var cache = struct {
sync.Mutex
mapping map[string]string
} {
mapping: make(map[string]string),
}
This looks like a struct with an embedded field sync.Mutex but I can't get my head around the second set of braces. It compiles and executes but what's up? Why does the label on the make instruction matter (it does) and the comma? Thanks...
The example you have is equivalent to:
type Cache struct {
sync.Mutex
mapping map[string]string
}
cache := Cache{
mapping: make(map[string]string),
}
Except in your example you do not declare a type of Cache and instead have an anonymous struct. In your example, as oppose to my Cache type, the type is the entire
struct {
sync.Mutex
mapping map[string]string
}
So think of the second pair of braces as the
cache := Cache{
mapping: make(map[string]string),
}
part.
make is a built in function that works similarly to C's calloc() which both initialize a data structure filled with 0'd values, in Go's case, certain data structures need to be initialized this way, other's (for the most part structs) are initialized with 0'd values automatically. The field there is needed so that the compiler now's cache.mapping is a empty map[string]string.
The comma there is part of Go's formatting, you can do Cache{mapping: make(map[string]string)} all on one line, but the moment the field's assignment is on a different line than the opening and closing braces, it requires a comma.
This is called a "struct literal" or an "anonymous struct" and is, in fact, how you always create structs in Go, it just may not be immediately obvious since you might be used to creating new types for struct types to make declaring them a bit less verbose.
An entire struct definition is actually a type in Go, just like int or []byte or string. Just as you can do:
type NewType int
var a NewType = 5 // a is a NewType (which is based on an int)
or:
a := 5 // a is an int
and both are distinct types that look like ints, you can also do the same thing with structs:
// a is type NewType (which is a struct{}).
type NewType struct{
A string
}
a := NewType{
A: "test string",
}
// a is type struct{A string}
a := struct{
A string
}{
A: "test string",
}
the type name (NewType) has just been replaced with the type of the struct itself, struct{A string}. Note that they are not the same type (an alias) for the purpose of comparison or assignment, but they do share the same semantics.
Related
Say I have a generic struct called foo and I create two objects from it. I can determine the concrete type of each using reflect.TypeOf(), like so:
package main
import (
"fmt"
"reflect"
)
type foo[T any] struct {
data T
}
func main() {
a := foo[string]{"cheese"}
b := foo[int]{42}
fmt.Println(reflect.TypeOf(a))
fmt.Println(reflect.TypeOf(b))
}
// main.foo[string]
// main.foo[int]
What I am interested in is determining just the generic type of these objects (i.e., foo) and not the concrete type (i.e., foo[string] and foo[int]). Is this possible or do I need to manually extract the generic type from these strings (e.g., with regex)?
Edit
Regex might look something like this:
func GetGenericType(x any) string {
// Get type as a string
s := reflect.TypeOf(x).String()
// Regex to run
r := regexp.MustCompile(`\.(.*)\[`)
// Return capture
return r.FindStringSubmatch(s)[1]
}
fmt.Println(GetGenericType(a))
fmt.Println(GetGenericType(b))
// foo
// foo
I've also seen this question but this doesn't answer this question because it gives the concrete type (i.e., main.foo[string]) rather than the generic type (i.e., foo).
Reflection doesn't see the name of the "base" generic type, because at run time that base type doesn't exist.
The relevant passage from the Go spec is Instantiations:
Instantiating a type results in a new non-generic named type; instantiating a function produces a new non-generic function.
So when you write:
b := foo[int]{42}
name := reflect.TypeOf(b).Name()
the name of that type is precisely foo[int].
It's worth noting that the identifier foo without the type parameter list is relevant at compile time, because it prevents you from redeclaring it in the same package. Type definitions:
A type definition creates a new, distinct type with the same
underlying type and operations as the given type and binds an
identifier, the type name, to it.
TypeDef = identifier [ TypeParameters ] Type .
But instantiations, as defined above, result in a new named type which is different than foo; and at run time when you can use reflection, you deal with instantiations only.
