c++ - Rewriting a program without using goto statements -

i had program similar problem using goto statements. asked rewrite our code without using goto statement. don't know how begin doing program. paste in previous program code using goto.

// 8 queens problem using 1 dimesional array , goto statement  #include "stdafx.h" #include <iostream> using namespace std;   int main() {     int q[8];     q[0] = 0;     int c = 0;     int count = 0;  nc: //cout  << "next column\n" << "column = " << c << endl;     c++;     if (c == 8) goto print;     q[c] = -1;  nr: //cout << "next row\n" << "row = " << q[c] << "\ncolumn = " << c << endl;     q[c]++;     if (q[c] == 8) goto backtrack;      for(int = 0; < c; i++){         if(q[i] == q[c] || abs(q[c] - q[i]) == (c - i))             goto nr;     }     goto nc;  backtrack:     //cout << "backtrack" << endl;     //cout <<"column = " << c << endl;     c--;     if(c == -1) return 0;     goto nr;  print:     //cout << "print" << endl;     ++count;     cout << count << endl;     for(int = 0; <= 7; i++){             cout << q[i];     }     cout << endl;     goto backtrack;       return 0; } 

this program hint professor posted class use.

 #include <iostream>     #include<cstdlib>     #include <cmath>     using namespace std;      bool ok(int q[], int col){     if configuration “bad” return false;     else     return true;     }      void backtrack(int &col){     col--;     if(col==-1) exit(1);     }      void print(int q[]){     static int count =0;     print array q     }      int main(){     int q[8]; q[0]=0;     int c=1;     // from_backtrack keeps track if need reset row     // top of current colum or not.      bool from_backtrack=false;     // outer loop keeps looking solutions     // program terminates function backtrack     // when forced backtack column -1     while(1){     while(c<8){ //this loop goes across columns     // if returned backtrack, use current value of row     // otherwise ready start @ top of column     if(!from_backtrack) // did not return backtrack     code goes here     from_backtrack=false;     while(q[c]<8){ // place queen in column     q[c]++;     // if row=8, there no valid square in column     // backtrack , continue loop in previous column     code goes here     //if position ok, place queen     // , move on (break) next column,     // otherwise keep looking in column     code goes here     }     c++; // placed ok, move next column     }     // 1 complete solution found, print it.     print(q); // board completed, print out     backtrack(c);     from_backtrack=true;     }     } 

and attempted i've made completed program

// nogoto.cpp.cpp : defines entry point console application. //  #include "stdafx.h" #include <iostream> using namespace std;    bool attack(int q[], int col){     if(q[i] == q[c] || abs(q[c] - q[i]) == (c - i)) return false;     else         return true; } // attack  void backtrack(int & col){     col--;     if(col == -1) exit(1); } // backtrack  void print(int q[]){     static int count = 0;     ++count;     cout << count << endl;     for(int = 0; < 8; i++)         cout << q[i];     cout << endl; } int main() {     int q[8];     q[0] = 0;     int c = 1;      bool from_backtrack = false;      while(1){         while(c < 8){ // loops across columns              if(!from_backtrack)                 attack(q[c],c)              from_backtrack = false;              while(q[c] < 8){ // place queen in column                 q[c]++;      return 0; } 

"i'm having problem writing code. how [can i] call each function make correctly find solutions?"

first, @ control flow stripping else , adding explicit gotos execution reaches label straight-line execution.

nc: if (…) goto print;     goto nr;  nr: if (…) goto backtrack;     if (…) goto nr;     goto nc;  backtrack:     if (…) return;     goto nr;  print:     goto backtrack; 

now, take unconditional gotos , try move blocks represent straight-line execution.

nr: if (…) goto backtrack;     if (…) goto nr;     goto nc;  nc: if (…) goto print;     goto nr;  print:     goto backtrack;  backtrack:     if (…) return;     goto nr; 

now eliminate straight-line gotos

nr: if (…) goto backtrack;     if (…) goto nr;  nc: if (…) goto print;     goto nr;  print:  backtrack:     if (…) return;     goto nr; 

note label backward-going gotos loop:

for (;;) { nr: if (…) goto backtrack;     if (…) continue;  nc: if (…) goto print;     continue;  print:  backtrack:     if (…) return; } 

hmm, can reverse sense of nc: if() , eliminate goto print. , goto backtrack jumps on statements, equivalent reversed if.

for (;;) { nr: if (! …) {         if (…) continue; nc:     if (! …) continue; print:     } backtrack:     if (…) return; } 

the loop has no condition, backtrack: … if(…) return; exits it, move block , condition loop.

for (;…; /* backtrack */ …) { nr: if (! …) {         if (…) continue; nc:     if (! …) continue; print:     } } 

looking pretty good, no more gotos , no "suspicious" structure! however, nc supposed entry point.

this blind, mechanistic, compiler-ish transformations fail. see 3 alternatives:

1. introduce variables force execution there first loop iteration. uglier goto, in opinion.
2. disguise goto nc; switch(0) , call label nc: case 0:. teacher not accept compromise.
3. copy blocks nc end of loop , paste above beginning of loop. can factor them out functions. actually, blind, mechanistic transformation. hooray. i'll represent nr , backtrack functions uniformity.

.

nc(); if (…) {     print(); }  while ( backtrack(), … ) { // <- note comma operator     nr();     if (! …) {         if (…) continue;         nc();         if (! …) continue;         print();     } } 

there more elegant solution involves looking @ contents of code, takes less thinking.