I encountered weird behaviour in go code today: when I append elements to slice in loop and then try to create new slices based on the result of the loop, last append overrides slices from previous appends.
In this particular example it means that sliceFromLoop j,g and h slice's last element are not 100,101 and 102 respectively, but...always 102!
Second example - sliceFromLiteral behaves as expected.
package main
import "fmt"
func create(iterations int) []int {
a := make([]int, 0)
for i := 0; i < iterations; i++ {
a = append(a, i)
}
return a
}
func main() {
sliceFromLoop()
sliceFromLiteral()
}
func sliceFromLoop() {
fmt.Printf("** NOT working as expected: **\n\n")
i := create(11)
fmt.Println("initial slice: ", i)
j := append(i, 100)
g := append(i, 101)
h := append(i, 102)
fmt.Printf("i: %v\nj: %v\ng: %v\nh:%v\n", i, j, g, h)
}
func sliceFromLiteral() {
fmt.Printf("\n\n** working as expected: **\n")
i := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
fmt.Println("initial slice: ", i)
j := append(i, 100)
g := append(i, 101)
h := append(i, 102)
fmt.Printf("i: %v\nj: %v\ng: %v\nh:%v\n", i, j, g, h)
}
link to play.golang:
https://play.golang.org/p/INADVS3Ats
After some reading, digging and experimenting I found that this problem is originated in slices referencing the same underlaying array values and can be solved by copying slice to new one before appending anything, however it looks quite... hesitantly.
What's the idomatic way for creating many new slices based on old ones and not worrying about changing values of old slices?
Don't assign append to anything other than itself.
As you mention in the question, the confusion is due to the fact that append both changes the underlying array and returns a new slice (since the length might be changed). You'd imagine that it copies that backing array, but it doesn't, it just allocates a new slice object that points at it. Since i never changes, all those appends keep changing the value of backingArray[12] to a different number.
Contrast this to appending to an array, which allocates a new literal array every time.
So yes, you need to copy the slice before you can work on it.
func makeFromSlice(sl []int) []int {
result := make([]int, len(sl))
copy(result, sl)
return result
}
func main() {
i := make([]int, 0)
for ii:=0; ii<11; ii++ {
i = append(i, ii)
}
j := append(makeFromSlice(i), 100) // works fine
}
The slice literal behavior is explained because a new array is allocated if the append would exceed the cap of the backing array. This has nothing to do with slice literals and everything to do with the internals of how exceeding the cap works.
a := []int{1,2,3,4,5,6,7}
fmt.Printf("len(a) %d, cap(a) %d\n", a, len(a), cap(a))
// len(a) 7, cap(a) 7
b := make([]int, 0)
for i:=1; i<8, i++ {
b = append(b, i)
} // b := []int{1,2,3,4,5,6,7}
// len(b) 7, cap(b) 8
b = append(b, 1) // any number, just so it hits cap
i := append(b, 100)
j := append(b, 101)
k := append(b, 102) // these work as expected now
If you need a copy of a slice, there's no other way to do it other than, copying the slice. You should almost never assign the result of append to a variable other than the first argument of append. It leads to hard to find bugs, and will behave differently depending on whether the slice has the required capacity or not.
This isn't a commonly needed pattern, but as with all things of this nature if you need to repeate a few lines of code multiple times, then you can use a small helper function:
func copyAndAppend(i []int, vals ...int) []int {
j := make([]int, len(i), len(i)+len(vals))
copy(j, i)
return append(j, vals...)
}
https://play.golang.org/p/J99_xEbaWo
There is also a little bit simpler way to implement copyAndAppend function:
func copyAndAppend(source []string, items ...string) []string {
l := len(source)
return append(source[:l:l], items...)
}
Here we just make sure that source has no available capacity and so copying is forced.
Related
I'm not sure how does append work, there are two case as follow:
case 1:
s := make([]int, 0)
s = append(s, 1)
s = append(s, 2)
s = append(s, 3)
s = append(s, 4)
s = append(s, 5)
fmt.Println(cap(s))
In this case output is 8 and this is make sense.
case 2:
s := make([]int, 0)
s = append(s, 1, 2, 3, 4, 5)
fmt.Println(cap(s))
But this case output is 6
Could anyone explain for me?
if you learn about the vector in C++,
you will know the capacity is Dynamically growing,
When you add elements, you will judge whether the capacity is enough. If not enough, the capacity will be increased(not one by one).
