Puzzle T23

Style 1: Variable Directly Contains a Triangle

Inspect the following incomplete program:

#include <stdio.h>
#include <stdlib.h>

/* Practice with various ways to declare things */

typedef struct
{
  int x, y;
} Point;

void printPoint( Point *p )
{
  printf("(%d, %d) ", p->x, p->y );
}

typedef struct
{
  int red, green, blue;
} Color;
 
void printColor( Color *c )
{
  printf("%3d red, %3d grn, %3d blu", c->red, c->green, c->blue );
}

typedef struct
{
  Point p0, p1, p2;
  Color color;
} Triangle;

void printTriangle( Triangle *t )
{
  printf("Points: ");
  printPoint( &t->p0 ); printPoint( &t->p1 ); printPoint( &t->p2 );
  printf("\nColor: ");
  printColor( &t->color );
  printf("\n");
}

int main ()
{
  /* Declare a variable here: variable directly contains a Triangle */
  ????
   
  /* Leave the following unchanged */
  tri.p0.x = 15;
  tri.p0.y = 15; 
  tri.p1.x = 85;
  tri.p1.y = 110; 
  tri.p2.x = 50;
  tri.p2.y = 50; 
  tri.color.red   = 123;
  tri.color.green = 50;
  tri.color.blue = 150;

  /* Insert the parameter in the function call */
  printTriangle( ??? ); 
 
} 

Notice how the parameter t in printTriangle() is used. It is a pointer to a Triangle:

void printTriangle( Triangle *t )

Accessing a member Point of the pointed-at Triangle is done through the pointer t:

t->p0

The function printPoint(Point *p) expects a pointer to a Point, so the address of the Point is supplied:

&( t->p0 )

Because of operator precedence, parentheses are not needed:

printPoint( &t->p0 );

The address-of operator & has lower precedence than the structure pointer member operator ->. Another way to do the same thing is:

printPoint( &( (*t).p0 ) );

The Puzzle: Edit main() in the following so that it contains the variable tri which works correctly with the rest of the program.