aga*_*row 4 c segmentation-fault
noob问题:我正在尝试编写一个简单的菜单界面,但我不断收到分段错误,我无法弄清楚原因.
#include <stdlib.h>
#include <stdio.h>
int flush(); int add(char *name, char *password, char *type); int delete(char *name);
int edit(char *name, char *password, char *type, char *newName, char *newPassword, char *newType);
int verify(char *name, char *password);
int menu(){
int input;
char *name, *password, *type, *newName, *newPassword, *newType;
printf("MAIN MENU \n ============\n");
printf("1. ADD\n");
printf("2. DELETE\n");
printf("3. EDIT\n");
printf("4. VERIFY\n");
printf("5. Exit\n");
printf("Selection:");
scanf("%d", &input);
flush();
switch (input){
case 1:
printf("%s\n", "Enter Name:");
scanf("%s", name);
flush();
printf("%s\n", "enter password" );
scanf("%s", password);
flush();
printf("%s\n","enter type" );
scanf("%s",type);
add(name, password, type);
menu();
break;
case 2:
printf("Enter Name:" );
scanf("%s",name);
flush();
delete(name);
menu();
break;
case 3:
printf("Enter Name:\n");
scanf("%s",name);
flush();
printf("Enter Password\n");
scanf("%s", password);
flush();
printf("enter type:\n");
scanf("%s", type);
flush();
printf("enter your new username:\n");
scanf("%s",newName);
flush();
printf("enter your new password\n");
scanf("%s", newPassword);
flush();
printf("enter your new type\n");
scanf("%s",newType);
flush();
edit(name, password, type, newName, newPassword, newType);
menu();
break;
case 4:
printf("Enter Name\n");
scanf("%s",name);
flush();
printf("Enter Password\n");
scanf("%s",password);
flush();
verify(name, password);
menu();
break;
case 5:
return 0;
default:
printf("invalid input, please select from the following:\n");
menu();
}
return 0;
}
int flush(){
int ch;
while ((ch = getchar()) != EOF && ch != '\n') ;
return 0;
}
Run Code Online (Sandbox Code Playgroud)
在任何菜单选项中输入两个字段后,我得到分段错误
你需要初始化你的指针.或者,使用堆栈分配的数组.
例如,而不是char *name,做char name[20].(请注意,这会将输入限制为19个字符;如有必要,请使用更大的缓冲区.)
现在,您正在传递未初始化的指针scanf(),这有效意味着scanf()将写入未定义的内存区域.它可能在一次执行时起作用,然后在下一次执行时失败.它可能会破坏进程地址空间中其他位置的内存.
不要使用未初始化的变量,并考虑将编译器警告尽可能高地调高; 编译器可以捕获这样的错误并发出警告.