C中的结构和功能

Question

以下代码中出现错误。如果我删除“结构点p2 ...”，错误将消失。 p1以相同的方式组装并且可以正常工作，这里有什么问题？

#include <stdio.h>

struct point {
    int x;
    int y;
};

struct point makepoint(int x, int y)
{
    struct point temp;

    temp.x = x;
    temp.y = y;
    return temp;
}
struct point addpoint(struct point p1, struct point p2)
{
    p1.x += p2.x;
    p1.y += p2.y;
    return p1;
}

void main()
{
    struct point p1 = makepoint(5, 7);
    printf("p1 = (%d, %d)\n", p1.x, p1.y);
    struct point p2 = makepoint(2, 9);
    printf("p2 = (%d, %d)\n", p2.x, p2.y);
}

Answer 1

• 为了避免对memcpy（）的许多隐藏调用，>
• 为了避免许多编译器分配的内存区域（不能被其他任何东西访问/使用。）

Answer 2

#include   <stdio.h>
#include   <string.h>
struct   arithmet   {
   char    eval;
   int       first_num,sec_num;
};

/*   function   declaration   */
void   eval_value(   struct   arithmet   *valyu   );
int   main(   )   {

   struct   arithmet   valyu1;               
   struct   arithmet   valyu2;
   char   oper;
   int   num1,num2;
   oper=getch();    //gets(oper);
   scanf("%d",&num1);
   scanf("%d",&num1);
   valyu1.first_num=num1;
   valyu2.sec_num=num2;
   valyu1.eval=oper;
   eval_value(&num1);
   eval_value(&num2);
   eval_value(&per);
   printBook(   &Book1   );
   printBook(   &Book2   );
   return   0;
}
void   eval_value(struct   arithmet   *valyu)
{
valyu->first_num;
valyu->sec_num;
valyu->oper;
if(oper=='+')
{
    printf("%d",addit(int   a,int   b));
}
if(oper=='-')
{
    printf("%d",subtractit(int   a,int   b));
}
if(oper=='*')
{
    printf("%d",multiplyit(int   a,int   b));
}
if(oper=='/')
{
    printf("%d",divideit(int   a,int   b));
}
if(oper=='@')
{
    printf("%d",intdivideit(int   a,int   b));
}
if(oper=='%')
{
    printf("%d",remdivideit(int   a,int   b));
}
if(oper=='~')
{
    printf("%d",exponeit(int   a,int   b));
}
}