I Go, I assumed slices were passed by reference, but this seems to work for values
but not for the array itself. For example, If I have this struct:
l := Line{
Points: []Point{
Point{3, 4},
},
}
I can define a variable, which gets passed a reference to the struct's slice
slice := l.Points
And then if I modify it, the original struct referenced by the variable
is going to reflect those modifications.
slice[0].X = 1000
fmt.Printf(
"This value %d is the same as this %d",
slice[0].X,
l.Points[0].X,
)
This differs from the behavior of arrays which, I assume, are passed by value.
So, for example, if I had defined the previous code using an array:
l := Line{
Points: [1]Point{
Point{3, 4},
},
}
arr := l.Points
arr[0].X = 1000
fmt.Println(arr.[0].X != s.Points[0].X) // equals true, original struct is untouched
Then, the l struct wouldn't have been modified.
Now, if I want to modify the slice itself I obviously cannot do this:
slice = append(slice, Point{99, 100})
Since that would only redefine the slice variable, losing the original reference.
I know I can simply do this:
l.Points = append(l.Points, Point{99, 100})
But, in some cases, it is more convenient to have another variable instead of having
to type the whole thing.
I tried this:
*slice = append(*slice, Point{99, 100})
But it doesn't work as I am trying to dereference something that apparently is not a pointer.
I finally tried this:
slice := &l.Points
*slice = append(l.Points, Point{99, 100})
And it works, but I am not sure what is happening. Why is the value of slice not overwritten? How does append works here?
Let's dispense first with a terminology issue. The Go language specification does not use the word reference the way you are using it. Go does however have pointers, and pointers are a form of reference. In addition, slices and maps are kind of special as there's some underlying data—the array underneath a slice, or the storage for a map—that may or may not already exist or be created by declaring or defining a variable whose type is slice of T or map[T1]T2 for some type T or type-pair T1 and T2.1
We can take your usage of the word reference to mean explicit pointer when talking about, e.g.:
func f1(p *int) {
// code ...
}
and the implied pointer when talking about:
func f2(m map[T1]T2) { ... }
func f3(s []T) { ... }
In f1, p really is a pointer: it thus refers to some actual int, or is nil. In f2, m refers to some underlying map, or is nil. In f3, s refers to some underlying array, or is nil.
But if you write:
l := Line{
Points: []Point{
Point{3, 4},
},
}
then you must have written:
type Line struct {
// ... maybe some fields here ...
Points []Point
// ... maybe more fields here ...
}
This Line is a struct type. It is not a slice type; it is not a map type. It contains a slice type but it is not itself one.
You now talk about passing these slices. If you pass l, you're passing the entire struct by value. It's pretty important to distinguish between that, and passing the value of l.Points. The function that receives one of these arguments must declare it with the right type.
For the most part, then, talking about references is just a red herring—a distraction from what's really going on. What we need to know is: What variables are you assigning what values, using what source code?
With all of that out of the way, let's talk about your actual code samples:
l.Points = append(l.Points, Point{99, 100})
This does just what it says:
Pass l.Points to append, which is a built-in as it is somewhat magically type-flexible (vs the rest of Go, where types are pretty rigid). It takes any value of type []T (slice of T, for any valid type T) plus one or more values of type T, and produces a new value of the same type, []T.
Assigns the result to l.Points.
When append does its work, it may:
receive nil (of the given type): in this case, it creates the underlying array, or
receive a non-nil slice: in this case, it writes into the underlying array or discards that array in favor of a new larger-capacity array as needed.2
So in all cases, the underlying array may have, in effect, just been created or replaced. It's therefore important that any other use of the same underlying array be updated appropriately. Assigning the result back to l.Points updates the—presumably one-and-only—slice variable that refers to the underlying array.
We can, however, break these assumptions:
s2 := l.Points
Now l.Points and s2 both refer to the (single) underlying array. Operations that modify that underlying array will, at least potentially, affect both s2 and l.Points.
Your second example is itself OK:
*slice = append(*slice, Point{99, 100})
but you haven't shown how slice itself was declared and/or assigned-to.