In
Where is the implementation of func append in Go?
we can see
CMPQ BX, DI // compare new length (2) with cap (1)
JHI $1, 225 // jump to grow code if len > cap
in offical website
func Append(slice []int, elements ...int) []int {
n := len(slice)
total := len(slice) + len(elements)
if total > cap(slice) {
// Reallocate. Grow to 1.5 times the new size, so we can still grow.
newSize := total*3/2 + 1
newSlice := make([]int, total, newSize)
copy(newSlice, slice)
slice = newSlice
}
slice = slice[:total]
copy(slice[n:], elements)
return slice
}
func Append(slice []int, items ...int) []int {
for _, item := range items {
slice = Extend(slice, item)
}
return slice
}
func Extend(slice []int, element int) []int {
n := len(slice)
if n == cap(slice) {
// Slice is full; must grow.
// We double its size and add 1, so if the size is zero we still grow.
newSlice := make([]int, len(slice), 2*len(slice)+1)
copy(newSlice, slice)
slice = newSlice
}
slice = slice[0 : n+1]
slice[n] = element
return slice
}
newSize := total*3/2 + 1
so first uses Append: 0->1->2->4->4->8(5x3/2+1)
second uses Extend: 0->1->2->4->4->6(3*2)
I was trying to make a clone of multidimensional slice, because when I have changed elements in the duplicated slice, the elements in the original one were overwritten also.
The only method that worked for me was:
duplicate := make([][]int, len(matrix))
for i := 0; i < len(matrix); i++ {
duplicate[i] = make([]int, len(matrix[0]))
for j := 0; j < len(matrix[0]); j++ {
duplicate[i][j] = matrix[i][j]
}
}
is there any other way - shorter or more efficient to achieve the same result? thanks
You can use copy for the inner loop (which should be more efficient) and range for the outer loop (which results in nicer code).
Result:
duplicate := make([][]int, len(matrix))
for i := range matrix {
duplicate[i] = make([]int, len(matrix[i]))
copy(duplicate[i], matrix[i])
}
If your goal is efficiency, it may make sense to do more allocation up front. This doesn't lead to more readable code but will lead to more efficient code if you are doing this often. This code assumes you have at least one row and that all rows are of the same length. You will need to add tests for that.
n := len(matrix)
m := len(matrix[0])
duplicate := make([][]int, n)
data := make([]int, n*m)
for i := range matrix {
start := i*m
end := start + m
duplicate[i] = data[start:end:end]
copy(duplicate[i], matrix[i])
}
Depending on what you are doing, it may make sense to make a "matrix type" that is implemented using only a single slice. A slice of slices is not the most efficient data structure, even if it is simpler to work with.
Before deciding if you need to be efficient, make sure that you are spending a lot of time doing copying using profiling. Then, after you have determined this is in fact a hotspot, start running benchmarks. See https://golang.org/pkg/testing/#hdr-Benchmarks for details.
You could roundtrip it through gob:
package main
import (
"bytes"
"encoding/gob"
"fmt"
)
func sliceCopy(in, out interface{}) {
buf := new(bytes.Buffer)
gob.NewEncoder(buf).Encode(in)
gob.NewDecoder(buf).Decode(out)
}
func main() {
a := [][]int{
{1, 2, 3}, {4, 5, 6}, {7, 8, 9},
}
var b [][]int
sliceCopy(a, &b)
fmt.Println(b)
}
https://golang.org/pkg/encoding/gob
I am using golang. Here is my code:
func main() {
quanPailie([]int{1, 2})
}
func quanPailie(nums []int) [][]int {
COUNT := len(nums)
//only one item
if COUNT == 1 {
return [][]int{nums}
}
insertItem(quanPailie(nums[:COUNT-1]), nums[COUNT-1])
return [][]int{}
}
func insertItem(res [][]int, insertNum int) {
fmt.Println("insertItem,res:", res, "insertNum", insertNum) //insertItem,res: [[1]] insertNum 2
for _, v := range res {
for i := 0; i < len(v); i++ {
fmt.Println("===before,v:", v)
c := append(v[:i], append([]int{insertNum}, v[i:]...)...)
fmt.Println("===after,v:", v)
fmt.Println("ccc", c)
}
}
}
What makes me very confused is the output:
===before,v: [1]
===after,v: [2]
Why did the value of v change? Hope someone can help me. Thanks a lot.
Go playground: https://play.golang.org/p/wITYsGpX7U
EDIT:
Thanks for icza's great help, I think I have understood this problem.
And, here is a simple code to show this issue.
func test1() {
nums := []int{1, 2, 3}
_ = append(nums[:2], 4)
fmt.Println("test1:", nums)
//nums changes because the cap is big enought, the original array is modified.
}
func test2() {
nums := []int{1, 2, 3}
c := append(nums[:2], []int{4, 5, 6}...)
fmt.Println("test2:", nums)
fmt.Println("cc:", c)
//nums dont't change because the cap isn't big enought.
//a new array is allocated while the nums still points to the old array.
//Of course, the return value of append points to the new array.
