Change golang slice in another function - go

I have a slice, if i remove one element from it directly in a main function the length of slice would be cut by one. But do the remove in another function and called it in main, the length of the slice is still keep origin. Who can explain it for me? Thanks!
package main
import "fmt"
func main() {
a := []int{1, 2, 3, 4}
i := 0
//copy(a[i:], a[i+1:])
//a[len(a)-1] = 0
//a = a[:len(a)-1]
//fmt.Println(a) //outputs: [2 3 4], this is correct
f(a, i)
fmt.Println(a) //outputs: [2 3 4 0], this is wrong!
}
func f(a []int, i int) {
copy(a[i:], a[i+1:])
a[len(a)-1] = 0
a = a[:len(a)-1]
fmt.Println(a) //outputs: [2 3 4], here still correct
}
Go Playground Link

The slice is passed by value, so changing it in your function f won't change it in function main. You can pass by pointer, like this:
package main
import "fmt"
func main() {
a := []int{1, 2, 3, 4}
i := 0
f(&a, i)
fmt.Println(a) //outputs: [2 3 4], correct
}
func f(a *[]int, i int) {
b := *a
copy(b[i:], b[i+1:])
// The following line seems pointless, but ok...
b[len(b)-1] = 0
b = b[:len(b)-1]
fmt.Println(b) //outputs: [2 3 4], here still correct
*a = b
}
Go Playground
As suggested by #zerkms in the comments, you could also return the new slice, avoiding the use of pointers:
package main
import "fmt"
func main() {
a := []int{1, 2, 3, 4}
i := 0
a = f(a, i)
fmt.Println(a)
}
func f(a []int, i int) []int {
copy(a[i:], a[i+1:])
// The following line seems pointless, but ok...
a[len(a)-1] = 0
a = a[:len(a)-1]
fmt.Println(a) //outputs: [2 3 4], here still correct
return a
}

Not providing new solution, just trying to explain why your program is behaving the way you asked:
Let us try to understand first how the built in function ‘copy’ works

Ref: [https://golang.org/pkg/builtin/#copy]
func copy(dst, src []Type) int
The copy built-in function copies elements from a source slice into a destination slice. (As a special case, it also will copy bytes from a string to a slice of bytes.) The source and destination may overlap. Copy returns the number of elements copied, which will be the minimum of len(src) and len(dst).
Two things:
1. 
First comment the line : //a[len(a)-1] = 0
Second: As you are using the same array i.e as source and destination you are getting [2,3,4,4] as output as the destination array is {1,2,3,4} which got overwritten to {2,3,4,4(which is already present)}

you can try with different array’s to make it more clear to you

Related

How to move elements from one slice to another

package main
import (
"fmt"
)
func main() {
arr0 := []int{
1,2,3,4,5,
}
arr1 := []int{}
fmt.Println(arr0)
fmt.Println(arr1)
fmt.Println("transferring...")
transfer(&arr0, &arr1)
fmt.Println(arr0)
fmt.Println(arr1)
}
func transfer(arr0 *[]int, arr1 *[]int) {
tmp := make([]int, 0)
for i:=0;i<len(*arr0);i++ {
tmp = append(tmp, (*arr0)[i])
}
arr1 = &tmp
s := make([]int, 0)
arr0 = &s
}
For function of transfer, I intented to transfer elements of slice arr0 to slice arr1 and empty slice arr0
But it is not successful
Here is my output
[1 2 3 4 5]
[]
transferring...
[1 2 3 4 5]
[]
After transferring, I need the result below.
[]
[1 2 3 4 5]
But actually, arr0, and arr1 in the main function remain as it was!
can someone tell me why this is not ok?
I thought in the memory, it should be like this
after running transfer function
#jacobsa has given an excellent answer.
Just keep that in mind. You can achieve the same effect but with better performance. Golang offers an excellent opportunity for this.
package main
import (
"fmt"
)
func main() {
arr0 := []int{
1, 2, 3, 4, 5,
}
arr1 := []int{}
fmt.Println(arr0)
fmt.Println(arr1)
fmt.Println("transferring...")
transfer(&arr0, &arr1)
fmt.Println(arr0)
fmt.Println(arr1)
}
func transfer(arr0 *[]int, arr1 *[]int) {
*arr0, *arr1 = *arr1, *arr0
}
These two lines:
arr1 = &tmp
arr0 = &s
change the local variables arr1 and arr0 within the function. Those variables happen to be pointers, but they are just copies of the input pointers provided by main—they are not references to the input pointers.
If you changed the things the arr1 and arr0 pointers point to, rather than the pointers themselves, then you would see a change to the values provided by main:
*arr1 = tmp
*arr0 = s

