i am reading the golang tutorial: https://tour.golang.org/moretypes/10
And i am confused about how fmt.Println prints the nil value, hope you could help me out.
package main
import "fmt"
func main() {
var z []int
fmt.Println("z: ", z)
if z == nil {
fmt.Println("z is nil!")
}
fmt.Println("nil:", nil)
}
the result is:
z: []
z is nil!
nil: <nil>
Since z is a nil, why is z printed as [] but not <nil>?
thanks!
The fmt package uses reflection to determine what to print. Since the type of z is a slice, fmt uses the [] notation.
Since slices, channels, interfaces and pointers can all be nil, it helpful if fmt prints something different when it can. If you want more context, use the %v format: http://play.golang.org/p/I1SAVzlv9f
var a []int
var b chan int
var c *int
var e error
fmt.Printf("a:%#v\n", a)
fmt.Printf("b:%#v\n", b)
fmt.Printf("c:%#v\n", c)
fmt.Printf("e:%#v\n", e)
Prints:
a:[]int(nil)
b:(chan int)(nil)
c:(*int)(nil)
e:<nil>
The word nil means : not initialized.
In this case you are initializing the slide but no values have bean assigned yet.
Remember PICS (Pointers, Interfaces, CHANNELS , and SLIDES )are all already
initialized.
Related
Consider the following function:
func NilCompare[T comparable](a *T, b *T) bool {
if a == nil && b == nil {
// if both nil, we consider them equal
return true
}
if a == nil || b == nil {
// if either is nil, then they are clearly not equal
return false
}
return *a == *b
}
This function works. However, when I call it, I must supply the type, as Go cannot infer (cannot infer T) it, e.g. NilCompare[string](a, b), where a and b are *string.
If I modify T to be *comparable and a and b to be T, I get this error instead:
cannot use type comparable outside a type constraint: interface is (or embeds) comparable
I am using Go 1.19.2.
$ go version
go version go1.19.2 linux/amd64
Ironically, my IDE (GoLand 2022.2.3) believes that the above function should be inferrable.
Is there a way to make a function that take nillable comparable and make it inferrable? Or am I doing it correct, but I need to help the go function along?
Type inference just works, in this case. You simply can't infer T using literal nil, as NilCompare(nil, nil) because that doesn't really carry type information.
To test your function with nils do this:
package main
import "fmt"
func main() {
var a *string = nil
var b *string = nil
// a and b are explicitly typed
res := NilCompare(a, b) // T inferred
fmt.Println(res) // true
}
this also would work:
func main() {
// literal nil converted to *string
res := NilCompare((*string)(nil), (*string)(nil)) // T inferred
fmt.Println(res) // true
}
I wrote a bit of code that in effect does this:
package main
import "fmt"
type SomeInterface interface {
Retrieve(identifier string)
}
type SomeStruct struct {}
func (r SomeStruct) Retrieve(identifier string) {
fmt.Println("identifier ", identifier)
}
type Handler struct {
Name string
SomeObject SomeInterface
}
func main() {
var someStruct *SomeStruct
var h = Handler{
Name: "helo",
SomeObject: someStruct,
}
fmt.Printf("before %+v\r\n", h.SomeObject)
if h.SomeObject == nil {
fmt.Printf("during %+v\r\n", h.SomeObject)
}
fmt.Printf("after %+v\r\n", h.SomeObject)
}
Please can someone explain to me why the output of the above is:
before <nil>
after <nil>
I've been reading about interfaces of type nil but in this case I have assigned the interface to a pointer that hasn't been assigned so I would have thought that the interface == nil and I would see during <nil> - alas it is not the case.
An interface value is a simple data structure with two parts, a type and an underlying value. So, the interface value itself can be nil, or the interface value can exist but the underlying value can be nil. For example:
var x interface{} = nil // x is nil
var y interface{} = (interface{})(nil) // y is a interface{}, which *contains* nil
This is in some ways conceptually similar to this difference:
var x []*int = nil // x is nil
var y []*int = []*int{nil} // y is a []*int, which *contains* nil
fmt.Printf obscures the difference in the case of an interface because of the way it formats the output; you could see the difference more clearly using reflection if you wanted to.
SomeObject is not nil, it just points to SomeStruct which is nil.
i think the confusion is fmt.Printf prints <nil> for this case cuz it's following the pointer and that end result is nil.
