In Go, when a variable is declared it is initialized with zero value as described in the specification.
http://golang.org/ref/spec#The_zero_value
But is it good coding practice to make use of this property and do not explicitly initialize your variable if it needs to initialized with the default value.
for example in the following example
http://play.golang.org/p/Mvh_zwFkOu
package main
import "fmt"
type B struct {
isInit bool
Greeting string
}
func (b *B) Init() {
b.isInit = true
b.Greeting = "Thak you for your time"
}
func (b *B) IsInitialized() bool {
return b.isInit
}
func main() {
var b B
if !b.IsInitialized(){
b.Init()
}
fmt.Println(b.Greeting)
}
The program relies on the default value of boolean to be false.
As everyone says, specification is clear here: all memory is initialised (zeroed). You should take advantage of this as standard packages do. In particular, it allows you to rely on "default constructor" for your own types and often skip New() *T kind of functions in favour of &T{}.
Many types in standard packages take advantage of this, some examples:
http.Client
A Client is an HTTP client. Its zero value (DefaultClient) is a usable client that uses DefaultTransport.
And then you will find var DefaultClient = &Client{} declared in the package.
http.Server
A Server defines parameters for running an HTTP server. The zero value for Server is a valid configuration.
bytes.Buffer
A Buffer is a variable-sized buffer of bytes with Read and Write methods. The zero value for Buffer is an empty buffer ready to use.
This is great, because you can just do var buf bytes.Buffer and start using it. As a consequence of this you will also often see boolean member variables to be used in a "negated" form – for example InsecureSkipVerify in tls.Config is not called Verify, because the default behaviour wouldn't then validate certificates (think I want the false – or zero – value to be used for desirable defaults).
Finally, answering your question:
But is it good coding practice to make use of this property and do not explicitly initialize your variable if it needs to be initialized with default value?
Yes, it is.
Initialize variables to their zero values only if you want to use the short declaration syntax.
//less verbose than ''var count int''
count := 0
empty := ""
Otherwise, initializing them explicitly is just noise.
You might think that there's something wrong with uninitialized variables... and you'd be right. Fortunately, there's no such thing in go. Zero values are
part of the spec and they're not going to suddenly change.
When a variable is declared it contains automatically the default zero or null value for its type: 0 for int, 0.0 for float, false for bool, empty string for string, nil for pointer, zero-ed struct, etc.
All memory in Go is initialized!.
For example: var arr [5]int in memory can be visualized as:
+---+---+---+---+
| | | | |
+---+---+---+---+
0 1 2 3
When declaring an array, each item in it is automatically initialized with the default zero-value of the type, here all items default to 0.
So preferably it's better to initialize without default value, in other cases than the situations when you explicitly want to declare a variable with a default value.
Related
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 was asked to declare a variable of integer type as:
var someInteger int8
Later when I printed this variable, it prints the value 0.
My Go Program looks like:
package main
import "fmt"
func main() {
var someInteger int
fmt.Println(someInteger) // Prints 0 in terminal
}
My question is since I haven't assigned any value, so it should return some Garbage value like C instead of behaving like static variables, which automatically initialize by value 0.
In Go you can't access uninitialized variables / memory. If you don't initialize a variable explicitly, it will be initialized implicitly to the zero value of its type.
This is covered in Spec: Variable declarations:
If a list of expressions is given, the variables are initialized with the expressions following the rules for assignments. Otherwise, each variable is initialized to its zero value.
Also mentioned at Spec: Variables:
If a variable has not yet been assigned a value, its value is the zero value for its type.
And also covered in the Go Tour: Zero Values which I highly recommend to take if you're learning the language.
Variables declared without an explicit initial value are given their zero value.
