C语言 (C 编程) 用户名和密码识别
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(C Programming) User name and Password Identification
提问by smnthLAW
Why do I get format '%s' expects argument of type 'char*'? How should I fix the problem?
为什么我得到format '%s' expects argument of type 'char*'？我应该如何解决问题？
Here are my codes:
这是我的代码：
char UserName[] = "iluvcake";
scanf("%s", &UserName);
printf("Please enter your password: \n");
char PassWord[] = "Chocolate";
scanf("%s", &PassWord);
//if...else statement to test if the input is the correct username.
if (UserName == "iluvcake")
{
if (PassWord == "Chocolate"){
printf("Welcome!\n");
}
}else
{
printf("The user name or password you entered is invalid.\n");
}
采纳答案by anishsane
- scanf for %s takes a char array/pointer, not pointer to it. drop the
&from thescanfstatements. - You cannot compare strings with
==. Usestrcmp.
- scanf for %s 需要一个字符数组/指针，而不是指向它的指针。
&从scanf语句中删除。 - 您不能将字符串与
==. 使用strcmp.
回答by K Scott Piel
&UserName is a pointer to an array of char (i.e., a char**). You should use
&UserName 是指向字符数组（即字符**）的指针。你应该使用
scanf( "%s", UserName );
回答by Maky
#include<stdio.h>
#include<conio.h>
#include<string.h>
main(){
char name[20];
char password[10];
printf("Enter username: ");
scanf("%s",name);
printf("Enter password: ");
scanf("%s",password);
if (strcmp(name, "Admin") == 0 && strcmp(password, "pass") == 0)
printf("Access granted\n");
else printf("Access denied\n");
getch();
}
:)
:)
回答by Johnny Mnemonic
Must be
必须是
scanf("%s", UserName);
scanf("%s", PassWord);
because UserNameand PassWordare pointers to chararrays.
因为UserName和PassWord是指向char数组的指针。