In Go, a variable referring to an implementer of some interface can have many types. It can be of type <nil> (yes, nil can describe a type as well as a value), it can be the type of one of its implementers, or it can be the type of a pointer to one of its implementers. By default, a variable referring to an interface is of type nil. Once you've assigned something to it (other than nil itself), it will then take on the type of the thing you've assigned to it (again, either the type of one of its implementers, or a pointer to one of those types).
You can print the type of an interface variable with %T, and its value with %v:
func main() {
var i SomeInterface
fmt.Printf("%T, %v\n", i, i) // Prints <nil>, <nil>
var someStruct SomeStruct
i = someStruct
fmt.Printf("%T, %v\n", i, i) // Prints main.SomeStruct, {}
var someStructPtr *SomeStruct
i = someStructPtr
fmt.Printf("%T, %v\n", i, i) // Prints *main.SomeStruct, <nil>
}
Now, whenever you compare h.SomeObject == nil, the comparison will only be evaluated as true if both the types and values of the two operands match. In your case, the value of h.SomeObject is clearly <nil> (after all, the value of someStruct is surely <nil>, and you store its value in h.SomeObject). The type of h.SomeObject, based on what I just explaind, must be *SomeStruct. The value of nil is obviously <nil>.
However, what is the type of nil?
Well, nil can take on many types, and the compiler has to decide what type it should take on for each usage. When it comes to comparisons and assignments, it simply takes on the type of the thing it is being compared to or assigned to. For instance, if you are comparing an integer pointer to nil, then the nil in such a case will be of type *int.
But all of this has to be decided at compile time, and a variable referring to an interface can change types during runtime. So when you compare a variable referring to an interface to nil, what type does the compiler give to the nil operand in such a case? Well, it gives it the only sensible type, <nil>.
For a final example, consider the following code:
func main() {
var p *SomeStruct = nil // = nil is optional; pointers default to nil
var i SomeInterface = p
printf("%t\n", p == nil) // Prints true
printf("%t\n", p == i) // Prints true
printf("%t\n", i == nil) // Prints false
}
p == nil is true, since p is of type *SomeStruct with value <nil>, and nil (in this case) is also of type *SomeStruct with value <nil>.
p == i is true, since i is also of type *SomeStruct with value <nil> (it is simply storing the type and value of p).
However, i == nil is false, because nil, in this case, takes on the type <nil> rather than *SomeStruct.
The solution around this problem is simply to never store something of value <nil> in an interface-referring variable, except for nil itself. That way, whenever the value of the interface-referring variable is <nil>, its type will also be <nil>, and comparisons against nil will work as expected. For this reason, you often see code that looks like this:
if p == nil {
i = nil
} else {
i = p
}
I'm trying to assign a value to a field, but my program panics with runtime error: invalid memory address or nil pointer dereference.
package main
type Node struct {
Value int
}
func (n *Node) SetValue(value int) {
n.Value = value
}
func main() {
var n *Node
n.SetValue(1)
}
This is reasonable since variable is nil.
But I've fount some Go internal structs are allowed to do this, e.g. bytes.Buffer
package main
import "bytes"
import "io"
import "os"
func main() {
var b bytes.Buffer
b.Write([]byte("Hello world"))
io.Copy(os.Stdout, &b)
}
Here is the `bytes.Buffer source code
func (b *Buffer) Write(p []byte) (n int, err error) {
b.lastRead = opInvalid
m := b.grow(len(p))
return copy(b.buf[m:], p), nil
}
Is it the thing only builtin structs can do or it's possible to accomplish this in my code?
EDIT
Here is the working example. Thanks #twotwotwo for suggestion.
package main
import "fmt"
type Node struct {
Value int
}
func (n *Node) SetValue(value int) {
n.Value = value
}
func main() {
var n Node
n.SetValue(1)
fmt.Println(n.Value)
}
The crucial thing is var b bytes.Buffer doesn't get you a nil pointer, it gets you a bytes.Buffer object with all its fields initialized with their zero values (in machine terms, with zero bytes). The spec says the zero value is "false for booleans, 0 for integers, 0.0 for floats, "" for strings, and nil for pointers, functions, interfaces, slices, channels, and maps"; follow that link for more detail.