Go subslice pointer reference

The result of s is [1, 2, 3], I thought that slices hold a reference to the underlying array. Isn't that the case?
package main
import (
"fmt"
)
func main() {
s := []int{1, 2, 3}
ss := s[1:]
ss = append(ss, 4)
for _, v := range ss {
v += 10
}
for i := range ss {
ss[i] += 10
}
fmt.Println(s)
}
I thought that slices hold a reference to the underlying array. Isn't that the case?
Yes it is. But you created an array with length 3 with this statement:
s := []int{1, 2, 3}
When you appended an element to ss, that required an allocation of a new, longer array. So you lost the link between ss and s with this statement:
ss = append(ss, 4)
You can verify that by running this example:
package main
import (
"fmt"
)
func main() {
s := []int{1, 2, 3}
ss := s[1:]
ss[0] += 5
ss = append(ss, 4)
ss[0] += 100
fmt.Println(s)
}
Which prints [1 7 3].
If you change your initialization of s to have a length greater than three, then no new array allocation will be required, and the link between s and ss will be maintained:
package main
import (
"fmt"
)
func main() {
s := make([]int, 3, 4)
s[0], s[1], s[2] = 1, 2, 3
ss := s[1:]
ss[0] += 5
ss = append(ss, 4)
ss[0] += 100
fmt.Println(s)
}
Output: [1 107 3]
The answer that theorizes that the problem is a range copy of the slice is incorrect, which can be shown with this example:
package main
import (
"fmt"
)
func main() {
s := make([]int, 3, 4)
s[0], s[1], s[2] = 1, 2, 3
ss := s[1:]
ss = append(ss, 4)
for i := range ss {
ss[i] += 10
}
fmt.Println(s)
}
Output: [1 12 13]
It seems like you've created a copy of slice s, and you've made changes to slice copy ss thinking that the changes would also be passed to the slice that the copy was made from.
The first for loop is also walking through the elements of slice ss, but is not actually doing anything with them, because range also creates a copy when providing you element values, so it doesn't actually do anything!
It does seem like you're trying to do the following:
Append the value 4 to slice s
Take each value from index 1 of slice to the end of the slice, and add 10
If that's the case, this should help you accomplish that:
package main
import (
"fmt"
)
func main() {
s := []int{1, 2, 3}
s = append(s, 4)
for i := range s {
if i == 0 {
continue
}
s[i] += 10
}
fmt.Println(s)
}
You can see this on the Go playground: Link

Truncate slice in place

Given a slice (not a pointer to a slice!) is there any way to truncate it in place?
The naive implementation doesn't work, because of scope:
package main
import (
"fmt"
)
func truncate(s []int, to int) []int{
s = s[:to] # <- has no effect outside this function
return s
}
func main() {
s := []int{0, 1, 2, 3, 4}
s1 := truncate(s, 3)
fmt.Printf("%v\n", s1)
fmt.Printf("%v\n", s)
}
prints
[0 1 2]
[0 1 2 3 4] # <- can we get [0 1 2] here?
Is there any way to modify the length or capacity of an existing slice, or are they immutable?
ETA: I thought this was obvious enough, but apparently not: when I ask whether it's possible to do this in place, I mean without reassigning s.
This is the way to go:
package main
import "fmt"
func main() {
s := []int{0, 1, 2, 3, 4}
s = truncate(s, 3)
fmt.Println(s) // [0 1 2]
}
func truncate(s []int, to int) []int {
return s[:to]
}
Slice is like a window to an underlying array.
The other way using pointer to the slice:
package main
import "fmt"
func main() {
s := []int{0, 1, 2, 3, 4}
truncate(&s, 3)
fmt.Println(s) // [0 1 2]
}
func truncate(s *[]int, to int) {
*s = (*s)[:to]
}
This can't be done, because slices are passed by value.
Note that as of April 2013, the Go language specification no longer refers to slices, maps, and channels as "reference types". (The behavior didn't change, just the language used to describe it.)
It can't be done with arrays either, because an array's length is part of its type.