Your third example is fine as well:
slice := &l.Points
*slice = append(l.Points, Point{99, 100})
The first of these lines declares-and-initializes slice to point to l.Points. The variable slice therefore has type *[]Point. Its value—the value in slice, that is, rather than that in *slice—is the address of l.Points, which has type []Point.
The value in *slice is the value in l.Points. So you could write:
*slice = append(*slice, Point{99, 100})
here. Since *slice is just another name for l.Points, you can also write:
l.Points = append(*slice, Point{99, 100})
You only need to use *slice if there's some reason that l.Points is not available,3 but you may use *slice if that's more convenient. Reading *slice reads l.Points and updating *slice updates l.Points.
1To see what I mean by may or may not be created here, consider:
var s []int
vs:
var s = []int{42}
The first leaves s == nil while the second creates an underlying array with the capacity to hold the one int value 42, holding the one int value 42, so that s != nil.
2It's not clear to me whether there is a promise never to write on an existing slice-array whose capacity is greater than its current length, but not sufficient to hold the final result. That is, can append first append 10 objects to the existing underlying array, then discover that it needs a bigger array and expand the underlying array? The difference is observable if there are other slice values referring to the existing underlying array.
3Here, a classic example would occur if you have reason to pass l.Points or &l.Points to some existing (pre-written) function:
If you need pass l.Points—the slice value—to some existing function, that existing function cannot change the slice value, but could change the underlying array. That's probably a bad plan, so if it does do this, make sure that this is OK! If it only reads the slice and underlying array, that's a lot safer.
If you need to pass &l.Points—a value that points to the slice value—to some existing function, that existing function can change both the slice, and the underlying array.
If you're writing a new function, it's up to you to write it in whatever manner is most appropriate. If you're only going to read the slice and underlying array, you can take a value of type []Point. If you intend to update the slice in place, you should take a value of type *[]Point—pointer to slice of Point.
Append returns a new slice that may modify the original backing array of the initial slice. The original slice will still point to the original backing array, not the new one (which may or may not be in the same place in memory)
For example (playground)
slice := []int{1,2,3}
fmt.Println(len(slice))
// Output: 3
newSlice := append(slice, 4)
fmt.Println(len(newSlice))
// Output: 4
fmt.Println(len(slice))
// Output: 3
While a slice can be described as a "fat pointer to an array", it is not a pointer and therefore you can't dereference it, which is why you get an error.
By creating a pointer to a slice, and using append as you did above, you are setting the slice the pointer points to to the "new" slice returned by append.
For more information, check out Go Slice Usage And Internals
Your first attempt didn't work because slices are not pointers, they can be considered reference types. Append will modify the underlying array if it has enough capacity, otherwise it returns a new slice.
You can achieve what you want with a combination of your two attempts.
playground
l := Line{
Points: []Point{
Point{3, 4},
},
}
slice := &l.Points
for i := 0; i < 100; i++ {
*slice = append(*slice, Point{99 + i, 100 + i})
}
fmt.Println(l.Points)
I know that this might be sacrilegious, but, for me, it is useful to think of slices
as structs.
type Slice struct {
len int
cap int
Array *[n]T // Pointer to array of type T
}
Since in languages like C, the [] operator is also a dereferencing operator, we can think that every time we are accessing a slice, we are actually dereferencing the underlying array and assigning some value to it. That is:
var s []int
s[0] = 1
Might be thought of as equivalent to (in pseudo-code):
var s Slice
*s.Array[0] = 1
That is why we can say that slices are "pointers". For that reason, it can modify its underlying array like this:
myArray := [3]int{1,1,1}
mySlice := myArray[0:1]
mySlice = append(mySlice, 2, 3) // myArray == mySlice
Modifying mySlice also modifies myArray, since the slice stores a pointer to the array and, on appending, we are dereferencing that pointer.