Go makes this thing easy by adding sensible default values based on the type of the variables. For example:
var someInteger int8 // will print 0 as default
var someFloat float32 // will print 0 as default
var someString string // will print nothing as it prints empty string
var someBoolean bool // will print false as default
As #icza mentioned in his answer you can read more about it here
Here is a sample of Go code which I do not really understand:
type MyClass struct {
field1 string
field2 string
}
...
objectinstance1 := MyClass{field1:"Hello", field2:"World"}
objectinstance2 := &MyClass{field1:"Hello", field2:"World"}
I can do exactly the same thing with objectinstance1 and objectinstance2 (method call for example) with the same syntax.
So I do not understand the role of the & operator. What I understand is that objectinstance1 contains an object whereas objectinstance2 contains a pointer.
It is for me the same thing in C than the difference between char and char*.
But in this case I should use -> instead of . (dot)?
The & operator gives you a pointer to a struct, while not using it gives you the struct value.
The biggest place this is relevant is when you pass this struct over to another function - if you pass the pointer that you made using the & operator, the other function has access to the same struct. If that function changes it, you've got the changed struct as well.
If you pass the variable that you made without the & operator, the function that you pass it to has a copy of the struct. There is nothing that that function or any other function can possibly do to change what you see in your variable.
This effectively makes the value variable safe for use across multiple go routines with no race conditions - everyone has their own copy of the struct.
If you pass the pointer made with & to other go routines, all have access to the same struct, so you really want that to be intentional and considered.
Difference is not visible because it’s hidden in 2 things:
Operator := which assigns value and type for a variable simultaneously. So it looks like objectinstance1 and objectinstance2 are the same. But in fact first is a MyClass instance and second is a pointer to it. It will be more palpable if use long-form operator:
var objectinstace1 MyClass = MyClass{}
var objectinstance2 *MyClass = &MyClass{}
If you omit * or & variable and type become incompatible and assignment fails.
Implicit pointer indirection. Go does it automatically in statements like ptr1.Field1 to access a specific field in struct by pointer to it.
Only on rare cases when there’s ambiguity you have to use full form:
*ptr1.Value1
or sometimes even:
(*ptr1).Value1
UPDATE:
Explicit pointer usage for disambiguation:
type M map[int]int
func (m *M) Add1() {
// this doesn't work - invalid operation: m[1] (type *M does not support indexing)
// m[1] = 1
// the same - invalid operation: m[1] (type *M does not support indexing)
// *m[1] = 1
// this works
(*m)[1] = 1
}
https://play.golang.org/p/JcXd_oNIAw
But in this case i should use -> instead of . (dot) ?
No. Golang is not C is not Golang. In Golang there is no ->. You use dot (.) for pointers aswell. Golang is meant to be simple, there is no point in introducing another operator if the intention is clear (what else would . on a pointer mean than calling a method?)
In go, the & operator takes the address of its argument and returns a pointer to the type of the argument.
Pointer indirection happens automatically so there is no -> operator, the dot operator handles all field (member) operations and accesses for all types, whether a pointer or a struct.
package main
import (
"fmt"
"reflect"
)
func main() {
type Foo struct{ name string }
foo1 := Foo{"Alpha"} // A plain struct instance.
foo2 := &Foo{"Bravo"} // A pointer to a struct instance.
fmt.Printf("foo1=%v, name=%v\n", reflect.TypeOf(foo1), foo1.name)
fmt.Printf("foo2=%v, name=%v\n", reflect.TypeOf(foo2), foo2.name)
// foo1=main.Foo, name=Alpha
// foo2=*main.Foo, name=Bravo
}
In Go you can initialise an empty string (in a function) in the following ways:
var a string
var b = ""
c := ""
As far as I know, each statement is semantically identical. Is one of them more idiomatic than the others?
You should choose whichever makes the code clearer. If the zero value will actually be used (e.g. when you start with an empty string and concatenate others to it), it's probably clearest to use the := form. If you are just declaring the variable, and will be assigning it a different value later, use var.
var a string
It's not immediately visible that it's the empty string for someone not really fluent in go. Some might expect it's some null/nil value.
var b = ""
Would be the most explicit, means everyone sees that it's a variable containing the empty string.
b := ""
The := assigment is so common in go that it would be the most readable in my opinion. Unless it's outside of a function body of course.
