为什么我会得到这个code例子警告？什么是正确的？ [英] Why am I getting warnings about this code example? What is proper?

问题描述

我学习一些C，并从的 scanf函数阅读了>其中包含下列code座本教程：

I'm learning some C, and was reading up on scanf from this tutorial where the following code block is included:

#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);
}

不过，我得到的警告：

However I get warnings that:

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

是教程错了？

推荐答案

这应该为你工作：

#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);
}

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