Pass slice as function argument, and modify the original slice

Question

I know everything is passed by value in Go, meaning if I give a slice to a function and that function appends to the slice using the builtin append function, then the original slice will not have the values that were appended in the scope of the function.

For instance:

nums := []int{1, 2, 3}

func addToNumbs(nums []int) []int {
    nums = append(nums, 4)
    fmt.Println(nums) // []int{1, 2, 3, 4}
}

fmt.Println(nums) // []int{1, 2, 3}

This causes a problem for me, because I am trying to do recursion on an accumulated slice, basically a reduce type function except the reducer calls itself.

Here is an example:

func Validate(obj Validatable) ([]ValidationMessage, error) {
    messages := make([]ValidationMessage, 0)

    if err := validate(obj, messages); err != nil {
        return messages, err
    }

    return messages, nil
}

func validate(obj Validatable, accumulator []ValidationMessage) error {
    // If something is true, recurse
    if something {
        if err := validate(obj, accumulator); err != nil {
            return err
        }
    }

    // Append to the accumulator passed in
    accumulator = append(accumulator, message)

    return nil
}

The code above gives me the same error as the first example, in that the accumulator does not get all the appended values because they only exist within the scope of the function.

To solve this, I pass in a pointer struct into the function, and that struct contains the accumulator. That solution works nicely.

My question is, is there a better way to do this, and is my approach idiomatic to Go?

Updated solution (thanks to icza):

I just return the slice in the recursed function. Such a facepalm, should have thought of that.

func Validate(obj Validatable) ([]ValidationMessage, error) {
    messages := make([]ValidationMessage, 0)
    return validate(obj, messages)
}

func validate(obj Validatable, messages []ValidationMessage) ([]ValidationMessage, error) {
    err := v.Struct(obj)

    if _, ok := err.(*validator.InvalidValidationError); ok {
        return []ValidationMessage{}, errors.New(err.Error())
    }

    if _, ok := err.(validator.ValidationErrors); ok {
        messageMap := obj.Validate()

        for _, err := range err.(validator.ValidationErrors) {
            f := err.StructField()
            t := err.Tag()

            if v, ok := err.Value().(Validatable); ok {
                return validate(v, messages)
            } else if _, ok := messageMap[f]; ok {
                if _, ok := messageMap[f][t]; ok {
                    messages = append(messages, ValidationMessage(messageMap[f][t]))
                }
            }
        }
    }

    return messages, nil
}

Answer 1

If you want to pass a slice as a parameter to a function, and have that function modify the original slice, then you have to pass a pointer to the slice:

func myAppend(list *[]string, value string) {
    *list = append(*list, value)
}

I have no idea if the Go compiler is naive or smart about this; performance is left as an exercise for the comment section.

For junior coders out there, please note that this code is provided without error checking. For example, this code will panic if list is nil.

Answer 2

Slice grows dynamically as required if the current size of the slice is not sufficient to append new value thereby changing the underlying array. If this new slice is not returned, your append change will not be visible.

Example:

package main

import (
    "fmt"
)

func noReturn(a []int, data ...int) {
    a = append(a, data...)
}

func returnS(a []int, data ...int) []int {
    return append(a, data...)
}

func main() {
    a := make([]int, 1)
    noReturn(a, 1, 2, 3)
    fmt.Println(a) // append changes will not visible since slice size grew on demand changing underlying array

    a = returnS(a, 1, 2, 3)
    fmt.Println(a) // append changes will be visible here since your are returning the new updated slice
}

Result:

[0]
[0 1 2 3]

Note:

1. You don't have to return the slice if you are updating items in the slice without adding new items to slice

Answer 3

Slice you passed is an reference to an array, which means the size is fixed. If you just modified the stored values, that's ok, the value will be updated outside the called function.

But if you added new element to the slice, it will reslice to accommodate new element, in other words, a new slice will be created and old slice will not be overwritten.

As a summary, if you need to extend or cut the slice, pass the pointer to the slice.Otherwise, use slice itself is good enough.

Update

---

I need to explain some important facts. For adding new elements to a slice which was passed as a value to a function, there are 2 cases:

A

the underlying array reached its capacity, a new slice created to replace the origin one, obviously the origin slice will not be modified.

B

the underlying array has not reached its capacity, and was modified. BUT the field len of the slice was not overwritten because the slice was passed by value. As a result, the origin slice will not aware its len was modified, which result in the slice not modified.

Answer 4

When appending data into slice, if the underlying array of the slice doesn't have enough space, a new array will be allocated. Then the elements in old array will be copied into this new memory, accompanied with adding new data behind