CODEWITHSUNDEEP
functiongoparameter-passingmutability
Omar Hayam
Omar Hayam

Reputation: 21

Strange mutation of map when passed value vs when passed by reference (Golang)

In the first case I pass a map to by value: package main

import (
    "fmt"
    "time"
)

func timeMap(z map[string]interface{}) {
    z["updated_at"] = time.Now()
}

func main() {
    foo := map[string]interface{}{
        "Matt": 42,
    }
    timeMap(foo)
    fmt.Println(foo)
}

The output is a muted map: 

map[updated_at:2009-11-10 23:00:00 +0000 UTC Matt:42]

In the second case the code is almost identical but for passing by reference:

package main

import (
    "fmt"
    "time"
)

func timeMap(z *map[string]interface{}) {
    (*z)["updated_at"] = time.Now()
}

func main() {
    foo := map[string]interface{}{
        "Matt": 42,
    }
    timeMap(&foo)
    fmt.Println(foo)
}

Obviously, the result differs:

map[Matt:42 updated_at:2009-11-10 23:00:00 +0000 UTC]

My expectations were the following:

Why does it happen so?

Upvotes: 0

Views: 1694

Answers (1)

Yerken
Yerken

Reputation: 1942

There is no such thing as passing by reference in Go. Whenever you pass anything (pointer, slice header, map) it is always passed by value. The question is what exactly is being passed by value (i.e. what is the actual value of the type).

When u pass a map, you pass a copy of the pointer to its header, which contains a set pointers to the buckets, as in the implementation of the HashTable. https://github.com/golang/go/blob/master/src/runtime/hashmap.go#L106

Therefore it rarely makes sense to pass a pointer to the map, because the operation of copying a map header pointer is extremely cheap.

Now why the order is different, this is simply due to internal implementation of the map, ranging over the keys occurs in a random fashion. Again this is just an implementation details.

EDIT:

As @icza correctly pointed out, passing a map is actually passing a copy of a pointer to the map header, not the map header itself. Sorry for the confusion

Upvotes: 5

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