It is possible to make your own structs whose zero values work and the Go team encourages it. struct Position { x, y int } is an easy example and Effective Go gives a more realistic one. But note that that doesn't make the nil pointer work; you would still need new(Node) or var n Node to allocate the zero Node. Same for bytes.Buffer.
Another common use of zero values: wherever your users create structs of your type directly (as folks do with, say, http.Server), the zero value is the default for any fields they don't specify. It's the default in a lot of other places: what you get for a not-found map key, if you receive from a closed channel, and probably others.
This should be a gimme for someone. Why do I not get what I expect ("Error is not nil") here?
http://play.golang.org/p/s8CWQxobVL
type Goof struct {}
func (goof *Goof) Error() string {
return fmt.Sprintf("I'm a goof")
}
func TestError(err error) {
if err == nil {
fmt.Println("Error is nil")
} else {
fmt.Println("Error is not nil")
}
}
func main() {
var g *Goof // nil
TestError(g) // expect "Error is nil"
}
This is, it turns out, a Frequently Asked Question about Go, and the short answer is that interface comparisons compare the type and the value, and (*Goof)(nil) and error(nil) have different types.
Since if err != nil is standard, you want a return value that'll work with it. You could declare var err error instead of var g *Goof: err's zero value is conveniently error(nil)
Or, if your func returns an error, return nil will return what you want.
For more background, here's the start of the FAQ's answer:
Under the covers, interfaces are implemented as two elements, a type and a value. The value, called the interface's dynamic value, is an arbitrary concrete value and the type is that of the value. For the int value 3, an interface value contains, schematically, (int, 3).
An interface value is nil only if the inner value and type are both unset, (nil, nil). In particular, a nil interface will always hold a nil type. If we store a pointer of type *int inside an interface value, the inner type will be *int regardless of the value of the pointer: (*int, nil). Such an interface value will therefore be non-nil even when the pointer inside is nil.
And == is strictly checking if the types are identical, not if a type (*Goof) implements an interface (error). Check out the original for more.
If it helps clarify, this doesn't only happen with nil: in this example, the data underlying the x and y variables is obviously 3, but they have different types. When you put x and y into interface{}s, they compare as unequal:
package main
import "fmt"
type Bob int
func main() {
var x int = 3
var y Bob = 3
var ix, iy interface{} = x, y
fmt.Println(ix == iy)
}
I am having the case in which a function with the following code:
func halfMatch(text1, text2 string) []string {
...
if (condition) {
return nil // That's the final code path)
}
...
}
is returning []string(nil) instead of nil. At first, I thought that perhaps returning nil in a function with a particular return type would just return an instance of a zero-value for that type. But then I tried a simple test and that is not the case.
Does anybody know why would nil return an empty string slice?
Nil is not a type. It is a description of the zero value for maps, chans, pointers, functions, slices, and interfaces.
When you put "nil" in your program, go gives it a type depending on the context. For the most part, you never need to explicitly type your nil. This is no exception, the compiler knows it must be a []string(nil) because the type returned is []string.
A nil string slice is a slice with no backing array and a length/capacity of zero. You may compare it to the literal "nil" and can get its length and capacity. It is a []string, just empty. If you wish to have an empty []string that is not nil, return []string{}. This creates a backing array (of length zero) and makes it no longer equivalent to nil.
I believe I know what's going on. The assert library I am using (github.com/bmizerany/assert) is using internally a reflect.DeepEqual.
The return value of func halfMatch(text1, text2 string) []string is always of type []string, but if it returns nil and is compared to a nil value via the == operator, it will return true. However, if reflect.DeepEqual is used, the type will matter and it won't consider both values the same.
playgound link with the test
I think maybe you are being confused by the output of print. (playground link)
package main
import "fmt"
func f() []string {
return nil // That's the final code path)
}
func main() {
result := f()
fmt.Printf("result = %v\n", result)
fmt.Printf("result = %v\n", result == nil)
}
Which produces
result = []
result = true
So I think the output really is nil
Go will return an enpty slice if condition is true.
There is a problem with your "test" because if you try to compare [] with nil, you get true.
I have modified your test to show you what I mean
package main
import "fmt"
func main() {
//fmt.Println(test1("") == nil)
fmt.Println(test1("")) // prints []
}
func test1(text1 string) []string {
if len(text1) > 0 {
return []string{text1}
}
return nil
}