c - 在输出平均数时，程序呈现段错误 11

Question

这里有两组代码看起来彼此没有什么不同，但其中一组生成警告和段错误11。

    //the program which turns out alright
    #include<stdio.h>
    int main()
    {int a,b,c,d;float f;
     scanf("%d %d %d",&a,&b,&c);
     d=a+b+c;
     f=(a+b+c)*1.0/3.0;
     printf("%d %.2f\n",d,f);
     return 0;
     }

   //the program that comes along small issues
   //format specifies type 'int *' but the argument has type 'int' [-Wformat]
   //After inputting 3 integers, in mac terminal console, it showed: segmentation fault:11
     #include<stdio.h>
    int main(void)
    {
      int a,b,c = 0;
      float average = 0.0f;
      printf("Please enter three integers you wish to calculate their average:");
      scanf("%d %d %d", a,b,c); //从input把数据放到arithmatic counter里面

      average = (float)(a+b+c)/3.0f;

      printf("\nThe average of the three numbers is:%.2f", average);
      return 0;
      }

Answer 1

Here are two sets of codes seem no different to each other,

您需要仔细查看。

参见scanf()语句:

• scanf("%d %d %d",&a,&b,&c); 正确
• scanf("%d %d %d",a,b,c); 是错误的。您的编译器已经就此警告您。

scanf() 期望针对 %d 提供的参数是指向整数的指针 - 为它们提供 int(基本上，任何其他与预期类型不兼容)将调用 undefined behavior 。段错误是副作用之一。