Checking the equality of two slices - go

Question

How can I check if two slices are equal, given that the operators == and != are not an option?
package main
import "fmt"
func main() {
s1 := []int{1, 2}
s2 := []int{1, 2}
fmt.Println(s1 == s2)
}
This does not compile with:
invalid operation: s1 == s2 (slice can only be compared to nil)

Answer 1

You should use reflect.DeepEqual()
DeepEqual is a recursive relaxation of Go's == operator.
DeepEqual reports whether x and y are “deeply equal,” defined as
follows. Two values of identical type are deeply equal if one of the
following cases applies. Values of distinct types are never deeply
equal.
Array values are deeply equal when their corresponding elements are
deeply equal.
Struct values are deeply equal if their corresponding fields, both
exported and unexported, are deeply equal.
Func values are deeply equal if both are nil; otherwise they are not
deeply equal.
Interface values are deeply equal if they hold deeply equal concrete
values.
Map values are deeply equal if they are the same map object or if they
have the same length and their corresponding keys (matched using Go
equality) map to deeply equal values.
Pointer values are deeply equal if they are equal using Go's ==
operator or if they point to deeply equal values.
Slice values are deeply equal when all of the following are true: they
are both nil or both non-nil, they have the same length, and either
they point to the same initial entry of the same underlying array
(that is, &x[0] == &y[0]) or their corresponding elements (up to
length) are deeply equal. Note that a non-nil empty slice and a nil
slice (for example, []byte{} and []byte(nil)) are not deeply equal.
Other values - numbers, bools, strings, and channels - are deeply
equal if they are equal using Go's == operator.

Answer 2

You need to loop over each of the elements in the slice and test. Equality for slices is not defined. However, there is a bytes.Equal function if you are comparing values of type []byte.
func testEq(a, b []Type) bool {
if len(a) != len(b) {
return false
}
for i := range a {
if a[i] != b[i] {
return false
}
}
return true
}

Answer 3

This is just example using reflect.DeepEqual() that is given in #VictorDeryagin's answer.
package main
import (
"fmt"
"reflect"
)
func main() {
a := []int {4,5,6}
b := []int {4,5,6}
c := []int {4,5,6,7}
fmt.Println(reflect.DeepEqual(a, b))
fmt.Println(reflect.DeepEqual(a, c))
}
Result:
true
false
Try it in Go Playground

Answer 4

If you have two []byte, compare them using bytes.Equal. The Golang documentation says:
Equal returns a boolean reporting whether a and b are the same length and contain the same bytes. A nil argument is equivalent to an empty slice.
Usage:
package main
import (
"fmt"
"bytes"
)
func main() {
a := []byte {1,2,3}
b := []byte {1,2,3}
c := []byte {1,2,2}
fmt.Println(bytes.Equal(a, b))
fmt.Println(bytes.Equal(a, c))
}
This will print
true
false

Answer 5

And for now, here is https://github.com/google/go-cmp which
is intended to be a more powerful and safer alternative to reflect.DeepEqual for comparing whether two values are semantically equal.
package main
import (
"fmt"
"github.com/google/go-cmp/cmp"
)
func main() {
a := []byte{1, 2, 3}
b := []byte{1, 2, 3}
fmt.Println(cmp.Equal(a, b)) // true
}

Answer 6

You cannot use == or != with slices but if you can use them with the elements then Go 1.18 has a new function to easily compare two slices, slices.Equal:
Equal reports whether two slices are equal: the same length and all elements equal. If the lengths are different, Equal returns false. Otherwise, the elements are compared in increasing index order, and the comparison stops at the first unequal pair. Floating point NaNs are not considered equal.
The slices package import path is golang.org/x/exp/slices. Code inside exp package is experimental, not yet stable. It will be moved into the standard library in Go 1.19 eventually.
Nevertheless you can use it as soon as Go 1.18 (playground)
sliceA := []int{1, 2}
sliceB := []int{1, 2}
equal := slices.Equal(sliceA, sliceB)
fmt.Println(equal) // true
type data struct {
num float64
label string
}
sliceC := []data{{10.99, "toy"}, {500.49, "phone"}}
sliceD := []data{{10.99, "toy"}, {200.0, "phone"}}
equal = slices.Equal(sliceC, sliceD)
fmt.Println(equal) // true
If the elements of the slice don't allow == and !=, you can use slices.EqualFunc and define whatever comparator function makes sense for the element type.

Answer 7

In case that you are interested in writing a test, then github.com/stretchr/testify/assert is your friend.
Import the library at the very beginning of the file:
import (
"github.com/stretchr/testify/assert"
)
Then inside the test you do:
func TestEquality_SomeSlice (t * testing.T) {
a := []int{1, 2}
b := []int{2, 1}
assert.Equal(t, a, b)
}
The error prompted will be:
Diff:
--- Expected
+++ Actual
## -1,4 +1,4 ##
([]int) (len=2) {
+ (int) 1,
(int) 2,
- (int) 2,
(int) 1,
Test: TestEquality_SomeSlice

Answer 8

Thought of a neat trick and figured I'd share.
If what you are interested in knowing is whether two slices are identical (i.e. they alias the same region of data) instead of merely equal (the value at each index of one slice equals the value in the same index of the other) then you can efficiently compare them in the following way:
foo := []int{1,3,5,7,9,11,13,15,17,19}
// these two slices are exactly identical
subslice1 := foo[3:][:4]
subslice2 := foo[:7][3:]
slicesEqual := &subslice1[0] == &subslice2[0] &&
len(subslice1) == len(subslice2)
There are some caveats to this sort of comparison, in particular that you cannot compare empty slices in this way, and that the capacity of the slices isn't compared, so this "identicality" property is only really useful when reading from a slice or reslicing a strictly narrower subslice, as any attempt to grow the slice will be affected by the slices' capacity. Still, it's very useful to be able to efficiently declare, "these two huge blocks of memory are in fact the same block, yes or no."

Answer 9

To have a complete set of answers: here is a solution with generics.
func IsEqual[A comparable](a, b []A) bool {
// Can't be equal if length differs
if len(a) != len(b) {
return false
}
// Empty arrays trivially equal
if len(a) == 0 {
return true
}
// Two pointers going towards each other at every iteration
left := 0
right := len(a) - 1
for left < right {
if a[left] != b[left] || a[right] != b[right] {
return false
}
left++
right--
}
return true
}
Code uses strategy of "two pointers" which brings runtime complexity of n / 2, which is still O(n), however, twice as less steps than a linear check one-by-one.