当我运行以下程序时:
#include <stdio.h>
#include <math.h>
int main()
{
double sum, increase;
long amount, j;
printf("sum = ");
scanf("%lf", &sum);
printf("increase = ");
scanf("%lf", &increase);
printf("amount = ");
scanf("%ld", &amount);
for (j = 1; j <= amount; j++)
{
sum += increase;
}
printf("%lf\n", sum);
return 0;
}
我对这些值获得以下响应:
MacBook:c benjamin$ ./test
sum = 234.4
increase = 0.000001
amount = 198038851
432.438851
MacBook:c benjamin$ ./test
sum = 234.4
increase = 0.000001
amount = 198038852
432.438851
MacBook:c benjamin$ ./test
sum = 234.4
increase = 0.000001
amount = 198038853
432.438852
我在每种情况下都将变量'amount'增加了1。
为什么会这样?
虽然代码看起来不是很有用,但我刚刚写了这个部分。我实际上想在更大的程序中使用它。
谢谢!
答案 0 :(得分:3)
这是格式化问题,而不是数字问题。如果您将printf
更改为
printf("%.12lf\n", sum);
结果看起来更符合您的预期:
sum = increase = amount = 432.438851000198
sum = increase = amount = 432.438852000198
sum = increase = amount = 432.438853000198
最后的“垃圾”是由于浮点数的精度有限。