用C表示If语句的指针

Question

我正在用C编写基于文本的游戏.在程序开始时,它会提示您是否要开始或结束游戏.但是,当我键入end或start时,会出现"Segmentation Fault"错误.这是我的代码:

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

//Variables
char *name, *answer ;

//Beginning of program
int main()
{
//Game starts and prompts you
printf ("--|| YOUR_ADVENTURE_HERE ||-- \n") ;
printf (" \n") ;
printf ("Type 'START' to start the game or 'END' to end the game. You can end the game at any point by typing it as well.") ;
scanf ("%59s, answer") ;

//If typed 'START'
if (answer = "START")
{
printf ("\n") ;
printf ("Starting game...") ;
sleep (5) ;

return 0 ;
}

//If they typed 'END' (this will be used in every scanf)
if (answer = "END")
{
system("exit") ;

return 0 ;  
}   

以下是我运行时的外观:

--|| YOUR_ADVENTURE_HERE ||-- 

Type 'START' to start the game or 'END' to end the game. You can end the game at any point by typing it as well.START
Segmentation fault

提前致谢!

Answer 1

在尝试存储任何内容之前,将内存分配给指针.

scanf ("%59s, answer") ;

应该先为内存分配 answer

answer = malloc(60);
scanf ("%59s", answer) ;

另外,在旁注中,用于strcmp比较字符串.

if (answer = "START")

应该

if (!strcmp(answer,"START"))

Answer 2

问题出在这里:

scanf ("%59s, answer");

首先,scanf需要一个参数列表来写入从输入中获得的值.你没有提供任何.相反,你只有一个长字符串.你的意思当然是:

scanf ("%59s", answer);

但即使这样也行不通.

它需要answer分配某种内存,而不是未初始化的指针.这会更好:

char answer [60];
scanf ("%59s", answer);

我没有进一步了解其他问题.