while循环中内容重复printf2次的解决办法
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while循环中内容之所以重复打印2次,是回车键(\n)作祟
c语言中输入字符后储存在缓冲区,输入完毕后按回车键后执行,回车键(\n)此时被视为字符,同样进入while循环验证真伪,造成printf重复执行了两次,下面以c primer plus 第七章7.12.11题为例演示解决办法
# include <stdio.h>
# define ARTICHOKE 2.05
# define BEETROOT 1.15
# define CARROT 1.09
# define DISCOUNT 0.05
# define PACKAGE1 6.5
# define PACKAGE2 14
# define PACKAGE3 0.5
//printf语句会执行两次,sum质量,sum1洋jian质量,sum2甜菜质量,sum3胡萝卜质量,sum4运费包装费,sum5未折扣钱数,sum6蔬菜钱数,sum7折后钱数
int main(void)
{
float i, j, k, sum, sum1, sum2, sum3, sum4, sum5, sum6, sum7;
char ch;
printf("please select the vegetable\na is artichoke\nb is beetroot\nc is carrot\nq is quit\n");
do
{
scanf("%c", &ch);
if(ch != 'q')
switch (ch)
{
case 'a':
{
printf("please enter the pound\n");
scanf("%f", &i);
sum1 += i;
break;
}
case 'b':
{
printf("please enter the pound\n");
scanf("%f", &j);
sum2 += j;
break;
}
case 'c':
{
printf("please enter the pound\n");
scanf("%f", &k);
sum3 += k;
break;
}
}
printf("please select the vegetable\na is artichoke\nb is beetroot\nc is carrot\nq is quit\n");
}while(ch != 'q');
sum = sum1 + sum2 + sum3;
if(sum <= 5)
sum4 = PACKAGE1;
else if(sum <= 20)
sum4 = PACKAGE2;
else if(sum > 20)
sum4 = PACKAGE2 + (sum - 20) * PACKAGE3;
sum6 = sum1 * ARTICHOKE + sum2 * BEETROOT + sum3 * CARROT;
sum5 = sum6 + sum4;
if(sum >= 100)
sum7 = sum5 - sum5 * DISCOUNT;
printf("artichoke is %f dollars/pound\nbeetroot is %f dollars/pound\ncarrot is %f dollars/pounds\n", ARTICHOKE, BEETROOT, CARROT);
if(sum > 0)
printf("the total weight is %.2f pound\n", sum);
if(sum6 > 0)
printf("the vegetable price is %.2f dollars\n", sum6);
if(sum7 > 0)
printf("the total price is %.2f dollars\n", sum7);
if(sum5 > 100)
printf("the discount dollars is %.2f\n", sum5 - sum7);
if(sum4 > 0)
printf("the package is %.2f dollars\n", sum4);
return 0;
}
运行结果示例
可见please select the vegetable执行了两次
解决问题的办法则是在do while循环中加入
while(getchar()!='\n')
continue;
//消耗掉\n(具体内容视情况修改)
以下是修改后的代码段以及运行结果
# include <stdio.h>
# define ARTICHOKE 2.05
# define BEETROOT 1.15
# define CARROT 1.09
# define DISCOUNT 0.05
# define PACKAGE1 6.5
# define PACKAGE2 14
# define PACKAGE3 0.5
//sum质量,sum1洋jian质量,sum2甜菜质量,sum3胡萝卜质量,sum4运费包装费,sum5未折扣钱数,sum6蔬菜钱数,sum7折后钱数
int main(void)
{
float i, j, k, sum, sum1, sum2, sum3, sum4, sum5, sum6, sum7;
char ch;
printf("please select the vegetable\na is artichoke\nb is beetroot\nc is carrot\nq is quit\n");
do
{
scanf("%c", &ch);
if(ch != 'q')
switch (ch)
{
case 'a':
{
printf("please enter the pound\n");
scanf("%f", &i);
sum1 += i;
break;
}
case 'b':
{
printf("please enter the pound\n");
scanf("%f", &j);
sum2 += j;
break;
}
case 'c':
{
printf("please enter the pound\n");
scanf("%f", &k);
sum3 += k;
break;
}
}
while(getchar() != '\n')
continue;
printf("please select the vegetable\na is artichoke\nb is beetroot\nc is carrot\nq is quit\n");
}while(ch != 'q');
sum = sum1 + sum2 + sum3;
if(sum <= 5)
sum4 = PACKAGE1;
else if(sum <= 20)
sum4 = PACKAGE2;
else if(sum > 20)
sum4 = PACKAGE2 + (sum - 20) * PACKAGE3;
sum6 = sum1 * ARTICHOKE + sum2 * BEETROOT + sum3 * CARROT;
sum5 = sum6 + sum4;
if(sum >= 100)
sum7 = sum5 - sum5 * DISCOUNT;
printf("artichoke is %f dollars/pound\nbeetroot is %f dollars/pound\ncarrot is %f dollars/pounds\n", ARTICHOKE, BEETROOT, CARROT);
if(sum > 0)
printf("the total weight is %.2f pound\n", sum);
if(sum6 > 0)
printf("the vegetable price is %.2f dollars\n", sum6);
if(sum7 > 0)
printf("the total price is %.2f dollars\n", sum7);
if(sum5 > 100)
printf("the discount dollars is %.2f\n", sum5 - sum7);
if(sum4 > 0)
printf("the package is %.2f dollars\n", sum4);
return 0;
}
问题得到解决
欢迎指正,如有其他更好的办法欢迎探讨交流
标签:解决办法,printf2,sum7,dollars,sum5,while,printf,sum,define 来源: https://blog.csdn.net/weixin_46409567/article/details/122514781