Flipping a slice with a for loop logic error

So I am trying to write a method that takes two slices, flips both of them and then gives them to each other.
Ex.
s1 = {1,2,3,4,5}
s2 = {6,7,8,9,10}
Should return:
s1 = {10,9,8,7,6}
s2 = {5,4,3,2,1}
Here is my code:
package main
import(
"fmt"
)
func main(){
f:= [5]int{1,2,3,4,5}
h:= [5]int{6,7,8,9,10}
var sliceF []int = f[0:5]
var sliceH []int = h[0:5]
fmt.Println(reverseReverse(sliceF,sliceH))
}
func reverseReverse(first []int, second []int) ([]int, []int){
//creating temp arrays to hold the traversed arrays before swapping.
var tempArr1 []int = first
var tempArr2 []int = second
//count is used for counting up the tempArrays in the correct order in the For loops
var count int= 0
//goes through the first array and sets the values starting from the end equal to the temp array
//which increases normally from left to right.
for i :=len(first)-1; i>=0;i--{
tempArr1[count] = first[i]
fmt.Println(i)
count++
}
count =0
//same as first for loop just on the second array
for i :=len(second)-1; i>=0;i--{
tempArr2[count] = second[i]
count++
}
//trying to replace the values of the param arrays to be equal to the temp arrays
first=tempArr2
second = tempArr1
//returning the arrays
return first,second
}
When run here is the output:
4
3
2
1
0
[10 9 8 9 10]
[5 4 3 4 5]
*Not I included a print statement in the for loop to check if the index is decreasing properly.
I understand there are better ways to do this but for proof of concept I want to use a for loop.
Any help appreciated. I am new to go and tend to have java habits so I assume somehow my problem is related to that.
This can be done much simpler by realizing there's no need to actually swap the individual elements. Instead, reverse each array and swap their order. Much simpler!
func reverseReverse( a, b []int ) ([]int, []int) {
return reverse(b), reverse(a)
}
func reverse( a []int ) []int {
end := len(a) - 1
// Allocate a new array slice of the same length to copy to.
ret := make( []int, len(a) )
// Copy each element of a into ret, reversed.
for i := range a {
ret[end-i] = a[i]
}
return ret
}
With that revelation, there's little need for the very specialized reverseReverse function. Swap the order yourself.
fmt.Println(reverse(sliceH), reverse(sliceF))
Note that if you just want to take a slice of an array, it's sufficient to write sliceH []int := h[:] without specifying the start and end. The start is assumed to be 0 and the end is the end. Also note there's no need to declare the type, := takes care of that for you.
Even better, you can declare and initialize them directly.
sliceF:= []int{1,2,3,4,5}
sliceH:= []int{6,7,8,9,10}
Short answer:
tempArr1[count] = first[i]
This line is logically identical to:
first[count] = first[i]
Detailed answer:
x := [5]int{} and x := []int{} are in fact two very different assignments. In the first case x is actually a static array. In the second case x is a slice which is in fact a data structure which has a length, capacity and a pointer to the underlying array. Therefore, var tempArr1 []int = first means copy the pointer to the underlying array of first into the tempArr1, so any modification to first[i] will be reflected in tempArr1 and vice versa
For example,
package main
import "fmt"
func reverse(s []int) []int {
for i := 0; i < len(s)/2; i++ {
s[i], s[len(s)-1-i] = s[len(s)-1-i], s[i]
}
return s
}
func main() {
s1, s2 := []int{1, 2, 3, 4, 5}, []int{6, 7, 8, 9, 10}
fmt.Println(s1, s2)
s1, s2 = reverse(s2), reverse(s1)
fmt.Println(s1, s2)
}
Output:
[1 2 3 4 5] [6 7 8 9 10]
[10 9 8 7 6] [5 4 3 2 1]

How to make sure slice is using a copy of another slice and not a reference to it?