There isn't a right way to declare empty variables, but there are some things to keep in mind, like you can't use the := shortcut outside of a function body, as they can with var:
var (
name string
age int64
)
I find var name = "" to be the clearest indication that you're declaring an empty variable, but I like the type explicitness of var name string. In any case, do consider where you are declaring variables. Sometimes all at once outside the function body is appropriate, but often nearest to where you actually use it is best.
rob (Pike?) wrote on a mailthread about top-level declaration
At the top level, var (or const or type or func) is necessary: each item must be introduced by a keyword for ur-syntactic reasons related to recognizing statement boundaries. With the later changes involving semicolons, it became possible, I believe, to eliminate the need for var in some cases, but since const, type, and func must remain, it's not a compelling argument.
There is a certain ambiguity in "short-variable" declarations (using :=), as to whether the variable is declared or redeclared as outlined in the specs:
Unlike regular variable declarations, a short variable declaration may redeclare variables provided they were originally declared earlier in the same block (or the parameter lists if the block is the function body) with the same type, and at least one of the non-blank variables is new. As a consequence, redeclaration can only appear in a multi-variable short declaration. Redeclaration does not introduce a new variable; it just assigns a new value to the original.
There is absolutely no difference in the generated code (with the current compiler – Go 1.7.4), and also gometalinter does not complain for any of those. Use whichever you like.
Some differences:
You can only use the short variable declaration in functions.
With short variable declaration, you can create variables of multiple types in one line, e.g.
a, b := "", 0
The following 3 apps generate identical code:
a.go
package main
import "fmt"
func main() { var a string; fmt.Println(a) }
b.go
package main
import "fmt"
func main() { var a = ""; fmt.Println(a) }
c.go
package main
import "fmt"
func main() { a := ""; fmt.Println(a) }
To verify, build each (e.g. with go build -o a), and use diff on Linux to compare the executable binaries:
diff a b && diff a c
I try to stick to the short declaration for a couple of reasons.
It's shorter
Consistency
Allocates memory for maps, slices and pointers to structs and types.
Although var a string and a := "" are the same, b := []T{} and var b []T are not the same. When dealing with slices and maps the shorter declaration will not be nil. More often then not (especially with maps) I want allocated memory.
There are few situations where var will be needed, for instance, you are calling a function that will populate a property of a type.
var err error
foo.Name, err = getFooName()
Using := syntax in the above situation will error out since foo.Name is already declared.
just
*new(string)
It's only stuff in stackoverf related to empty strings in go. So it should be here
What's the cleanest way to handle a case such as this:
func a() string {
/* doesn't matter */
}
b *string = &a()
This generates the error:
cannot take the address of a()
My understanding is that Go automatically promotes a local variable to the heap if its address is taken. Here it's clear that the address of the return value is to be taken. What's an idiomatic way to handle this?
The address operator returns a pointer to something having a "home", e.g. a variable. The value of the expression in your code is "homeless". if you really need a *string, you'll have to do it in 2 steps:
tmp := a(); b := &tmp
Note that while there are completely valid use cases for *string, many times it's a mistake to use them. In Go string is a value type, but a cheap one to pass around (a pointer and an int). String's value is immutable, changing a *string changes where the "home" points to, not the string value, so in most cases *string is not needed at all.
See the relevant section of the Go language spec. & can only be used on:
Something that is addressable: variable, pointer indirection, slice indexing operation, field selector of an addressable struct, array indexing operation of an addressable array; OR
A composite literal
What you have is neither of those, so it doesn't work.
I'm not even sure what it would mean even if you could do it. Taking the address of the result of a function call? Usually, you pass a pointer of something to someone because you want them to be able to assign to the thing pointed to, and see the changes in the original variable. But the result of a function call is temporary; nobody else "sees" it unless you assign it to something first.
