Rounding-Error: Use proper banker's rounding

[orgua]

Isn't that comparison (first line) wrong considering the following comment?

    if ((!(remainder < 0.5) || (remainder > 0.5)) && (number_.integral & 1)) {
      // exactly 0.5 and ODD, then round up
      // 1.5 -> 2, but 2.5 -> 2
      ++number_.integral;
    }

Link

I think it should be

    if (!((remainder < 0.5) || (remainder > 0.5)) && (number_.integral & 1)) {

[eyalroz]

So, this is code from Marco Paland's original repository, which I haven't touched. It does seem a bit weird and I need to take another look.

... but I don't quite understand your suggestion, e.g. why do you suggest we avoid the equality operator.

[orgua]

yeah, this is inherited from the original repo, but it's still wrong compared to the embedded comment. it came up during linting of a code-base.

Doing it like that with an extra step might fix some weird float-behavior with very small remainders in the fraction-part?!

[eyalroz]

So, I just made it Banker's rounding.

[HenryLin2000]

In fact, I found that when executing this block, there is always remainder<=0.5, because the remainder here is updated (in line 605)

printf/src/printf/printf.c

Lines 604 to 611 in f8ed5a9

	if (precision == 0U) {
	remainder = abs_number - (double) number_.integral;
	if ((!(remainder < 0.5) || (remainder > 0.5)) && (number_.integral & 1)) {
	// exactly 0.5 and ODD, then round up
	// 1.5 -> 2, but 2.5 -> 2
	++number_.integral;
	}
	}

The abs_number here is the absolute value of the original number, and the value of integral depends on whether the number is carried.
There are 3 situations at this time

1. If the previous remainder>0.5 (carry), then the current remainder is a negative number
2. If the previous remainder<0.5 (no carry), then the current remainder is between the interval [0,0.5)
3. If the previous remainder=0.5 (special case), then the current remainder is also 0.5, and rounding processing is executed.

So, there is never remainder>0.5, and the condition here is really a bit strange.