This behavior, nonetheless, is not always like this. If we exceed the capacity of the original array, a new array is created and the original array is left untouched.
myArray := [3]int{1,1,1}
mySlice := myArray[0:1]
mySlice = append(mySlice, 2, 3, 4, 5) // myArray != mySlice
The confusion arises when we try to treat the slice itself as an actual pointer. Since we can modify an underlying array by appending to it, we are led to believe that in this case:
sliceCopy := mySlice
sliceCopy = append(sliceCopy, 6)
both slices, slice and sliceCopy are the same, but they are not. We have to explicitly pass a reference to the memory address of the slice (using the & operator) in order to modify it. That is:
sliceAddress := &mySlice
*sliceAddress = append(mySlice, 6) // or append(*sliceAddress, 6)
See also
https://forum.golangbridge.org/t/slice-pass-as-value-or-pointer/2866/4
https://blog.golang.org/go-slices-usage-and-internals
https://appliedgo.net/slices/
As far as I understand, types slice and map are similar in many ways in Go. They both reference (or container) types. In terms of abstract data types, they represent an array and an associative array, respectively.
However, their behaviour is quite different.
var s []int
var m map[int]int
While we can use a declared slice immediately (append new items or reslice it), we cannot do anything with a newly declared map. We have to call make function and initialize a map explicitly. Therefore, if some struct contains a map we have to write a constructor function for the struct.
So, the question is why it is not possible to add some syntaсtic sugar and both allocate and initialize the memory when declaring a map.
I did google the question, learnt a new word "avtovivification", but still failing to see the reason.
I am not talking about struct literal. Yes, you can explicitly initialize a map by providing values such as m := map[int]int{1: 1}. However, if you have some struct:
package main
import (
"fmt"
)
type SomeStruct struct {
someField map[int]int
someField2 []int
}
func main() {
s := SomeStruct{}
s.someField2 = append(s.someField2, -1) // OK
s.someField[0] = -1 // panic: assignment to entry in nil map
fmt.Println(s)
}
It is not possible to use a struct immediately (with default values for all fields). One has to create a constructor function for SomeStruct which has to initialize a map explicitly.
While we can use a declared slice immediately (append new items or reslice it), we cannot do anything with a newly declared map. We have to call make function and initialize a map explicitly. Therefore, if some struct contains a map we have to write a constructor function for the struct.
That's not true. Default value–or more precisely zero value–for both slices and maps is nil. You may do the "same" with a nil map as you can do with a nil slice. You can check length of a nil map, you can index a nil map (result will be the zero value of the value type of the map), e.g. the following are all working:
var m map[int]int
fmt.Println(m == nil) // Prints true
fmt.Println(len(m)) // Prints 0
fmt.Println(m[2]) // Prints 0
Try it on the Go Playground.
What you "feel" more about the zero-value slice is that you may add values to it. This is true, but under the hood a new slice will be allocated using the exact make() builtin function that you'd have to call for a map in order to add entries to it, and you have to (re)assign the returned slice. So a zero-value slice is "no more ready for use" than a zero-value map. append() just takes care of necessary (re)allocation and copying over. We could have an "equivalent" addEntry() function to which you could pass a map value and the key-value pairs, and if the passed map is nil, it could allocate a new map value and return it. If you don't call append(), you can't add values to a nil slice, just as you can't add entries to a nil map.
The primary reason that the zero value for slices and maps is nil (and not an initialized slice or map) is performance and efficiency. It is very often that a map or slice value (either variable or a struct field) will never get used, or not right away, and so if they would be allocated at declaration, that would be a waste of memory (and some CPU) resources, not to mention it gives more job to the garbage collector. Also if the zero value would be an initialized value, it would often be insufficient (e.g. a 0-size slice cannot hold any elements), and often it would be discarded as you add new elements to it (so the initial allocation would be a complete waste).
Yes, there are cases when you do want to use slices and maps right away, in which cases you may call make() yourself, or use a composite literal. You may also use the special form of make() where you supply the (initial) capacity for maps, avoiding future restructuring of the map internals (which usually requires non-negligible computation). An automatic non-nil default value could not guess what capacity you'd require.
You can! What you're looking for is:
package main
import "fmt"
func main() {
v := map[int]int{}
v[1] = 1
v[2] = 2
fmt.Println(v)
}
:= is declare and assign, where as var is simply declare.
What's the difference about below ?
type Demo struct {s string}
func getDemo1()([]*Demo) // 1
func getDemo2()([]Demo) // 2
Is there any memory difference between getDemo1 and getDemo2?
I'm going to answer this, despite my better judgement to just send OP to the tour and documentation/specification. Mostly because of this:
Is there any memory difference between getDemo1 and getDemo2?
