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Why Does Go's `s[lo:hi]` Slice Exclude the Element at Index `hi`?

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Release: 2024-12-09 19:36:21
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Why Does Go's `s[lo:hi]` Slice Exclude the Element at Index `hi`?

Why Does Go Slice s[lo:hi] End at Element hi-1?

The Go slice expression s[lo:hi] creates a slice containing elements from index lo to index hi-1, inclusive. This behavior is explained by several design principles:

Pointer Arithmetic

In Go, slices are implemented as pointers to the underlying array. Using 0-indexed arrays and inclusive-exclusive slicing allows for straightforward pointer arithmetic. The element at index i in the slice is simply the element pointed to by the slice's pointer plus i.

Length of Slice

The length of a Go slice is also the position at which to slice it. This means that arr[0:len(arr)] returns arr itself. This property is convenient for extracting data from a slice after a read or write operation.

Non-Overlapping Indices

Inclusive-exclusive slicing ensures that contiguous sub-slices of an array do not overlap. For example, if arr = [1, 2, 3, 4], then arr[0:2] = [1, 2], arr[2:4] = [3, 4], and arr[0:4] = arr. This property simplifies operations that require partitioning or manipulation of sub-slices.

Example

Consider the following Go code:

func consecutiveSlices(ints []int) [][]int {
    ret := make([][]int, 0)
    i, j := 0, 1
    for j < len(ints) {
        if ints[j] != ints[j-1] + 1 {
            ret = append(ret, ints[i:j])
            i = j
        }
    }
    ret = append(ret, ints[i:j])
    return ret
}
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This function splits a slice of integers into consecutive sub-slices based on the difference between each element and its predecessor. The inclusive-exclusive slicing technique allows for a simpler and more efficient implementation compared to using inclusive-inclusive slicing.

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