I would like to increase my float value to the first multiple of 0.5 after it.
For example, if I have: float x = 3.397; I would like it to become: x = 3.5;. Or, if I have: float x = 7.895; it should become x = 8.0;.
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Adrian Mole
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ThatsSamu
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1Yes, your requirement is clear. What is not clear is what problems you have or what question you want to ask. – Seelenvirtuose Oct 21 '20 at 15:38
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3Multiply by 2, round up, divide by 2 – QBrute Oct 21 '20 at 15:39
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@Seelenvirtuose I do not know how to do what I would like to be done – ThatsSamu Oct 21 '20 at 15:41
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Perhaps `ceil(d * 2)/2`? – chux - Reinstate Monica Oct 21 '20 at 15:57
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1What should be the result of 0.5? of large values in the range where xxxx.5 is not representable? – chux - Reinstate Monica Oct 21 '20 at 15:58
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@chux-ReinstateMonica That worked! Thank you! – ThatsSamu Oct 21 '20 at 16:13
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lots of duplicates: [round BigDecimal to nearest 5 cents](https://stackoverflow.com/q/2106615/995714), [How can I round up to the nearest multiple of the specified number?](https://stackoverflow.com/q/38116385/995714), [Rounding up a number to nearest multiple of 5](https://stackoverflow.com/q/9303604/995714), [Rounding numbers to specific multiples](https://stackoverflow.com/q/10456564/995714)... – phuclv Oct 22 '20 at 16:07
1 Answers
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Quite simple, but in every detail errors are hidden.
float nextHalf(float x) {
// CORRECTED THIS: long n = Math.round(x * 2.0) + 1;
return (float) (Math.ceil(x * 2.0) / 2);
}
With corrections due to @chux-ReinstateMonica.
As rounding methods round/floor/ceil target integer units of 1, for 0.5 one must first scale the numbers by a factor 2 and afterwards divide by 2.
Of course because of the incorporated exponent in a floating point number at some large numbers the difference between successive numbers is no longer below 1.0 so then
nextHalf(x) == nextHalf(x + 1.0)
Joop Eggen
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@chux-ReinstateMonica my fault half heartedly reading the question. Corrected as no other answer – Joop Eggen Oct 22 '20 at 07:14
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With `ceil()`, which returns `double`, `long n` is unnecessarily restrictive. Perhaps `double n` and avoid the "float might overflow long" problem? – chux - Reinstate Monica Oct 22 '20 at 10:28