In conclusion, I think your solution with regex is acceptable, until some helper function is added to the stdlib (if ever). Reposting it here for clarity:
func GetGenericType(x any) string {
// Get type as a string
s := reflect.TypeOf(x).String()
// Regex to run
r := regexp.MustCompile(`\.(.*)\[`)
// Return capture
return r.FindStringSubmatch(s)[1]
}
Just keep in mind the difference between Type.String() and Type.Name(): any type can have a string representation, but only named types have a name. (Obviously, right?). So for example if you wrote:
b := &foo[int]{42}
then the type of b is *foo[int], which is an anonymous composite type, and Name() returns an empty string.
Consider the following code in Go
type A struct {
f int
}
type B struct {
f int `somepkg:"somevalue"`
}
func f() {
var b *B = (*B)(&A{1}) // <-- THIS
fmt.Printf("%#v\n", b)
}
Will the marked line result in a memory copy (which I would like to avoid as A has many fields attached to it) or will it be just a reinterpretation, similar to casting an int to an uint?
EDIT: I was concerned, whether the whole struct would have to be copied, similarly to converting a byte slice to a string. A pointer copy is therefore a no-op for me
It is called a conversion. The expression (&A{}) creates a pointer to an instance of type A, and (*B) converts that pointer to a *B. What's copied there is the pointer, not the struct. You can validate this using the following code:
a:=A{}
var b *B = (*B)(&a)
b.f=2
fmt.Printf("%#v\n", a)
Prints 2.
The crucial points to understand is that
First, unlike C, C++ and some other languages of their ilk, Go does not have type casting, it has type conversions.
In most, but not all, cases, type conversion changes the type but not the internal representation of a value.
Second, as to whether a type conversion "is a no-op", depends on how you define the fact of being a no-op.
If you are concerned with a memory copy being made, there are two cases:
Some type conversions are defined to drastically change the value's representation or to copy memory; for example:
Type-converting a value of type string to []rune would interpret the value as a UTF-8-encoded byte stream, decode each encoded Unicode code point and produce a freshly-allocated slice of decoded Unicode runes.
Type-converting a value of type string to []byte, and vice-versa, will clone the backing array underlying the value.
Other type-conversions are no-op in this sense but in order for them to be useful you'd need to either assign a type-converted value to some variable or to pass it as an argument to a function call or send to a channel etc — in other words, you have to store the result or otherwise make use of it.
All of such operations do copy the value, even though it does not "look" like this; consider:
package main
import (
"fmt"
)
type A struct {
X int
}
type B struct {
X int
}
func (b B) Whatever() {
fmt.Println(b.X)
}
func main() {
a := A{X: 42}
B(a).Whatever()
b := B(a)
b.Whatever()
}
Here, the first type conversion in main does not look like a memory copy, but the resulting value will serve as a receiver in the call to B.Whatever and will be physically copied there.
The second type conversion stores the result in a variable (and then copies it again when a method is called).
Reasonong about such things is easy in Go as there everything, always, is passed by value (and pointers are values, too).
It may worth adding that variables in Go does not store the type of the value they hold, so a type conversion cannot mutate the type of a variable "in place". Values do not have type information stored in them, either. This basically means that type conversions is what compiler is concerned with: it knows the types of all the participating values and variables and performs type checking.
I learnt that it is not allowed to embed slices or maps into Go structs. But I found two workarounds to this:
Declare the slice or map outside the struct and embed it into the struct
var mySlice []int
type myStruct struct {
mySlice
}
I don't really understand this second workaround, but I found it's called Composition and it's done by just omitting the var keyword when declaring the slice or map within the struct
type myStruct struct {
mySlice []int
}
My first question is, can anyone explain why we can't directly embed slices and maps into a struct?
Second question: using the second workaround, is there any negative performance effect to that?
Third question: why do the first and second workaround work?
Spec: Struct types:
A field declared with a type but no explicit field name is called an embedded field. An embedded field must be specified as a type name T or as a pointer to a non-interface type name *T, and T itself may not be a pointer type. The unqualified type name acts as the field name.
You can't embed a type like map[int]string, that's an anonymous type. Spec requires a type name. Even if you could embed that, it contains brackets which would make it disqualify from being an identifier, and again, spec says the unqualified type name acts as the field name.
Simply create a new type for it which you can then embed:
type mymap map[int]string
type s struct {
mymap
}
When you embed a type in a struct, you omit the field name. In your second attempt:
type myStruct struct {
mySlice []int
}
You did not omit the field name (it's mySlice), so this is not embedding, it's a regular, named field.
