Question

我正在编写一个基于文本的冒险游戏，以改进编码，我用C语言编写，因为那是我在学校使用的。我一直得到错误game.c：28：15：错误：被叫对象'room2'不是函数或函数指针。有关如何改进代码的任何提示？

#include <stdio.h>

int c;
void start(), begin(), room2();

int main(){
   start();
   return 0;
}

void start(){
   printf("Welcome to the Adventure Game! Press s to start\n");
   c = getchar();
   if(c=='s'){begin();}
}

void begin(){
   printf("Text here involving describing a room and directions the player can take, if they press n they will go north\n");
   c = getchar();
   if(c=='n'){room2();}/*This is where I get the error*/
}

void room2(){
   printf("Describes room2 and the player's surroundings, etc\n");
}

Answer 1

尝试将每个函数的定义移动到它所调用的行之上，如下所示：

#include <stdio.h>

int c;
void start(), begin(), room2();

void room2(){
   printf("Describes room2 and the player's surroundings, etc\n");
}

void begin(){
   printf("Text here involving describing a room and directions the player can take, if they press n they will go north\n");
   c = getchar();
   if(c=='n'){room2();}/*This is where I get the error*/
}

void start(){
   printf("Welcome to the Adventure Game! Press s to start\n");
   c = getchar();
   if(c=='s'){begin();}
}

int main(){
   start();
   return 0;
}

仍有问题将c更改为char

C错误：被调用对象不是函数或函数指针

1 个答案: