Go 1.21
Use the builtin clear
.
func main() {
m := map[string]int{"foo": 1}
clear(m)
}
It's a single function call, and has the minor but important advantage that it will delete irreflexive keys (see below for details about what this means). The standard library maps
package doesn't have maps.Clear
anymore.
Go 1.18 to 1.20
You can use maps.Clear
. The function belongs to the package golang.org/x/exp/maps
(experimental and not covered by the compatibility guarantee)
Clear removes all entries from m, leaving it empty.
Example usage:
func main() {
testMap := map[string]int{"gopher": 1, "badger": 2}
maps.Clear(testMap)
fmt.Println(testMap)
testMap["zebra"] = 2000
fmt.Println(testMap)
}
Playground: https://go.dev/play/p/qIdnGrd0CYs?v=gotip
If you don't want to depend on experimental packages, you can copy-paste the source, which is actually extremely simple:
func Clear[M ~map[K]V, K comparable, V any](m M) {
for k := range m {
delete(m, k)
}
}
IMPORTANT NOTE: just as with the builtin delete
— which the implementation of maps.Clear
uses —, this does not remove irreflexive keys from the map. The reason is that for irreflexive keys, by definition, x == x
is false. Irreflexive keys are NaN floats and every other type that supports comparison operators but contains NaN floats somewhere.
See this code to understand what this entails:
func main() {
m := map[float64]string{}
m[1.0] = "foo"
k := math.NaN()
fmt.Println(k == k) // false
m[k] = "bar"
maps.Clear(m)
fmt.Printf("len: %d, content: %v\n", len(m), m)
// len: 1, content: map[NaN:bar]
a := map[[2]float64]string{}
a[[2]float64{1.0, 2.0}] = "foo"
h := [2]float64{1.0, math.NaN()}
fmt.Println(h == h) // false
a[h] = "bar"
maps.Clear(a)
fmt.Printf("len: %d, content: %v\n", len(a), a)
// len: 1, content: map[[1 NaN]:bar]
}
Playground: https://go.dev/play/p/LWfiD3iPA8Q