标签:形式 ext line 错误 插入 应用 问题分析 匹配 else
#include<stdio.h> main() { int l,w,h; printf("请输入箱子的长,宽,高;\n"); scanf("%d%d%d",&l,&w,&h); if(l==w&&l==h&&w==h) printf("该箱子是正方体\n"); else printf("该箱子是长方体\n"); }
#include<stdio.h>
main()
{
int a,b,c,x;
float sum;
printf("打印纸 18 yuan/本 墨盒 132yuan/个 光盘 4.5yuan/张\n");
scanf("%d%d%d",&a,&b,&c);
sum=18*a+132*b+4.5*c;
x=sum/100;
switch(x)
{
case 0:printf("sum=%.2f\n",sum); break;
case 1:printf("sum=%.2f\n",sum*0.95); break;
case 2:printf("sum=%.2f\n",sum*0.94); break;
case 3:printf("sum=%.2f\n",sum*0.93); break;
case 4:printf("sum=%.2f\n",sum*0.92); break;
case 5:printf("sum=%.2f\n",sum*0.90); break;
}
}
#include<stdio.h> main() { int year,month,days; printf("please enter year and month:\n"); scanf("%d%d",&year,&month); switch(month) { case 2:if((year%4==0&&(year%100==!0))||year%400==0) days=29; else days=28; break; case 1: case 3: case 5: case 7: case 8: case 10: case 12:days=31; break; case 4: case 6: case 9: case 11:days=30; break; } printf("天数为:%d",days); }
#include<stdio.h> int main() { int x,n,y; float sum=0.0; printf("请选择:1,日用品 2,文具 3,食品\n"); scanf("%d",&x); switch(x) { case 1:printf("请选择:1.牙刷(3.5YUAN/支) 2.牙膏(6.2yuan/支)\n"); printf(" 3.肥皂(2yuan/块) 4.毛巾(8.6yuan/条)\n"); scanf("%d",&y); printf("数量?"); scanf("%d",&n); switch(y) { case 1:sum=3.5*n;break; case 2:sum=6.2*n;break; case 3:sum=2*n;break; case 4:sum=8.6*n;break; } break; case 2:printf("请选择:1.笔(3YUAN/支) 2.笔记本(1.2yuan/支)\n"); printf(" 3.文件夹(12yuan/块) 4.文具盒(8.6yuan/条)\n"); scanf("%d",&y); printf("数量?"); scanf("%d",&n); switch(y) { case 1:sum=3*n;break; case 2:sum=1.2*n;break; case 3:sum=12*n;break; case 4:sum=8.6*n;break; } break; case 3:printf("请选择:1.白糖(3.6YUAN/支) 2.盐(1yuan/支)\n"); printf(" 3.饼(2yuan/块) 4.方便面(3.6yuan/条)\n"); scanf("%d",&y); printf("数量?"); scanf("%d",&n); switch(y) { case 1:sum=3.6*n;break; case 2:sum=1*n;break; case 3:sum=2*n;break; case 4:sum=3.6*n;break; } break; } printf("总计:%.2f yuan\n",sum); return 0; }
#include<stdio.h> int main() { double x=1000/3.0; double y=x-333.0; double z=3*y-1.0; printf("x=%lf\n",x); printf("y=%lf\n",y); printf("z=%lf\n",z); if(z==0) printf("z==0.\n"); else printf("z不等于0.\n"); return 0; }
#include<stdio.h> int main() { int num=20; if(5<num&&num<10) printf("%d in range (5,10)!\n",num); else printf("%d out of range (5,10)!\n",num); }
不能够实现重复的运算。。。。使用了 do...while 结构实现了循环
#include<stdio.h> main() { int x,y,z,a,b,c,d,m; do{ printf("选择计算类型:\n 1--------加法\n 2--------减法\n 3--------乘法\n 4--------除法\n"); scanf("%d",&z); printf("z=%d\n",z); printf("请输入计算的两个数:"); scanf("%d%d",&x,&y); a=x+y; b=x-y; c=x*y; d=x/y; switch(z) { case 1:printf("%d+%d=%d",x,y,a); break; case 2:printf("%d-%d=%d",x,y,b); break; case 3:printf("%d*%d=%d",x,y,c); break; case 4: printf("%d/%d=%d",x,y,d); break; default:printf("输出错误!"); } printf("继续输入请按 1 \n"); scanf("%d",&m); } while(m==1); }
通过以上几个实验,让我对switch,,case结构的使用更加熟练,更加理解switch,,case结构的特点,在编写实验的过程中,也暴露出了自己一些常见的错误,打错没有,小错不断,这也说明了自身的编写习惯还有待进步,在最后的计算机的编写中,更是体现了这一点,在最开始的编写中,经过了很多遍的调试才成功,而且最初的计算器只有最基础的计算能力,经过不断的改写和自己找资料,使得计算器实现了重复的计算,在这个过程中,感觉收获了很多,这种类似的实验很有意义
标签:形式 ext line 错误 插入 应用 问题分析 匹配 else
原文地址:https://www.cnblogs.com/wace2020/p/12593615.html