如何解决此 C 程序中的类型重定义错误

Question

I'm a new C programmer,

the program I was writing has to return 0 if the points are colinear, otherwise it has to return 1. I've split the code in .h and .c. This is the code: [geometry.c]

struct point {
    int x; 
    int y; 
} p1; 

struct point {
    int x;
    int y;
} p2; 

struct point {
    int x;
    int y;
} p3; 


int colinear(struct point* p1, struct point* p2, struct point* p3) {

    if (((p1->x - p2->x)* (p1->y - p2->y) == ((p3->y - p2->y) * (p1->x - p2->x))))

    return 0;
    
    else
        return 1;
        }

and: [geometry.h]

#ifndef geometry.h 
#define geometry.h 
#endif 
#include "geometry.c"


extern int colinear(struct point *p1, struct point *p2, struct point *p3); 

Using the debugger: "C2011: 'point': 'struct' type redefinition".

Where are the errors?

Answer 1

No Need to define 3 times

struct point {
    int x; 
    int y; 
} p1; 

struct point {
    int x;
    int y;
} p2; 

struct point {
    int x;
    int y;
} p3; 

define only once, and create variables as you wish.

struct point {
    int x;
    int y;
};
struct point p1,p2,p3; 

Answer 2

The other answer addresses the problem well. (ie why the code resulted in multiply defined compiler error.) But as a possible enhancement to your corrected code, the following is a related construct using an array of a typedef struct . This may make some of what you need to do a little easier to pass around and manipulate the struct member values:

typedef struct {
   int x;
   int y;
} point_s;

point_s p[3];

This effectively allows the prototype:

int colinear(struct point* p1, struct point* p2, struct point* p3) {

To be refactored into:

int colinear(point_s *p) { //where p is an pointer to 3 instances of point_s

Usage example:

int main(void)
{
    point_s p[3] = {{4,6}, {-2, 5}, {-0.4, 15}}; 
    
    int result = colinear(p);
    
    return 0;
}

int colinear(point_s *p)
{
    if (((p[0].x - p[1].x)* (p[0].y - p[1].y) == ((p[2].y - p[1].y) * (p[0].x - p[1].x))))
    {
        return 0;
    }

    else
    {
        return 1;
    }
}