}
Go playground: https://play.golang.org/p/jBNFsCqUn3
This is the code in question:
fmt.Println("===before,v:", v)
c := append(v[:i], append([]int{insertNum}, v[i:]...)...)
fmt.Println("===after,v:", v)
You ask why v changes between the 2 Println() statements.
Because you are using the builtin append() function, quoting from its doc:
The append built-in function appends elements to the end of a slice. If it has sufficient capacity, the destination is resliced to accommodate the new elements. If it does not, a new underlying array will be allocated. Append returns the updated slice.
So if the slice you append to has enough room (capacity) to accomodate the elements you want to append, no new slice will be allocated, instead the destination slice will be re-sliced (which will use the same underlying array) and append will happen in that.
Let's check the capacity:
fmt.Println("===before,v:", v, cap(v))
c := append(v[:i], append([]int{insertNum}, v[i:]...)...)
fmt.Println("===after,v:", v, cap(v))
Output:
===before,v: [1] 2
===after,v: [2] 2
The v slice has a capacity of 2. When for loop starts, i=0, v[:i] is v[:0] which is an empty slice (but has capacity 2) and so appending 1 or 2 elements will not allocate a new array/slice, it will be done "in place". This "in place" is the 0th element of v, since v[:i] is shorthand for v[0:i]. Hence the elements will be appended starting from v[0] in the underlying array which is shared, so the element denoted by v[0] will change.
Note that slicing a slice results in a slice which shares its underlying backing array with the original slice (does not make a copy of the elements).
If you want to avoid this, use or allocate a new slice, copy original content and append to the new slice, e.g.:
src := []int{1, 2}
c := make([]int, len(src))
copy(c, src)
// Append something:
c = append(c, 3, 4)
fmt.Println(src) // [1 2] - src doesn't change
fmt.Println(c) // [1 2 3 4]
In Go, what is the difference between var s []int and s := make([]int, 0)?
I find that both works, but which one is better?
Simple declaration
var s []int
does not allocate memory and s points to nil, while
s := make([]int, 0)
allocates memory and s points to memory to a slice with 0 elements.
Usually, the first one is more idiomatic if you don't know the exact size of your use case.
In addition to fabriziom's answer, you can see more examples at "Go Slices: usage and internals", where a use for []int is mentioned:
Since the zero value of a slice (nil) acts like a zero-length slice, you can declare a slice variable and then append to it in a loop:
// Filter returns a new slice holding only
// the elements of s that satisfy f()
func Filter(s []int, fn func(int) bool) []int {
var p []int // == nil
for _, v := range s {
if fn(v) {
p = append(p, v)
}
}
return p
}
It means that, to append to a slice, you don't have to allocate memory first: the nil slice p int[] is enough as a slice to add to.
Just found a difference. If you use
var list []MyObjects
and then you encode the output as JSON, you get null.
list := make([]MyObjects, 0)
results in [] as expected.
A bit more complete example (one more argument in .make()):
slice := make([]int, 2, 5)
fmt.Printf("length: %d - capacity %d - content: %d", len(slice), cap(slice), slice)
Out:
length: 2 - capacity 5 - content: [0 0]
Or with a dynamic type of slice:
slice := make([]interface{}, 2, 5)
fmt.Printf("length: %d - capacity %d - content: %d", len(slice), cap(slice), slice)
Out:
length: 2 - capacity 5 - content: [<nil> <nil>]
What is the best way to remove items from a slice while ranging over it?
For example:
type MultiDataPoint []*DataPoint
func (m MultiDataPoint) Json() ([]byte, error) {
for i, d := range m {
err := d.clean()
if ( err != nil ) {
//Remove the DP from m
}
}
return json.Marshal(m)
}
As you have mentioned elsewhere, you can allocate new memory block and copy only valid elements to it. However, if you want to avoid the allocation, you can rewrite your slice in-place:
i := 0 // output index
for _, x := range s {
if isValid(x) {
// copy and increment index
s[i] = x
i++
}
}
// Prevent memory leak by erasing truncated values
// (not needed if values don't contain pointers, directly or indirectly)
for j := i; j < len(s); j++ {
s[j] = nil
}
s = s[:i]
Full example: http://play.golang.org/p/FNDFswPeDJ
Note this will leave old values after index i in the underlying array, so this will leak memory until the slice itself is garbage collected, if values are or contain pointers. You can solve this by setting all values to nil or the zero value from i until the end of the slice before truncating it.
I know its answered long time ago but i use something like this in other languages, but i don't know if it is the golang way.
Just iterate from back to front so you don't have to worry about indexes that are deleted. I am using the same example as Adam.
m = []int{3, 7, 2, 9, 4, 5}
for i := len(m)-1; i >= 0; i-- {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
}
}
There might be better ways, but here's an example that deletes the even values from a slice:
m := []int{1,2,3,4,5,6}
deleted := 0
for i := range m {
j := i - deleted
if (m[j] & 1) == 0 {
m = m[:j+copy(m[j:], m[j+1:])]
deleted++
}
}
Note that I don't get the element using the i, d := range m syntax, since d would end up getting set to the wrong elements once you start deleting from the slice.
