为什么我会收到有关此代码示例的警告？什么才是正确的呢？

我正在学习一些 C，并且正在阅读scanf from 本教程 http://www.tutorialspoint.com/c_standard_library/c_function_scanf.htm其中包含以下代码块：

#include <stdio.h>

int main()
{
   char str1[20], str2[30];

   printf("Enter name: ");
   scanf("%s", &str1);

   printf("Enter your website name: ");
   scanf("%s", &str2);

   printf("Entered Name: %s\n", str1);
   printf("Entered Website:%s", str2);

   return(0);
}

但是我收到警告：

"Format specifies type 'char *' but the argument has type 'char (*)[20]'

教程有错吗？

这应该适合你：

#include <stdio.h>

int main()
{
   char str1[20], str2[30];

   printf("Enter name: ");
   scanf("%19s", str1);
         //^^   ^ Removed address operator
         //So only the right amount of characters gets read   

   printf("Enter your website name: ");
   scanf(" %29s", str2);
        //^ Added space to catch line breaks from the buffer

   printf("Entered Name: %s\n", str1);
   printf("Entered Website:%s", str2);

   return(0);
}