CODEWITHSUNDEEP
go
Megidd
Megidd

Reputation: 7938

How builtin function "append" works: appending to a slice whose elements are of type interface

The section of appending to slices on the specification, mentions the following example:

var t []interface{}
t = append(t, 42, 3.1415, "foo")   // t == []interface{}{42, 3.1415, "foo"}

I'm confused here, why can we append values of int, float and string to a slice whose elements are of interface type? And why is the result of the append like that? I tried hard/long, but I don't get it.

Upvotes: 0

Views: 59

Answers (1)

user1529891
user1529891

Reputation:

Because:

all types implement the empty interface

For details read over the ref spec for interfaces.

interface is similar to Object in java where all types/classes/etc are also an Object.

You can see this effect by using reflect:

package main

import (
    "fmt"
    "reflect"
)

func main() {
    var t []interface{}
    z := append(t, "asdf", 1, 2.0)
    fmt.Println(z)

    for i := range z {
        fmt.Println(reflect.TypeOf(z[i]))
    }
}

Output:

[asdf 1 2]
string
int
float64

Why is it like this? Well, think about serialization especially of JSON objects; the types can be string, int, object, etc. This allows for deserialization without specifying a fully mapped struct (maybe you don't care and want only some of the data, etc). Basically, it allows you to have a form of "weak typing" in Go; while still being able to have strong typing.

As mentioned below, reflect identifies the type for you. When programming though, you may have to do it manually.

Upvotes: 5

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