为什么我的代码返回的数字比应返回的数字小？

Question

So I am doing an assignment for a programming course and I am having trouble with one line of my code.

the answer should be (if my math is correct) 40 but the program returns 39.

example: 18.33 * 40 *52 = 38126.4

I then use this code

double wage = 0;
wage = wage * 40 * 52;
wage = wage - floor(wage);
wage = wage * 100;
printf(" %d cents", (int)wage);

But the program returns 39. If I don't change it to int it is 40.0000000

What is happening?

Answer 1

Floating point numbers are incapable of representing every real number to infinite precision. And 40.0 is no exception. Instead, if your calculation is done in a correct and numerically stable way, you'll get the closest number a double can represent.

Functions like printf make the output pretty by considering how close the operand they received is to a given number.

So your result may be 39.99998 , the closest number a double can represent to your correct result. And it is shown as 40.0000000 .

However casting to an int performs truncation. So it becomes 39 .

For your purposes (you seem to be dealing in money), an approach which will work best is to get rid of floating point numbers completely. You only need two integers. One for cents (which grows only up to 100), and one for dollars ¹ .

_{¹ I assume dollars and cents because of the multiplication by 100 in your code. Naturally it applies for any currency}