c - 为什么我会收到有关此代码示例的警告？什么是适当的？

Question

我正在学习一些 C，并且正在阅读来自 this tutorial 的 scanf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);
}

但是我收到的警告是:

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

教程有错吗？

Answer 1

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);
}