嘿,我很难理解结构如何作为指针传递给函数,因为我想稍后在项目中修改结构。因此,我正在创建一个有房间的游戏,此功能将创建一个二维数组。我要传递struct room和struct World,因为我将每个房间都存储在2d数组World struct中的绳子中。当我打印出每个房间时,它什么也没打印出来,我认为问题出在我的指针上。
所以我的第一个函数在struct数组中,它将保存所有房间,并将每个房间初始化为数组。然后,我尝试打印出每个房间号,但是什么也没打印。
第二个功能创建了我的2d世界,它将是3x3世界。在世界电线上放了一个房间
struct room
{
int roomNum;
int creaturesTotal;
char description[50];
char cleanOrdirty[20];
};
int main()
{
int MaxRooms = 8;
int rows = 2;
int cols = 2;
//creaste each room
struct room Room1;
struct room Room2;
struct room Room3;
struct room Room4;
struct room Room5;
struct room Room6;
struct room Room7;
struct room Room8;
struct room Room9;
struct PC player;
player.respect = 40;
//Store them into a array
struct room AllRooms[MaxRooms];
AllRooms[0] = Room1;
AllRooms[1] = Room2;
AllRooms[2] = Room3;
AllRooms[3] = Room4;
AllRooms[4] = Room5;
AllRooms[5] = Room6;
AllRooms[6] = Room7;
AllRooms[7] = Room8;
AllRooms[8] = Room9;
struct room World[rows][cols];
createRooms(AllRooms);
//create the world
createWorld(AllRooms,rows,cols,World);
struct PC cords[rows][cols];
player = startPointPC(&player,rows,cols,&World,&cords);
}
void createRooms(struct room* AllRooms[])
{
int MaxRooms = 8;
int number = 1;
for(int i = 0;i <= MaxRooms; i++)
{
AllRooms[i]->roomNum = number;
number++;
}
for(int i = 0;i <= MaxRooms; i++)
{
printf("%d",AllRooms[i]->roomNum);
}
}
//creates the 2d world
void createWorld(struct room* AllRooms[], int rows, int cols, struct room* World[rows][cols])
{
int counter = 0;
for(int i = 0; i <= rows;i++)
{
for(int j = 0; j <= cols; j++)
{
World[i][j] = AllRooms[counter];
counter++;
printf("\nWorld %d", World[i][j]->roomNum);
}
}
}
struct PC startPointPC(struct PC* player,int rows, int cols, struct room*
World[rows][cols],struct PC* cords[rows][cols])
{
int x = 0;
int y = 0;
player->x = x;
player->y = y;
//cords[x][y] = player;
printf("\nYou are starting in Room %d",World[x][y]->roomNum);
return player;
}
答案 0 :(得分:2)
此声明
AllRooms[8] = Room9;
访问数组之外的内存,该内存的元素范围为[0, MaxRooms).
此外,像这样的任务
AllRooms[0] = Room1;
可以调用未定义的行为,因为对象Room1尚未初始化,并且其整数数据成员可以具有陷阱值。
您创建了一个一维数组
struct room AllRooms[MaxRooms];
并将其作为参数传递给函数createRooms
createRooms(AllRooms);
表达式中的数组指示符(极少数例外)将转换为指向其第一个元素的指针。
所以实际上这个电话
createRooms(AllRooms);
等同于
createRooms(&AllRooms [0]);
并且参数的类型为struct room *
。
但是,您声明的函数具有结构类型room* AllRooms[]
的参数,该参数由编译器调整为类型struct room **
。
编译器应发出一条错误消息,表明没有从一种类型的参数到另一种类型的函数参数的隐式转换。
此外,该函数应在main中使用之前声明。
也在函数中这样循环
for(int i = 0; i <= MaxRooms; i ++)
您使用的索引范围不正确。有效范围是[0, MaxRooms)
因此,至少应通过以下方式声明和定义函数
void createRooms( struct room AllRooms[] )
{
int MaxRooms = 8;
int number = 1;
for(int i = 0;i < MaxRooms; i++)
{
AllRooms[i].roomNum = number;
number++;
}
for(int i = 0;i < MaxRooms; i++)
{
printf("%d",AllRooms[i].roomNum);
}
}
使用函数的局部变量MaxRooms
int MaxRooms = 8;
而不是传递指定数组大小的相应参数。
函数createWorld存在类似问题。
应该像这样声明和定义
void createWorld( struct room AllRooms[], int rows, int cols, struct room World[rows][cols])
{
int counter = 0;
for(int i = 0; i < rows;i++)
{
for(int j = 0; j < cols; j++)
{
World[i][j] = AllRooms[counter];
counter++;
printf("\nWorld %d", World[i][j].roomNum);
}
}
}