Avoid possible overflow (theoretical) in the example #28
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Avoid possible overflow if array size is greater than (MAX_INT / 2) and searched item is one of the lasts.
Bolshem
changed the title
| @@ -3,7 +3,7 @@ def binarySearch[T: Ordering](sorted: IndexedSeq[T], target: T): Boolean = { | |||
| // If the sequence you are trying to search has no items, the item cannot be found | |||
| if (start == end) false | |||
| else { | |||
| val middle = (start + end) / 2 | |||
| val middle = start + (end - start) / 2 | |||
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A more concise way used in JDK: val middle = (start + end) >>> 1
JDK16 source code:
https://github.com/openjdk/jdk16/blob/4de3a6be9e60b9676f2199cd18eadb54a9d6e3fe/src/java.base/share/classes/java/util/Collections.java#L228
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Avoid possible overflow if the array size is greater than (MAX_INT / 2) and the searched item is one of the lasts.
I understand this is totally theoretical, but this seems as the right way to find a middle.