I'm learning GO and I have a theoretical problem.
How do I use a copy of slice rather than a reference to it?
package main
import "fmt"
func main() {
// slice containing 3 items
slice1 := []int{1, 2, 3}
// make an empty slice
slice2 := make([]int, 2, 5)
// create slice3 by appending int 4 to slice2
slice3 := append(slice2, 4)
// print [0 0 4]
fmt.Println(slice3)
// copy elements of slice1 onto slice2
copy(slice2, slice1)
// print [1 2 3] [1 2] [1 2 4]; how to make sure slice3 is using a copy [0 0 4]?
fmt.Println(slice1, slice2, slice3)
}
problem playground link
I have came up with a potential solution, but it is pointless as it relies on slice3 being created empty and slice2 being copied via copy() onto slice3. Is there no shortcut?
package main
import "fmt"
func main() {
// slice containing 3 items
slice1 := []int{1, 2, 3}
// make an empty slice
slice2 := make([]int, 2, 5)
// create slice3, copy slice2 and append int 4 to slice3
slice3 := make([]int, 2)
copy(slice3, slice2)
slice3 = append(slice3, 4)
// print [0 0 4]
fmt.Println(slice3)
// copy elements of slice1 onto slice2
copy(slice2, slice1)
// print [1 2 3] [1 2] [0 0 4];
fmt.Println(slice1, slice2, slice3)
}
solution playground link
EDIT:
I've read that there is a peculiar behaviour that in this naive example would work as a solution (see below). However, in any other case it would not work.
Basically, if the empty slice is created without the size of an underlying array being specified, GO's append function provides a copy of that array, otherwise, if there is room to grow, append will return a slice that references the original array.
Note: the only change is slice2 := make([]int, 2, 5) into slice2 := make([]int, 2)
package main
import "fmt"
func main() {
// slice containing 3 items
slice1 := []int{1, 2, 3}
// make an empty slice
slice2 := make([]int, 2)
// create slice3 by appending int 4 to slice2
slice3 := append(slice2, 4)
// print [0 0 4]
fmt.Println(slice3)
// copy elements of slice1 onto slice2
copy(slice2, slice1)
// print [1 2 3] [1 2] [1 2 4]; how to make sure slice3 is using a copy [0 0 4]?
fmt.Println(slice1, slice2, slice3)
}
playground with a wanted behaviour
So the question becomes: Is it possible to replicate the behaviour of the above, when the slice we are appending to points to an array with specified size and room to grow?
EDIT 2:
I think there is some confusion as to what I want to achieve.
How to get the result of the second call while passing a slice in the format used in the first call?
package main
import "fmt"
func main() {
fmt.Println("s3 references an array of s1")
worker(make([]int, 2, 5))
fmt.Println("\ns3 copies an array of s1")
worker(make([]int, 2))
}
func worker(s1 []int) {
s2 := []int{1, 2, 3}
fmt.Println(s1)
s3 := append(s1, 4)
fmt.Println(s3)
copy(s1, s2)
fmt.Println(s3)
}
playground
A few people commented that I was not clear enough last night. So I would like to clarify and provide an answer that I arrived at with the help of #CoreyOgburn and #JimB
I was learning about slices in GO and I found an inconsistency, that led me to believe I'm doing something wrong. While hardly a real life example I found the following a good example of the copy and append functionality.
package main
import "fmt"
func main() {
fmt.Println("s3 references an array of s1")
// we pass a slice of length 2 and capacity 5
worker(make([]int, 2, 5))
fmt.Println("\ns3 copies an array of s1")
// we pass a slice of lenght 2 and capacity 2
worker(make([]int, 2))
}
func worker(s1 []int) {
// create new slice for future use
s2 := []int{1, 2, 3}
fmt.Println(s1)
// create a new slice by appending a value to a slice passed into this function
s3 := append(s1, 4)
// s3 holds whatever was passed into this function + int 4, that we just appended
fmt.Println(s3)
// copy contents of s2 onto s1
copy(s1, s2)
// if s1 had spare capacity when it was passed i.e. make([]int, 2, 5) s3 will be referencing the same array as s1, hence s3 will now hold the same values as s1
// if s1 capacity was the same as its length i.e. make([]int, 2) s3 will be referencing a new array after append(), hence copy has no effect on the values of s3
fmt.Println(s3)
}
#JimB posted a comment with a link to a blog post explaining how slices work, which is a great read if you are learning the language.
What is most important in the section A possible "gotcha" there is an explanation of a 'fix' to a real life scenario, that can be extrapolated as a fix the inconsistency in my example. (create a copy of a passed slice and use that instead)
Playground

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