Improve algorithm to count digits in Long #1548
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🎉 |
Egorand
commented
Nov 14, 2024
Comment on lines
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| private val DigitCountToLargestValue = longArrayOf( | ||
| -1, // Every value has more than 0 digits. | ||
| 9L, // For 1 digit (index 1), the largest value is 9. | ||
| 99L, | ||
| 999L, | ||
| 9999L, | ||
| 99999L, | ||
| 999999L, | ||
| 9999999L, | ||
| 99999999L, | ||
| 999999999L, | ||
| 9999999999L, | ||
| 99999999999L, | ||
| 999999999999L, | ||
| 9999999999999L, | ||
| 99999999999999L, | ||
| 999999999999999L, | ||
| 9999999999999999L, | ||
| 99999999999999999L, | ||
| 999999999999999999L, // For 18 digits (index 18), the largest value is 999999999999999999. | ||
| Long.MAX_VALUE, // For 19 digits (index 19), the largest value is MAX_VALUE. | ||
| ) |
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This is a slight deviation from the original table proposed in the article:
private val PowersOfTen = longArrayOf(
0,
10,
100,
1000,
10000,
100000,
1000000,
10000000,
100000000,
1000000000,
10000000000,
100000000000,
1000000000000,
10000000000000,
100000000000000,
1000000000000000,
10000000000000000,
100000000000000000,
1000000000000000000
)@swankjesse & I discovered that the original table wouldn't work for the Long.MAX_VALUE test case, as it has more digits than 10^18, and 10^19 is outside of the range of Long. The code instead compares against largest numbers that can be represented with a given digit count, which allowed us to fit another entry into the table!
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I saw that and it's a great improvement!
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The original code made it up into kotlinx.io. Do you want to send them a PR, too? |
Sure, done! Kotlin/kotlinx-io#413 |
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Based on "Down Another Rabbit Hole" by @romainguy - thanks!