If the purpose of creating the pointer is to create something with a dynamic lifetime, similar to new() or taking the address of a composite literal, then you can assign the result of the function call to a variable and take the address of that.
In the end you are proposing that Go should allow you to take the address of any expression, for example:
i,j := 1,2
var p *int = &(i+j)
println(*p)
The current Go compiler prints the error: cannot take the address of i + j
In my opinion, allowing the programmer to take the address of any expression:
Doesn't seem to be very useful (that is: it seems to have very small probability of occurrence in actual Go programs).
It would complicate the compiler and the language spec.
It seems counterproductive to complicate the compiler and the spec for little gain.
I recently was tied up in knots about something similar.
First talking about strings in your example is a distraction, use a struct instead, re-writing it to something like:
func a() MyStruct {
/* doesn't matter */
}
var b *MyStruct = &a()
This won't compile because you can't take the address of a(). So do this:
func a() MyStruct {
/* doesn't matter */
}
tmpA := a()
var b *MyStruct = &tmpA
This will compile, but you've returned a MyStruct on the stack, allocated sufficient space on the heap to store a MyStruct, then copied the contents from the stack to the heap. If you want to avoid this, then write it like this:
func a2() *MyStruct {
/* doesn't matter as long as MyStruct is created on the heap (e.g. use 'new') */
}
var a *MyStruct = a2()
Copying is normally inexpensive, but those structs might be big. Even worse when you want to modify the struct and have it 'stick' you can't be copying then modifying the copies.
Anyway, it gets all the more fun when you're using a return type of interface{}. The interface{} can be the struct or a pointer to a struct. The same copying issue comes up.
You can't get the reference of the result directly when assigning to a new variable, but you have idiomatic way to do this without the use of a temporary variable (it's useless) by simply pre-declaring your "b" pointer - this is the real step you missed:
func a() string {
return "doesn't matter"
}
b := new(string) // b is a pointer to a blank string (the "zeroed" value)
*b = a() // b is now a pointer to the result of `a()`
*b is used to dereference the pointer and directly access the memory area which hold your data (on the heap, of course).
Play with the code: https://play.golang.org/p/VDhycPwRjK9
Yeah, it can be annoying when APIs require the use of *string inputs even though you’ll often want to pass literal strings to them.
For this I make a very tiny function:
// Return pointer version of string
func p(s string) *string {
return &s
}
and then instead of trying to call foo("hi") and getting the dreaded cannot use "hi" (type string) as type *string in argument to foo, I just wrap the argument in a call to to p():
foo(p("hi"))
a() doesn't point to a variable as it is on the stack. You can't point to the stack (why would you ?).
You can do that if you want
va := a()
b := &va
But what your really want to achieve is somewhat unclear.
At the time of writing this, none of the answers really explain the rationale for why this is the case.
Consider the following:
func main() {
m := map[int]int{}
val := 1
m[0] = val
v := &m[0] // won't compile, but let's assume it does
delete(m, 0)
fmt.Println(v)
}
If this code snippet actually compiled, what would v point to!? It's a dangling pointer since the underlying object has been deleted.
Given this, it seems like a reasonable restriction to disallow addressing temporaries
guess you need help from More effective Cpp ;-)
Temp obj and rvalue
“True temporary objects in C++ are invisible - they don't appear in your source code. They arise whenever a non-heap object is created but not named. Such unnamed objects usually arise in one of two situations: when implicit type conversions are applied to make function calls succeed and when functions return objects.”
And from Primer Plus
lvalue is a data object that can be referenced by address through user (named object). Non-lvalues include literal constants (aside from the quoted strings, which are represented by their addresses), expressions with multiple terms, such as (a + b).
In Go lang, string literal will be converted into StrucType object, which will be a non-addressable temp struct object. In this case, string literal cannot be referenced by address in Go.
Well, the last but not the least, one exception in go, you can take the address of the composite literal. OMG, what a mess.