Problem with float and int multiplication in C

Question

I'm using the online compiler https://www.onlinegdb.com/ and in the following code when I multiply 2.1 with 100 the output becomes 209 instead of 210.

#include<stdio.h>
#include <stdint.h>

int main() 
{
    float x = 1.8;

    x = x + 0.3;

    int coefficient = 100;

    printf("x: %2f\n", x);

    uint16_t y = (uint16_t)(x * coefficient);

    printf("y: %d\n", y);

    return 0;
}

Where am I doing wrong? And what should I do to obtain 210?

I tried to all different type casts still doesn't work.

Casting floating point to integer rounds down. Use `round` function explicitly. — hyde, Dec 11 '22 at 20:25
Floating point numbers are stored in a binary format. 1.8 and .3 are both repeating fractions in binary (1.11001100... and .0100110011..., respectively) and so lose precision being stored slightly less than the wanted decimal value. — Mark Tolonen, Dec 11 '22 at 20:26
`float x = 1.8;` gave you 1.7999999523162841796875. And then adding 0.3 (which is really 0.300000011920928955078125) gave you 2.099999904632568359375. And then multiplying that by 100, and throwing away the fraction, gave you 209. Try `printf(%.20f\n", x);` — Steve Summit, Dec 11 '22 at 20:40

score 1 · Answer 1 · answered Dec 11 '22 at 20:28

Floating point numbers are not exact, when you add 1.8 + 0.3, the FPU might generate a slightly different result from the expected 2.1 (by margin smaller then float Epsilon) read more about floating-point numbers representation in wiki https://en.wikipedia.org/wiki/Machine_epsilon

what happens to you is:

1.8 + 0.3 = 209.09999999... then you truncate it to int resulting in 209

you might find this question also relevant to you Why float.Epsilon and not zero? might be

Actually, floating point numbers can be perfectly exact — they're just not exact *in decimal*. — Steve Summit, Dec 11 '22 at 20:43

Eric Postpischil · Accepted Answer · 2022-12-11T22:01:44.193

The following assumes the compiler uses IEEE-754 binary32 and binary64 for float and double, which is overwhelmingly common.

float x = 1.8;

Since 1.8 is a double constant, the compiler converts 1.8 to the nearest double value, 1.8000000000000000444089209850062616169452667236328125. Then, to assign it to the float x, it converts that to the nearest float value, 1.7999999523162841796875.

x = x + 0.3;

The compiler converts 0.3 to the nearest double value, 0.299999999999999988897769753748434595763683319091796875. Then it adds x and that value using double arithmetic, which produces 2.09999995231628400205181605997495353221893310546875.

Then, to assign that to x, it converts it to the nearest float value, 2.099999904632568359375.

uint16_t y = (uint16_t)(x * coefficient);

Since x is float and coefficient is int, the compiler converts the coefficient to float and performs the multiplication using float arithmetic. This produces 209.9999847412109375.

Then the conversion to uint16_t truncates the number, producing 209.

One way to get 210 instead is to use uint16_t y = lroundf(x * coefficient);. (lroundf is declared in <math.h>.) However, to determine what the right way is, you should explain what these numbers are and why you are doing this arithmetic with them.

picchiolu · Answer 3 · 2022-12-12T00:50:56.313

0

#include<stdio.h>
#include <stdint.h>
#include <inttypes.h>

int main() 
{
    float x = 1.8;

    x = x + 0.3;

    uint16_t coefficient = 100;

    printf("x: %2f\n", x);

    uint16_t y = round(x * coefficient);

    printf("y: %" PRIu16 "\n", y);

    return 0;
}

edited Dec 12 '22 at 00:50

answered Dec 11 '22 at 20:33

picchiolu

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Problem with float and int multiplication in C

3 Answers3