C math calculation not working as expected

Question

I have the following in a program (part of a much larger function, but this is the relevant testing bit):

int test = 100 + (100 * (9 / 100));
sprintf (buf, "Test: %d\n\r", test);
display_to_pc (buf, player);

Basically it amounts to:

x = a + (a * (b / 100))

Where a is a given figure, b is a percentage modifier, and x is the result (the original plus a percentage of the original)... I hope that makes sense.

It gives me:

Test: 100

I thought the math in my head might be wrong, but I've checked several calculators and even the expression evaluator in my IDE, and they all give me the expected result of 109 for the first expression.

Can anyone enlighten me as to what I'm missing here?

Thanks much. :)

Answer 1

int test = 100 + (100 * (9 / 100));

9/100 = 0

0 * 100 = 0

100 + 0 = 100

Answer 2

The answer of 9/10 gets truncated to 0, and then you multiply that by 100, and then add 100 to it.

Answer 3

You're using integer math.

9/100 = 0.

100 + (100 * (0) = 100.

Your calculators use floating point math, and can handle decimals appropriately.

Answer 4

you are using integer division so 9 / 100 is zero so test = 100 + 0 = 100

Answer 5

Replace

int test = 100 + (100 * (9 / 100));

with

int test = 100 + (100 * 9 / 100);
// x = a + (a * b / 100)

and it will work as expected. 9 / 100 is performed using integer division; the nearest integer to .09 is 0 (and 0 * 100 is still 0).

Doing the multiplication first results in 900 / 100, which gives you the 9 that you were expecting.

If you need more than integer precision, you may need to go the floating point route.

Answer 6

Your mistake is that 9 / 100 is interpreted as integer division, and evaluates to 0 and not 0.09.

You can write 9 / 100.0 instead, or rearrange the expression.

Answer 7

As others have pointed out, the problem is that you are doing integer arithmethic. You need to do the calculation as floating point by casting the variable to double or using a double constant, then allow truncation to give you just the integer portion.

x = a + (a * (b / 100.0));

or

x = a + (a * ((double)b / 100)));

Answer 8

Change:

int test = 100 + (100 * (9 / 100));

to

int test = (int)(100 + (100 * (9 / 100.0)));

and see if it works as expected.

EDIT: wow. Lots of answers while I was typing. Most of them will work, too.

Answer 9

You should be using floating point arithmetic so 9/100 becomes 0.09.

int test = 100.0 + (100.0 * 9.0 / 100.0);

Answer 10

Use Daniel L's answer if you will only be working with expressions that will result in whole integer values. If the values you'll be working with are less clean, use double literals instead of integer literals:

int test = 100.0 + (100.0 * (9.0 / 100.0));

Answer 11

You are using integers for your variable types, so the inner most expression (9 / 100) will result in a 0. Then the next expression would be 100 * 0, which is 0, and finally 100 + 0...

To get your desired result try using float instead:

float test = 100.0 + (100.0 * (9.0 / 100.0));

printf("result: %0.2f\n", test);