• 作业三:代码复审


       我审查的刘彦麟同学的代码,我已经用CFREE运行,可以成功运行也发符合条件。并且每一行代码都很清晰,可以很清晰的看出每一行代码的意义,有很好的规范性,在复查过程中,我也学习了很多,包括对代码的规范性。我觉得代码复查时件很有用,很能学东西的事情!

    #include <stdio.h>
    #include <stdlib.h>
    #include <time.h>
    int add(int num1,int num2)
    {
     return num1+num2;
    }
    int subtraction(int num1,int num2)
    {
     return num1-num2;
    }
    int mul(int num1,int num2)
    {
     return num1*num2;
    }
    float divis(int num1,int num2)
    {
     return ((int)(((float)num1/num2)*100+0.5))/100.0;
    }
    int main()
    { 
     int type = -1;
     int data = -1;
     int choice,num1,num2,results;
     float div_result;
     char ch;
     int i,j,num3,num4; 
     printf("请输入四则运算题目的数量:
    ");
     scanf("%d",&j); 
     srand((unsigned int)time(NULL));
     for(i=0;i<j;i++){
         num3=rand()%100+1;
         num4=rand()%100+1;
           choice=rand()%8;
      switch (choice)
      {
      case 1:
        num1 = rand()%100+1;
        num2 = rand()%100+1;
        printf("%d+%d=
    ",num1,num2);
       break;
      case 2:
        num1 = rand()%100+1;
        num2 = rand()%100+1;
        printf("%d-%d=
    ",num1,num2);
       break;
      case 3:
        num1 = rand()%10;
        num2 = rand()%10;
        printf("%d*%d=
    ",num1,num2);
       break;
      case 4:
        num1 = rand()%10;
        num2 = rand()%10+1;
        printf("%d/%d=
    ",num1,num2);
       break;
        case 5:
        num1 = rand()%100+1;
        num2 = rand()%100+1;
        printf("%d/%d+%d/%d=
    ",num1,num3,num2,num4);
       break;
      case 6:
        num1 = rand()%100+1;
        num2 = rand()%100+1;
        printf("%d/%d-%d/%d=
    ",num1,num3,num2,num4);
       break;
      case 7:
        num1 = rand()%10;
        num2 = rand()%10;
        printf("%d/%d*%d/%d=
    ",num1,num3,num2,num4);
       break;
      case 8:
        num1 = rand()%10;
        num2 = rand()%10+1;
        printf("%d/%d/%d/%d=
    ",num1,num3,num2,num4);
       break;
      }
    }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/cq130201201/p/5304882.html
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