Here is a more idiomatic Go way to remove elements from slices.
temp := s[:0]
for _, x := range s {
if isValid(x) {
temp = append(temp, x)
}
}
s = temp
Playground link: https://play.golang.org/p/OH5Ymsat7s9
Note: The example and playground links are based upon #tomasz's answer https://stackoverflow.com/a/20551116/12003457
One other option is to use a normal for loop using the length of the slice and subtract 1 from the index each time a value is removed. See the following example:
m := []int{3, 7, 2, 9, 4, 5}
for i := 0; i < len(m); i++ {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
i-- // -1 as the slice just got shorter
}
}
I don't know if len() uses enough resources to make any difference but you could also run it just once and subtract from the length value too:
m := []int{3, 7, 2, 9, 4, 5}
for i, s := 0, len(m); i < s; i++ {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
s--
i--
}
}
Something like:
m = append(m[:i], m[i+1:]...)
You don't even need to count backwards but you do need to check that you're at the end of the array where the suggested append() will fail. Here's an example of removing duplicate positive integers from a sorted list:
// Remove repeating numbers
numbers := []int{1, 2, 3, 3, 4, 5, 5}
log.Println(numbers)
for i, numbersCount, prevNum := 0, len(numbers), -1; i < numbersCount; numbersCount = len(numbers) {
if numbers[i] == prevNum {
if i == numbersCount-1 {
numbers = numbers[:i]
} else {
numbers = append(numbers[:i], numbers[i+1:]...)
}
continue
}
prevNum = numbers[i]
i++
}
log.Println(numbers)
Playground: https://play.golang.org/p/v93MgtCQsaN
I just implement a method which removes all nil elements in slice.
And I used it to solve a leetcode problems, it works perfectly.
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNil(lists *[]*ListNode) {
for i := 0; i < len(*lists); i++ {
if (*lists)[i] == nil {
*lists = append((*lists)[:i], (*lists)[i+1:]...)
i--
}
}
}
You can avoid memory leaks, as suggested in #tomasz's answer, controlling the capacity of the underlying array with a full slice expression. Look at the following function that remove duplicates from a slice of integers:
package main
import "fmt"
func removeDuplicates(a []int) []int {
for i, j := 0, 1; i < len(a) && j < len(a); i, j = i+1, j+1 {
if a[i] == a[j] {
copy(a[j:], a[j+1:])
// resize the capacity of the underlying array using the "full slice expression"
// a[low : high : max]
a = a[: len(a)-1 : len(a)-1]
i--
j--
}
}
return a
}
func main() {
a := []int{2, 3, 3, 3, 6, 9, 9}
fmt.Println(a)
a = removeDuplicates(a)
fmt.Println(a)
}
// [2 3 3 3 6 9 9]
// [2 3 6 9]
For reasons #tomasz has explained, there are issues with removing in place. That's why it is practice in golang not to do that, but to reconstruct the slice. So several answers go beyond the answer of #tomasz.
If elements should be unique, it's practice to use the keys of a map for this. I like to contribute an example of deletion by use of a map.
What's nice, the boolean values are available for a second purpose. In this example I calculate Set a minus Set b. As Golang doesn't have a real set, I make sure the output is unique. I use the boolean values as well for the algorithm.
The map gets close to O(n). I don't know the implementation. append() should be O(n). So the runtime is similar fast as deletion in place. Real deletion in place would cause a shifting of the upper end to clean up. If not done in batch, the runtime should be worse.
In this special case, I also use the map as a register, to avoid a nested loop over Set a and Set b to keep the runtime close to O(n).
type Set []int
func differenceOfSets(a, b Set) (difference Set) {
m := map[int]bool{}
for _, element := range a {
m[element] = true
}
for _, element := range b {
if _, registered := m[element]; registered {
m[element] = false
}
}
for element, present := range m {
if present {
difference = append(difference, element)
}
}
return difference
}
Try Sort and Binary search.
Example:
package main
import (
"fmt"
"sort"
)
func main() {
// Our slice.
s := []int{3, 7, 2, 9, 4, 5}
// 1. Iterate over it.
for i, v := range s {
func(i, v int) {}(i, v)
}
// 2. Sort it. (by whatever condition of yours)
sort.Slice(s, func(i, j int) bool {
return s[i] < s[j]
})
// 3. Cut it only once.
i := sort.Search(len(s), func(i int) bool { return s[i] >= 5 })
s = s[i:]
// That's it!
fmt.Println(s) // [5 7 9]
}
https://play.golang.org/p/LnF6o0yMJGT