#include<stdio.h>
int main()
{
int income,tax;
printf("Please input your income:");
while(scanf("%d",&income)!=-1)
{
    printf("\nIncome = %d greenbacks\n",income);
    if(income<6000)
         tax=(int)income*3/10.0;
     else
        tax=(int)income*6/10.0;
    printf("Tax = %d greenbacks\n",tax);
    printf("\nPlease input your income:");
}
return 0;
}

什么问题，是计算结果精度问题吗！

#include<stdio.h>

int main()
{
int income;
double tax;
printf("Please input your income:");
scanf("%d",&income);
while(income!=-1)
{
    printf("\nIncome = %d greenbacks\n",income);
    if(income<6000)
         tax=income*0.3;
     else
        tax=income*0.6;
    printf("Tax = %f greenbacks\n",tax);
    printf("\nPlease input your income:");
}
return 0;
}
只共参考

if(income<6000)
    printf("Tax = %d greenbacks\n", income*0.3);
else
    printf("Tax = %d greenbacks\n", income*0.6);

这错在你输出的结果是以整型输出的  带income*0.6的结果为FLOAT型 所以答案错了
改法：
1：if(income<6000)
    printf("Tax = %f greenbacks\n", income*0.3);
else
    printf("Tax = %f greenbacks\n", income*0.6);


2:  if(income<6000)
    printf("Tax = %d greenbacks\n", (int)(income*0.3));
else
    printf("Tax = %d greenbacks\n", (int)(income*0.6));

#include<stdio.h>

int main()
{
int income,tax;
printf("Please input your income:");
while(scanf("%d",&income)!=EOF)
{
    printf("\nIncome = %d greenbacks\n",income);
    if(income<6000)
    printf("Tax = %d greenbacks\n", (int)(income*0.3));
else
    printf("Tax = %d greenbacks\n", (int)(income*0.6));
    printf("\nPlease input your income:");
}
return 0;
}
问题的主要在于强制类型的转换！在此我们强制类型转换的目的是将表达式income*0.3或者income*0.6得值转换成int型，所以必须将整个表达式用圆括号括起来。