How to pass an Array to AngularJS?

How to pass a 2d array to a function..?

• I am trying to write a c program for solving sudoku..and at a point I came across this situation.. I have to pass an array to a function and when I pass it normally.. i get some warnings and segmentation fault.. when executed.. ill try to put up my program here soon.. i have used gdb to find the bug..and it stopped executing at the line a[q][w]==expval; now this a[q][w] is inside a function and when i asked to print value mof q,w it prints 6,6 and when i asked to print the value of a[q][w] it says it cant access that memory location.. i ll try to post up the code soon .. please try to fix it..

• Answer:

  Examine the following code. It's a chess puzzle solver for the Knight's Tour on a 5x5 grid. It passes a 2D array around. The declaration "int (*c)[5]" says "c is a pointer to an array of 5 int's." So c[0] would be the first array of 5 int's, c[1] would be the second, c[2] would be the third, etc. Tris's reply is screwed up. C doesn't create 2D arrays as an array of pointers to arrays. You can do that in C if you work hard at it, but it's not how C sets up a 2D array using the usual syntax. Defining "int x[8][8];" does NOT create ANY pointers at all. It simply allocates 64 integers. That's all. And sets them up in a certain order. If you want to pass x to a function, you do NOT declare the parameter as "int **p" for example. But as "int (*p)[8]". Do NOT follow Tris's advice. #include <stdio.h> static int s[50], t[50], z; int iscompleted( int (*c)[5] ) {     int i, j;         for ( i= 0; i < 5; ++i )             for ( j= 0; j < 5; ++j )                 if ( c[i][j] == 0 )                     return 0;     return 1; } int move( int a, int b, int (*c)[5] ) {     s[z]= a;     t[z++]= b;     c[a][b]= 1;     if ( iscompleted( c ) ) {         int i, count= 0;         for ( i= 0; i < z-1; ++i ) {             printf( "<%d %d>", s[i], t[i] );             count++;             if ( count == 10 ) {                 printf( "\n      " );                 count= 0;             } else                 printf( "  " );         }                 return printf( "<%d %d>\n", s[z-1], t[z-1] ), 1;     }     if ( ( a < 4 && ( ( b < 3 && c[a+1][b+2] == 0 && move( a+1, b+2, c ) ) ||                    ( b > 1 && c[a+1][b-2] == 0 && move( a+1, b-2, c ) )) ) ||          ( a < 3 && ( ( b < 4 && c[a+2][b+1] == 0 && move( a+2, b+1, c ) ) ||                    ( b > 0 && c[a+2][b-1] == 0 && move( a+2, b-1, c ) )) ) ||          ( a > 0 && ( ( b < 3 && c[a-1][b+2] == 0 && move( a-1, b+2, c ) ) ||                    ( b > 1 && c[a-1][b-2] == 0 && move( a-1, b-2, c ) )) ) ||          ( a > 1 && ( ( b < 4 && c[a-2][b+1] == 0 && move( a-2, b+1, c ) ) ||                    ( b > 0 && c[a-2][b-1] == 0 && move( a-2, b-1, c ) )) ) )         return 1;     c[a][b]= 0;     --z;     return 0; } int main( int argc, char *argv[] ) {     int board[5][5], i, j, x, y;         for ( x= 0; x < 5; ++x )             for ( y= 0; y < 5; ++y ) {                 int r;                 z= 0;                 for ( i= 0; i < 5; ++i )                     for ( j= 0; j < 5; ++j )                         board[i][j]= 0;                 printf( "%d %d = ", x, y );                 r= move( x, y, board );                 if ( !r )                     printf( "failure\n" );             }     return 0; }