My first solution with out the base case

spiral_thr_matrix.c

#include<stdio.h>
#include<conio.h>
//int w[20][20];
int w[3][3]={1,2,3,4,5,6,7,8,9};
int N=3;

void printer(int n) {
 int i,j;
 
 printf("\n ");
 for ( i=N-n;i<n;i++) {
    printf(" %d",w[N-n][i]);
  }


 for ( j=N-n+1;j<n;j++) {
 printf("\n");
 for ( i=N-n;i<n;i++) {
    printf("  ");
  }
    printf("%d",w[j][n-1]);
  }

 for ( j=N-n+1;j<n;j++) {
    printf("\n %d",w[j][N-n]);
  }
  for ( i=N-n+1;i<n-1;i++) {
    printf(" %d",w[n-1][i]);
  }

  //printer(n-1);
  
}
int main()
{
int i,j;
/*
printf("enter n:");
scanf("%d",&N);
for ( i=0;i<N;i++) {
  for ( j=0;j<N;j++) {
    scanf("%d",&w[i][j]);
  }
}*/
for ( i=0;i<N;i++) {
printf("\n");
  for ( j=0;j<N;j++) {
    printf(" %d",w[i][j]);
  }
} 
printer(N);

return 0;
}

Spiraling through a matrix half answer.