SliceTricks

Helmut Kemper edited this page Jun 28, 2017 · 23 revisions
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Since the introduction of the append built-in, most of the functionality of the container/vector package, which was removed in Go 1, can be replicated using append and copy.

Here are the vector methods and their slice-manipulation analogues:

AppendVector

a = append(a, b...)

Copy

b = make([]T, len(a))
copy(b, a)
// or
b = append([]T(nil), a...)

Cut

a = append(a[:i], a[j:]...)

Delete

a = append(a[:i], a[i+1:]...)
// or
a = a[:i+copy(a[i:], a[i+1:])]

Delete without preserving order

a[i] = a[len(a)-1] 
a = a[:len(a)-1]

NOTE If the type of the element is a pointer or a struct with pointer fields, which need to be garbage collected, the above implementations of Cut and Delete have a potential memory leak problem: some elements with values are still referenced by slice a and thus can not be collected. The following code can fix this problem:

Cut

copy(a[i:], a[j:])
for k, n := len(a)-j+i, len(a); k < n; k++ {
	a[k] = nil // or the zero value of T
}
a = a[:len(a)-j+i]

Delete

copy(a[i:], a[i+1:])
a[len(a)-1] = nil // or the zero value of T
a = a[:len(a)-1]

Delete without preserving order

a[i] = a[len(a)-1]
a[len(a)-1] = nil
a = a[:len(a)-1]

Expand

a = append(a[:i], append(make([]T, j), a[i:]...)...)

Extend

a = append(a, make([]T, j)...)

Insert

a = append(a[:i], append([]T{x}, a[i:]...)...)

NOTE The second append creates a new slice with its own underlying storage and copies elements in a[i:] to that slice, and these elements are then copied back to slice a (by the first append). The creation of the new slice (and thus memory garbage) and the second copy can be avoided by using an alternative way:

Insert

s = append(s, 0)
copy(s[i+1:], s[i:])
s[i] = x

InsertVector

a = append(a[:i], append(b, a[i:]...)...)

Pop

x, a = a[len(a)-1], a[:len(a)-1]

Pop Back

x, a = a[0], a[1:]

Push

a = append(a, x)

Push Back

a = append([]T{ x }, a...)

Shift

x, a := a[0], a[1:]

Unshift

a = append([]T{x}, a...)

Additional Tricks

Filtering without allocating

This trick uses the fact that a slice shares the same backing array and capacity as the original, so the storage is reused for the filtered slice. Of course, the original contents are modified.

b := a[:0]
for _, x := range a {
	if f(x) {
		b = append(b, x)
	}
}

Reversing

To replace the contents of a slice with the same elements but in reverse order:

for i := len(a)/2-1; i >= 0; i-- {
	opp := len(a)-1-i
	a[i], a[opp] = a[opp], a[i]
}

The same thing, except with two indices:

for left, right := 0, len(a)-1; left < right; left, right = left+1, right-1 {
	a[left], a[right] = a[right], a[left]
}