Solved

# function procedure design

Posted on 2006-11-12

hi,

I have a program like below:

the problem is the backtrack() will return me more than one set of (newt[N][i][j] != 0), but only first set goes to addpath(N, i, j). how can I let the program run every set of (newt[N][i][j] != 0)?

---------------------------------------

main ()

{

backtrack();

system("pause");

return 0;

}

void backtrack(void)

{

for (i=0; i<3;i++){

for (j=0; j<3; j++) {

if (newt[N][i][j] != 0)

{

addpath(N, i, j);

}//end if

}// 3rd for

} // 2nd for

}

----------------------------------------------------------------------------

<ps> the addpath(N, i, j) will continue call other functions.....but I am sure that the final function is looks like

void addpath (int s, int i, int j)

{

if( s == 0 )

{

printf("final path is %d%d%d : \n",s,i,j);

printf("finish!!\n");-------------------------------------------->> end here

return;

}

call other functions....

}