Say we have a struct like so:
type Foo struct {
one string
two int
}
is it possible to declare literal values for this, something like:
type Foo struct {
one "foobar"
two int
}
or
type Foo struct {
one string
two 5678
}
basically for some objects we might have a hardcoded value for a field.
No, Go does not support literal types (a la TypeScript). Moreover, it actually can't, because of zero values.
Every type has a zero value which always exists and is always assignable to it. When variables of a type are declared, they are implicitly assigned to their type's zero value.
For integer (int, uint, int32, uint32, int64, uin64) or float (float32, float64) or complex (complex64 or complex128) types, this is just 0 (0.0 respectively).
For string types, this is the empty string "".
For slices, maps, pointers, channels, and interfaces, the zero value is nil.
For arrays (not slices: arrays are value-types with statically-known length); their zero value is just the zero value of the element type repeated to fill every slot
The zero value of a struct type is a struct all of whose fields are zero values
In any case, because it is always possible for any type to have a zero value, it would be incompatible to create a type which only allows for any particular non-zero value.
The best a literal type "foo" could possibly represent is that the value is either "foo" or the zero value "" (and no, Go doesn't support this anyway).
The closest you'll be able to do is a constant declaration, or a receiver function that just-so-happens to return a fixed value instead of an actual field.
Go is statically typed language, meaning every variable need to be declared with specific data type.
Setting default value for each fields inside a struct on declaration is not possible in Go. it's not supported.
But maybe you can do something like this.
type Foo struct {
one string
two int
}
func NewFoo() *Foo {
foo := new(Foo)
foo.one = "default value for one"
foo.two = 2
return foo
}
func main() {
objectWithDefaultValueForItsField := NewFoo()
}
What I did is basically just created a function with name isNew<struct name>(). This function returns a new object with default value for each of the fields defined.
Notable exception, for object that created from anonymous struct, the default value of the fields can be specified on declaration. i.e.
obj := struct {
one string
two int
}{
"default value for one",
2,
}
I would like to optimize my code, I have then the following situation:
I have a general struct where only one field gives the specification, let say a cache struct example:
# the main cache struct
type Cache struct {
name string
memory_cache map[string]interface{}
mutex *sync.Mutex
... ...
# common fields
}
# an element stored in the Cache.memory_cache map
type ElementA {
name string
count int64
}
# an element stored in the Cache.memory_cache map
type ElementB {
name string
tags []string
}
My current solution follow the previously definition and I create a cache for each element (it must be so: one cache per element):
var cache_for_element_A Cache{}
var cache_for_element_B Cache{}
But in this way I must always cast the memory_cache when reading, even if I know already what is the content (then no cast-case should be needed).
The following code do what I would like to have, but it defines twice a lot of redundants/commons fields, for this reason I would like to find another solution.
type CacheForA struct {
name string
memory_cache map[string]ElementA{}
mutex *sync.Mutex
... ...
# common fields
}
type CacheForB struct {
name string
memory_cache map[string]ElementB{}
mutex *sync.Mutex
... ...
# common fields
}
Then, is it possible to define a field in the struct (more precisely Cache.memory_cache) that can be further defined when the declaration occurs and without using interface?
Go doesn't have generics, so there's no simple way of doing this, like you would in Java for instance (class Cache<T>()....).
One thing you can do is wrap your cache with a small typed function that just fetches objects from a generic cache and converts the interface to the right type. This just saves you from writing the interface conversion over and over in your code.
type ElemACache struct {
Cache
}
func (c *ElemeACache)Get(key string) ElemeA {
return c.Cache.Get(key).(ElemeA) //of course add checks here
}
Struct embedding is the main thing you are looking for I think:
type Cache struct {
name string
mutex *sync.Mutex
}
type CacheA struct {
Cache
memory_cache map[string]ElementA{}
}
Then you make a type of interface, say "Cacher" that has a set of methods for the things you need to do with your various caches (cacheA, CacheB). Create those methods for CacheA, CacheB, and assertions are only needed only for the return type:
type Cacher interface {
GetItem(string) (interface{}, error)
}
If all your CacheFor types have that GetItem method, the interface will be fulfilled.
Still a fair amount of boilerplate, but this reduces the problem with redundancy in struct definitions. There are code generation tools if you don't want to type the boiler plate.