Why pointer of Array/Slice change in Go if len doesnt change?

Question

I made a program to learn more about arrays and slices, in this example I just change the value of a fix position. When I look to the pointer address it changed after each change. Why this happen?

Source:

func main() {
    tstSlice2()

}

func tstSlice2() {
    var meuSlice = make([]int, 1, 2)
    meuSlice[0] = 1
    printSlice(meuSlice, "meuSlice")
    meuSlice[0] = 2
    printSlice(meuSlice, "meuSlice")
    meuSlice[0] = 3
    printSlice(meuSlice, "meuSlice")
}

func printSlice(meuSlice []int, sliceName string) {
    fmt.Printf("%v> %v - %v - %v - %p \n", sliceName, len(meuSlice), cap(meuSlice), meuSlice, &meuSlice)
}

Output:

▶ go run .\src\array_e_slices\slice.go    
meuSlice> 1 - 2 - [1] - 0xc0420503e0  
meuSlice> 1 - 2 - [2] - 0xc042050440  
meuSlice> 1 - 2 - [3] - 0xc042050480

MichalG · Accepted Answer · 2018-07-19 12:11:47Z

2

What you are passing in printSlice function is not a slice but a slice header, that is describing the underlying array. When you pass a slice to a function, a copy of the header is made, since in go elements are always passed by value.

In a function call, the function value and arguments are evaluated in the usual order. After they are evaluated, the parameters of the call are passed by value to the function and the called function begins execution. The return parameters of the function are passed by value back to the calling function when the function returns.

If you want to inspect what's inside of this header, check reflect.SliceHeader type.

An example of the code:

package main

import (
    "fmt"
    "reflect"
    "unsafe"
)

func main() {
    tstSlice2()

}

func tstSlice2() {
    var meuSlice = make([]int, 1, 2)
    meuSlice[0] = 1
    printSlice(meuSlice, "meuSlice")
    meuSlice[0] = 2
    printSlice(meuSlice, "meuSlice")
    meuSlice[0] = 3
    printSlice(meuSlice, "meuSlice")
}

func printSlice(meuSlice []int, sliceName string) {
    hdr := (*reflect.SliceHeader)(unsafe.Pointer(&meuSlice))
    fmt.Printf("%v> %v - %v - %v - %p - array: %v\n", sliceName, len(meuSlice), cap(meuSlice), meuSlice, &meuSlice, hdr.Data)

}

Output:

meuSlice> 1 - 2 - [1] - 0xc42000a060 - array: 842350559504
meuSlice> 1 - 2 - [2] - 0xc42000a0a0 - array: 842350559504
meuSlice> 1 - 2 - [3] - 0xc42000a0e0 - array: 842350559504

Go playground: https://play.golang.org/p/x2OoIHgU9St

You can find more information in the blog post Go Slices: usage and internals (especially Slice internals section)

A slice is a descriptor of an array segment. It consists of a pointer to the array, the length of the segment, and its capacity (the maximum length of the segment).

What you are passing to your function is a value of this struct, not the underlying array itself. To do that, you'd need to use *[]int argument, as mentioned in other answers.

edited Jul 19, 2018 at 12:11

answered Jul 18, 2018 at 13:55

MichalG

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2 Comments

Peter Over a year ago

"elements are by default passed by value". Not by default, always. golang.org/ref/spec#Calls

MichalG Over a year ago

true, that was a bit of a mental shortcut. Indeed, in Go arguments are always passed as values. What I meant was that we may choose to pass the pointer, that will be passed as a value, but it will point to the same address. I edited the answer to add this clarification.

Alexander Trakhimenok · Accepted Answer · 2018-07-18 13:47:02Z

1

Because you are passing each time a copy of slice. While underlying array is not copied the slice does. I think in principal you can treat slice as a struct:

type Slice struct {
   pointer PointerToArray
   StartIndex int
   Len int
}

Code below should print the same pointer:

func tstSlice2() {
    var meuSlice = make([]int, 1, 2)
    meuSlice[0] = 1
    printSlice(&meuSlice, "meuSlice")
    meuSlice[0] = 2
    printSlice(&meuSlice, "meuSlice")
    meuSlice[0] = 3
    printSlice(&meuSlice, "meuSlice")
}

func printSlice(meuSlice *[]int, sliceName string) {
    fmt.Printf("%v> %v - %v - %v - %p \n", sliceName, len(*meuSlice), cap(*meuSlice), *meuSlice, meuSlice)
}

answered Jul 18, 2018 at 13:47

Alexander Trakhimenok

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