删除/添加元素和印刷 - 结构的数组 [英] Array of structs - deleting/adding elements and printing

问题描述

对于低于code

struct orderSlip
{
    char source[64];
    char destination[64];
    char item[64];  
    int position;
};
//global
struct orderSlip data[100];

是否有比下面这些方法的其他各元素打印出来的数据的另一种方法：

Is there another way of printing out the data for each element other than these methods below:

printf("%s\n",data[0].source);
    printf("%s\n",data[0].destination);
    printf("%s\n",data[0].item);
    printf("%i\n\n", data[0].position);

    printf("%s\n",data[1].source);
    printf("%s\n",data[1].destination);
    printf("%s\n",data[1].item);
    printf("%i\n", data[1].position);

等

for(int n = 0; n< 3; n++)
{
    printf("%s\n",data[n].source);
    printf("%s\n",data[n].destination);
    printf("%s\n",data[n].item);
    printf("%i\n\n", data[n].position);
}

有关删除和添加，做我必须做结构的动态数组？如果是这样，这将是做到这一点的最简单的语法？
事情是这样的C ++ code

For deleting and adding, do I have to make a dynamic array of structs? If so, what would be the simplest syntax to do that? Something like this c++ code

int * bobby;
bobby = new int [5];
delete bobby[5];

但在C？我猜它做malloc和free

but in C? I'm guessing it has do with malloc and free

推荐答案

的删除和添加，我必须作出结构的动态数组？如果是这样，这将是最简单的语法来做到这一点？事情是这样的C ++ code 的

如果你不知道，你永远不会比项目X数量更多或至少检查，以确保你没有超过你的计划是最大的。然后你可以使用你的静态数组。

Not if you know that you will never have more than x amount of items or at least check to make sure you aren't exceeding what you planned was the max. Then you can use your static array.

添加只需要你有一个跟踪有多少项目是数组中的变量：

Adding only requires you to have a variable that keeps track of how many items are in the array:

void add_item(struct orderSlip *p,struct orderSlip a,int * num_items)
{
   if ( *num_items < MAX_ITEMS )
   {
      p[*num_items] = a;
      *num_items += 1;
   }
}

这一个静态数组删除将需要一个用于循环，将移动项目它上面向下和递减的项目数的整型保持轨道

Deleting from a static array would require a for loop that would move the items above it down one and decrementing the int keeping track of the number of items.

void delete_item(struct orderSlip *p,int *num_items, int item)
{
   if (*num_items > 0 && item < *num_items && item > -1)
   {
      int last_index = *num_items - 1;
      for (int i = item; i < last_index;i++)
      {
         p[i] = p[i + 1];
      }
      *num_items -= 1;
   }
}

您可以简化通过将其传递到执行工作的函数打印结构。

You could simplify printing the struct by passing it to a function that does the work.

void print(const struct orderSlip  *p);

或

void print(const struct orderslip s);

可选

void print(const struct orderslip s, FILE *fp);

或

void print(const struct orderslip *p, FILE *fp)
{
   fprintf(fp,"%s\n",p->source);
    ...
}

和

void print_all(const struct orderSlip *p, int num_items)



//global
struct orderSlip data[MAX_ITEMS];
int num_items = 0;



int main(void)
{
...
       print_all(data,num_items);                                       
       strcpy(a.source,"source 2");
       strcpy(a.destination,"destination 20");
       strcpy(a.item,"item xyz");
       a.position = 99;
       add_item(data,a,&num_items);
       print_all(data,num_items);
       delete_item(data,&num_items,0);
       print_all(data,num_items);

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