Need a bit of help with pointers

[tmac619619]

int main()
  {
  int form;
  int x1,x2,y1,y2;
  double m,b;
 
  if(get_problem(form)==1)   //if get_problem return value is 1
  {
  get2_pt(&x1,&x2,&y1,&y2); //call the next function
  display2_pt(&x1,&x2,&y1,&y2);
  }
  return 0;
 }
 
  int get_problem(int form)
  {
  printf("Select the form that you like to convert to slope-intercept form:\n1    ) Two-point form (you know two points on the line)\n2) Point-slope form (you     know the line's slope and one point)\n=> ");
  scanf("%d",&form);
  return form;  //user either enters 1 or 2
  }
 
 
   void get2_pt(int *x1,int *x2, int *y1,int *y2) //potential problem source
  {
  printf("Enter the x-y coordinates of the first point seperated by a space=>      ");
  scanf("%d %d",&x1,&y1);
  printf("Enter the x-y coordinates of the second point seperated by a space=>      ");
 

void display2_pt(int *x1, int *x2, int *y1, int *y2)
  {
  printf("Two-point form\n");
  printf("%d%d",x2,x1); //instead of printing out the value, it prints out the address.
}

My question is. Where did i go wrong? Print only prints only the addresses of the variable, and not the value

[laserlight]

Well, if you want to print the value, print *x2 and *x1.

[tmac619619]

Hmm printf("%d%d",*x2,*x1);

still prints out the addresses. This is the deref operator, correct?

[laserlight]

Hmm printf("%d%d",*x2,*x1);

still prints out the addresses.

No, it doesn't.

This is the deref operator, correct?

Yes.

Take a look at this line in the get2_pt function:

scanf("%d %d",&x1,&y1);

x1 and y1 are pointers, so why do you take their address when you want to read into an int? You should have written:

scanf("%d %d", x1, y1);

[tmac619619]

Thank you based god!