我可以省略make的类型吗？

When making a slice or map...

type myType []map[string]someType

v = make(myType, 1)

v[0] = make(map[string]someType)

^{https://play.golang.org/p/Af9b0n0ykT}

... I always have to specify the type twice. Can make not infer it?

This would be especially helpful when the type is an anonymous struct.

For example in Java the second mention of the type can be omitted:

HashMap<String, SomeType> v = new HashMap<>();

No, it's not possible. make requires a type as its first argument, and there are only two ways to specify a type: either as a name (eg: myType) or as a type literal (eg: []int).

In principle, you can work around this using reflect:

package main

import (
    "fmt"
    "reflect"
)

type someType struct{}

type myType []map[string]someType

func main() {
    var v = make(myType, 1)
    reflect.ValueOf(v).Index(0).Set(reflect.MakeMap(reflect.TypeOf(v).Elem()))
    fmt.Println(v)
}

But this doesn't look like a good idea at all to me.

with type myType []map[string]someType you define a slice of maps and it is different type

https://play.golang.org/p/rElNLx0H0S

package main

type someType int
type mymap map[string]someType
type myType []mymap 

func main() {
    v := make(myType, 1)
    v[0] = make(mymap)

    // HashMap<String, SomeType> v = new HashMap<>();
    var _ mymap = make(mymap)
}

runs totally fine