C 中的结构和函数

Question

I am getting errors in the following code. The errors disappear if I take out "struct point p2...". p1 is assembled the same way and works fine, what is the catch here?

#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

#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));
}
}

Answer 2

#include <stdio.h>

struct arithmetic {
 int x;
 int y;
 char oper;
 };

 int calcul(int a,int b)
{
 struct arithmetic temp;
int x,y;
char proc;
temp.x=x;
temp.y=y;
temp.proc=proc;
if(proc=='+')
{
  int addit(int x,int y)
    {
        return(x+y);
    }
 }
 if(proc=='-')
 {
  int devit(int x,int y)
    {
        return(x-y);
    }
}
  if(proc=='*')
 {
  int cumlatit(int x,int y)
     {

            return(x*y);
     }
   }
  if(proc=='/')
   {
    int dividit(int x,int y)
     {

            return(x/y);
     }
   }
  if(proc=='@')
  {
  int intdividit(int x,int y)
      {

          return(x/y);
      }
  }
  if(proc=='%')
   {
  int remdivit(int x,int y)
    {

        return(x%y);
    }
  }
  if(proc=='`')
  {
  int exponit(int x,int y)
    {
        int i=1,k=1;
        for(i=1;i<=y;i++)
            {
                k*=x;
            }
        return(k);
       }
  }
  void main()
  {
  struct arithmetic v1;
  int num1,num2;
  char proc;
  scanf(&num1);
  scanf(&num1);
  proc=getch();
  v1.x=num1;
  v1.y=num2;
  v1.oper=proc;
  struct calcul();
  }

Answer 3

• To avoid the many hidden calls to memcpy()
• To avoid the many compiler allocated memory areas (that cannot be accessed/used by anything else.)

I suggest the following code, which compiles cleanly and exercises each function

#include <stdio.h>

struct point
{
    int x;
    int y;
};

void makepoint(int x, int y, struct point* newPoint)
{
    newPoint->x = x;
    newPoint->y = y;
} // end function: makepoint

void addpoint(struct point* p1, struct point* p2)
{
    p1->x += p2->x;
    p1->y += p2->y;
} // end function: addpoint

int main()
{
    struct point p1;
    struct point p2;

    makepoint(5, 7, &p1);
    printf("p1 = (%d, %d)\n", p1.x, p1.y);

    makepoint(2, 9, &p2);
    printf("p2 = (%d, %d)\n", p2.x, p2.y);

    addpoint( &p1, &p2 );
    printf("p1 = (%d, %d)\n", p1.x, p1.y);

    return(0);
} // end function: main