#include<math.h>
#include<stdio.h>
//定义点结构体 //
struct point
{
    float x;
    float y;
};
/*定义矩形结构体*/
struct rect
{
    struct point pt1;
    struct point pt2;
    struct point pt3;
    struct point pt4;
};
/*声明函数*/
float GetDis(struct point,struct point);
struct point makepoint(float x,float y);
//定义构造点函数//
struct point makepoint(float x,float y)
{
    struct point temp;
    temp.x=x;
    temp.y=y;
    return temp;
}
//定义构造矩形函数//
struct rect makerect(struct point pt1,struct point pt2,struct point pt3,struct point pt4,struct rect *screen)//struct rect *screen ？????//
{
    (*screen).pt1=pt1;
    (*screen).pt2=pt2;
    (*screen).pt3=pt3;
    (*screen).pt4=pt4;
}  
//定义判断函数//
int PtRect(struct point p,struct rect r)
{
    return p.x>=r.pt1.x && p.x<=r.pt3.x && p.y>=r.pt1.y && p.y<=r.pt3.y;

}

main()
{
    struct point pt1=makepoint(0,0);
    struct point pt2=makepoint(50,0);
    struct point pt3=makepoint(50,10);
    struct point pt4=makepoint(0,10);
   
    struct rect screen;
    makerect(pt1,pt2,pt3,pt4,&screen);//???????//
    struct point middle=makepoint((screen.pt1.x+screen.pt2.x)/2,(screen.pt2.y+screen.pt4.y)/2);
    printf("\n the middle point is(%f,%f)\n",middle.x,middle.y);
    printf("%f\n",screen.pt1.x);
    printf("%f\n",screen.pt1.y);
    printf("%f\n",screen.pt2.x);
    printf("%f\n",screen.pt2.y);
    printf("%f\n",screen.pt3.x);
    printf("%f\n",screen.pt3.y);
    printf("%f\n",screen.pt4.x);
    printf("%f\n",screen.pt4.y);
   
    if (PtRect(middle,screen))
    printf("the middle point is in screen!");
    else
    printf("the middle point is not in screen");
   
    getchar();
    getchar();
}
结构体在定义时和引用是格式为什么不一致，前面定义时是：
struct rect makerect(struct point pt1,struct point pt2,struct point pt3,struct point pt4,struct rect *screen)
后面引用时是：
makerect(pt1,pt2,pt3,pt4,&screen)
请问这是为什么？

定义跟引用当然不同的格式，这有什么问题？比如你定义int x，引用的时候是写x还是仍然写int x？

你这问题跟以下帖子的是异曲同工之妙：https://bbs.bccn.net/redirect.php?tid=433058&goto=lastpost#lastpost

其实你这个makerect()函数无需那样写的，传入四个角点，直接返回一个矩形即可，不需要用传址返回构造的矩形，这样定义反而不清晰。

struct rect makerect(struct point pt1,struct point pt2,struct point pt3,struct point pt4,struct rect screen)//定义//
{
    screen.pt1=pt1;
    screen.pt2=pt2;
    screen.pt3=pt3;
    screen.pt4=pt4;
}  
//引用
struct rect juxing;
makerect(pt1,pt2,pt3,pt4,juxing);
这样定义和引用的话，结果会出错。跪求解答！！！

[ 本帖最后由 suv121 于 2014-6-20 10:55 编辑 ]

把报错时的编辑画面截出来看看，尽量把相关代码放到一起。

screen.pt1=pt1;
 screen.pt2=pt2;
 screen.pt3=pt3;
 screen.pt4=pt4;
把这个改成memcpy试试吧。

#include<math.h>
#include<stdio.h>
//定义点结构体 // 
struct point
{
    float x;
    float y;
};
/*定义矩形结构体*/
struct rect
{
    struct point pt1;
    struct point pt2;
    struct point pt3;
    struct point pt4; 
};
/*声明函数*/
float GetDis(struct point,struct point);
struct point makepoint(float x,float y);
//定义构造点函数//
struct point makepoint(float x,float y)
{
    struct point temp;
    temp.x=x;
    temp.y=y;
    return temp;
}
//定义构造矩形函数// 
struct rect makerect(struct point pt1,struct point pt2,struct point pt3,struct point pt4,struct rect screen)//struct rect *screen ？????//
{
    screen.pt1=pt1;
    screen.pt2=pt2;
    screen.pt3=pt3;
    screen.pt4=pt4;
}  
//定义判断函数//
int PtRect(struct point p,struct rect r)
{
    return p.x>=r.pt1.x && p.x<=r.pt3.x && p.y>=r.pt1.y && p.y<=r.pt3.y;

}

main()
{
    struct point pt1=makepoint(0,0);
    struct point pt2=makepoint(50,0);
    struct point pt3=makepoint(50,10);
    struct point pt4=makepoint(0,10); 
    
    struct rect juxing;
    makerect(pt1,pt2,pt3,pt4,juxing);//???????//
    struct point middle=makepoint((juxing.pt1.x+juxing.pt2.x)/2,(juxing.pt2.y+juxing.pt4.y)/2);
    printf("\n the middle point is(%f,%f)\n",middle.x,middle.y);
    printf("%f\n",pt1.x);
    printf("%f\n",pt1.y);
    printf("%f\n",pt2.x);
    printf("%f\n",pt2.y);
    printf("%f\n",pt3.x);
    printf("%f\n",pt3.y);
    printf("%f\n",pt4.x);
    printf("%f\n",pt4.y);
    
    if (PtRect(middle,juxing))
    printf("the middle point is in screen!");
    else 
    printf("the middle point is not in screen");
    
    getchar();
    getchar();
} 

是这个效果么

这样定义的话，矩形的四个点都是0，而且你看中点输出结果大的惊人