n');
        if(d < 11)  {
            for(tab = 0; tab < d; tab++) 
              putchar('\t');
        }
      }
  }
}
int isSolved(int d)  {
  int i;
  int done = 0;
  if(b[0][0][d]==1 && b[0][1][d]==2 && b[0][2][d]==3 && b[0][3][d]==4)
      if(b[1][0][d]==8 && b[1][1][d]==7 && b[1][2][d]==6 && b[1][3][d]==5)  {
        done = 1;
        printf("\n\n\t The Solution is:");
        for(i = 0; i < d; i++)
            printf(" %d", line[i]);
        showB(d);
      }
     
  return done;
}

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