我正在尝试从多个 dyanimc 数组在 C 中创建一个对象(typedef 结构),但是我在为成员赋值时遇到了一些问题,我的代码如下:
#define MAX_SHIPS 200
typedef struct enemy {
int enemyX[MAX_SHIPS];
int enemyY[MAX_SHIPS];
int enemyDistance[MAX_SHIPS];
int enemyHealth[MAX_SHIPS];
int enemyType[MAX_SHIPS];
}enemy;
^ 定义 MAX_SHIPS 并创建结构敌人。
number_of_friends = 0;
number_of_enemies = 0;
if (number_of_ships > 1)
{
for (i=1; i<number_of_ships; i++)
{
if (IsaFriend(i))
{
friendX[number_of_friends] = shipX[i];
friendY[number_of_friends] = shipY[i];
friendHealth[number_of_friends] = shipHealth[i];
friendFlag[number_of_friends] = shipFlag[i];
friendDistance[number_of_friends] = shipDistance[i];
friendType[number_of_friends] = shipType[i];
number_of_friends++;
}
else
{
int x;
for (x = 0; x < number_of_ships; x++)
{
enemy[x].enemyX = shipX[i];
enemy[x]. enemyY = shipY[i];
enemy[x].enemyDistance = shipDistance[i];
enemy[x].enemyHealth = shipHealth[i];
enemy[x].enemyType = shipType[i];
}
此刻我得到错误int x expected an identifier
。
enemyX[number_of_enemies] = shipX[i];
enemyY[number_of_enemies] = shipY[i];
enemyHealth[number_of_enemies] = shipHealth[i];
enemyFlag[number_of_enemies] = shipFlag[i];
enemyDistance[number_of_enemies] = shipDistance[i];
enemyType[number_of_enemies] = shipType[i];
number_of_enemies++;
}
}
}
^ 我想删除/替换为创建敌人结构的代码。
最佳答案
结构是一种船舶矩阵,既笨拙又浪费。你不断地摆弄阵列只是为了与个别船只一起工作。您不必分配一大块内存来容纳最大数量的船只。您需要为敌人和友军复制整个结构。您必须将所有内容复制进和复制出结构才能使用一艘船...
相反,创建单个 Ship
结构。然后你可以只传递指针并为友军和敌舰使用相同的结构。您可以在 Ship
指针列表中保留友军和敌军舰船,而无需复制所有数据。
#include <stdio.h>
#include <stdlib.h>
/* A structure to store ships */
typedef struct {
int x;
int y;
int distance;
int health;
int type;
} Ship;
/* No matter how simple the struct, always write functions to
create and destroy it. This makes using it simpler, and it
shields your code from making future changes to the struct. */
Ship *Ship_new() {
/* Use calloc(), rather than malloc(), to guarantee everything
is initialized to 0 rather than dealing with garbage. */
return calloc(1, sizeof(Ship));
}
void Ship_destroy( Ship *ship ) {
free(ship);
}
/* Constants are easier to debug than macros */
const int MAX_FRIENDLIES = 200;
const int MAX_ENEMIES = 200;
int main() {
/* Store just a list of pointers to Ships. This is more
flexible and saves a lot of memory. */
Ship *friendlies[MAX_FRIENDLIES];
Ship *enemies[MAX_ENEMIES];
/* Make new ships for demonstration purposes */
Ship *enemy = Ship_new();
Ship *friendly = Ship_new();
/* Just to demonstrate setting values */
enemy->x = 5;
enemy->y = 10;
enemy->health = 100;
enemy->type = 5;
friendly->x = 99;
friendly->y = 23;
friendly->health = 50;
friendly->type = 10;
/* Assign them to their lists. Since it's a list of Ship *
we only need to copy the pointer, not all the data. */
friendlies[0] = friendly;
enemies[0] = enemy;
/* Make use of them. friendlies[0] and friendly point to the
same ship, not a copy. */
printf("Friendly #1 health: %d\n", friendlies[0]->health);
printf("Enemy #1 health: %d\n", enemies[0]->health);
}
关于在 C 中使用动态数组创建对象/typedef 结构,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41685147/