The answer to this specific question depends on how you utilize the slice. Go is Pass by value, so passing struct values around copies them. For instance, consider the following example.
https://play.golang.org/p/VzjYXwUy0EI
d1 := getDemo1()
d2 := getDemo2()
for _, v := range d1 {
// v is of type *Demo, so this modifies the value in the slice
v.s = "same"
}
fmt.Println(d1)
for _, v := range d2 {
// v is of type Demo, and is a COPY of the struct in the slice, so the original is not modified
v.s = "same"
}
So as to the memory question, obviously using *Demo, which returns a copy of the pointer in the range (effectively a uint64) as opposed to returning a copy of a Demo (the entire struct and all it's fields) would use less memory. BUT, you can still index directly to the array to avoid copies, except when you pass individual items in the slice around.
That said, passing the slice itself around, the two types have no difference in overhead. A slice is an abstraction of an array, and the slice itself that gets passed around is merely a slice header, which would be the same memory footprint regardless what type the slice contains.
BTW, the paradigm for modifying the values in the case of []Demo is:
for i, _ := range d2 {
d2[i].s = "same"
}
In Go, reflect.SliceOf() makes a Type representing a slice of the given Type:
SliceOf returns the slice type with element type t. For example, if t represents int, SliceOf(t) represents []int.
However, I already have a Type for []int but want to get a Type for int. Is there an easy way to do that? (Note, I'm using int as an example. In reality, all I know is I have a slice, and I need to find what Type each element of the slice is.)
I'm trying to populate a slice of bool, int, float, or string, from a []string using reflection... here's the relevant piece:
numElems := len(req.Form["keyName"])
if structField.Kind() == reflect.Slice && numElems > 0 {
slice := reflect.MakeSlice(structField.Type(), numElems, numElems)
for i := 0; i < numElems; i++ {
// I have some other code here to fill out the slice
}
}
But in order to fill out the slice, I need to know the type of the slice I'm filling out...
In your case, you already have the element type: structField.Type(). You can use reflect.New(t).Elem() to get an "editable" reflect.Value of a type. Once you've populated that value, you can call slice.Index(i).Set(...) to assign that value into the resulting slice.
To answer the letter of your question though, if you do have a slice and need to populate it, say you have a reflect.Type of a []int, then you can call .Elem() to get the reflect.Type for int.
See the Type documentation for the methods you can call on a Type.
I'm getting this return value from a function call in the "reflect" package:
< map[string]string Value >.
Wondering if I can access the actual map inside the return value and if so, how?
EDIT:
So this is where I'm making the call which returns the Value object.
It returns [< map[string]string Value >] to which I grab the first object in that array. However, I'm not sure how to convert [< map[string]string Value >] into a regular map.
view_args := reflect.ValueOf(&controller_ref).MethodByName(action_name).Call(in)
Most reflect Value objects can be converted back to a interface{} value using the .Interface() method.
After obtaining this value, you can assert it back to the map you want. Example (play):
m := map[string]int{"foo": 1, "bar": 3}
v := reflect.ValueOf(m)
i := v.Interface()
a := i.(map[string]int)
println(a["foo"]) // 1
In the example above, m is your original map and v is the reflected value. The interface value i, acquired by the Interface method is asserted to be of type map[string]int and this value is used as such in the last line.
To turn the value in a reflect.Value into an interface{}, you use iface := v.Interface(). Then, to access that, you use a type assertion or type switch.
If you know you're getting a map[string]string the assertion is simply m := iface.(map[string]string). If there's a handful of possibilities, the type switch to handle them all looks like:
switch item := iface.(type) {
case map[string]string:
fmt.Println("it's a map, and key \"key\" is", item["key"])
case string:
fmt.Println("it's a string:", item)
default:
// optional--code that runs if it's none of the above types
// could use reflect to access the object if that makes sense
// or could do an error return or panic if appropriate
fmt.Println("unknown type")
}
Of course, that only works if you can write out all the concrete types you're interested out in the code. If you don't know the possible types at compile time, you have to use methods like v.MapKeys() and v.MapIndex(key) to work more with the reflect.Value, and, in my experience, that involves a long time looking at the reflect docs and is often verbose and pretty tricky.