Explanation for floating point rounding effect

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#include <stdio.h>
int main() {
    printf("%.14f\n", 0.0001f * 10000000.0f);  // 1
    printf("%.14f\n", 0.001f * 1000000.0f);  // 2
    printf("%.14f\n", 0.01f * 100000.0f);  // 3
    return 0;
}

Godbolt

The output of this code is:

1000.00000000000000
1000.00006103515625
1000.00000000000000

I know, decimal fractions cannot be represented exactly with floats. But why are lines 1 and 3 calculated correctly and 2 is not? Do you have a clear explanation of what is going on here in detail?

floating-point c
2 Answers

Sometimes the cumulative roundings (3 steps each in OP samples) result in the same as the mathematical/decimal one, sometimes not. @Steve Summit, @Steve Summit

a clear explanation of what is going on here in detail?

There are 3 steps of potential rounding for each line of code:

  • Source code to float. Recall common float are of the form: some_limited_integer * 2some_power.

  • float multiplication with a rounding

  • Printing of a float rounded to 14 decimal places*1.

For printf("%.14f\n", 0.0001f * 10000000.0f); // 1

  • Code 0.0001f to a float with a value of 0.0000999999974737875163555145263671875

  • 0.0000999999974737875163555145263671875 * 10000000.0 --> 999.999974737875163555145263671875 --> rounded to nearest float --> 1000.0

  • 1000.0 --> "1000.00000000000000".

For printf("%.14f\n", 0.001f * 1000000.0f); // 2

  • Code 0.001f to a float with a value of 0.001000000047497451305389404296875

  • 0.001000000047497451305389404296875 * 1000000.0 --> 1000.000047497451305389404296875 --> rounded to nearest float --> 1000.00006103515625

  • 1000.00006103515625 --> "1000.00006103515625".

In #1 the rounding was down, then up - tending to cancel.
In #2, the rounding was up and up - resulting in noticeable double rounding effect.

Roughly, each step may inject up to a 1/2 ULP error.

Other considerations: 1) alternative rounding mode. Above uses round to nearest. 2) Weak libraries. Above assumes a quality printf().

*1 In OP's samples, there was no rounding error. In general, printing float with "%f" can round.

user_24103590

The other way of answering this is to say that you did not get one "wrong" answer and two "right" answers. You actually got three "right" answers, where "right" means, "as good as could have been expected".

Type float only gives you about 7 decimal digits of precision. So for numbers in the range of 1000, that's three places past the decimal. So change the program like this:

printf("%.3f\n", 0.0001f * 10000000.0f);
printf("%.3f\n", 0.001f * 1000000.0f);
printf("%.3f\n", 0.01f * 100000.0f);

The output is:

1000.000
1000.000
1000.000

No discrepancy, all answers apparently correct.

Or, do it in exponential notation, with one digit before the decimal and 6 after.

printf("%.6e\n", 0.0001f * 10000000.0f);
printf("%.6e\n", 0.001f * 1000000.0f);
printf("%.6e\n", 0.01f * 100000.0f);

gives

1.000000e+03
1.000000e+03
1.000000e+03

Again, all answers the same.

This might seem like "cheating": we happen to know there's some "interesting" things going on in the digits off to the right of what we can see, so isn't it wrong to suppress them in this way, and make it look like all the answers were the same? I'm not going to answer that question, other than to point out that when you're doing floating-point work, there's almost always something off there to the right, that you're rounding off and suppressing -- it's just a matter of degree.

user_39238960
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