I have a function I have written and I would like it to return a string. I have declared the variable before my for loop, but when I try and return it at the end of the function it claims the variable is unused. I do not seem to have the problem when returning other types. I am trying to figure out what I am doing wrong.
func GetInstanceAlarms(instance *Instance) string {
params := &cloudwatch.DescribeAlarmsInput{}
c := cloudwatch.New(session.New())
resp, err := c.DescribeAlarms(params)
var Arn string
if err != nill {
panic(err.Error())
}
for idx := range resp.MetricAlarms{
for _, alarm := range resp.MetricAlarms[idx].Dimensions{
if *alarm.Name == "InstanceId" && *alarm.Value == instance.InstanceId{
log.Println(resp.MetricAlarms[idx].AlarmArn)
//claims this variable is unused
Arn := resp.MetricAlarms[idx].AlarmArn
}
}
}
return Arn
}
Inside your for loop you have recreated a variable Arn which shadows the outer variable. Remember that := creates new variables, and = assigns to existing variables.
Arn := resp.MetricAlarms[idx].AlarmArn
This Arn variable is only valid in its containing block, and once you exit the block, the outer variable Arn, created just before the loop, becomes visible again.
Instead of creating a new variable, assign to the existing one.
Arn = resp.MetricAlarms[idx].AlarmArn
Read more about scope in Go.
Running go vet should catch this issue.
In you code you have err!=nill, change to err != nil
I check your code, I don't see any issue. Once you declared var Arn string, by default it has zero value i.e empty string "".
I tested with (Mac Os) with the following code and I don't see any error. Also I have added in code to print go version and OS.
package main
import (
"fmt"
"runtime"
)
func main(){
fmt.Printf("Checking on %s OS with Go version %s\n",runtime.GOOS, runtime.Version())
fmt.Println(test())
}
func test() string {
var a string
for i:=0;i<5;i++{
a:="string"
return a
}
return a
}
Related
This code:
package main
import (
"fmt"
)
func main() {
fmt.Println("Hello, playground")
var a bool
var b interface{}
b = true
if a, ok := b.(bool); !ok {
fmt.Println("Problem!")
}
}
Yields this error in the golang playground:
tmp/sandbox791966413/main.go:11:10: a declared and not used
tmp/sandbox791966413/main.go:14:21: a declared and not used
This is confusing because of what we read in the short variable declaration golang docs:
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.
So, my questions:
Why can't I redeclare the variable in the code snippet above?
Supposing I really can't, what I'd really like to do is find
a way to populate variables with the output of functions while
checking the error in a more concise way. So is there any way
to improve on the following form for getting a value out of an
error-able function?
var A RealType
if newA, err := SomeFunc(); err != nil {
return err
} else {
A = newA
}
This is happening because you are using the short variable declaration in an if initialization statement, which introduces a new scope.
Rather than this, where a is a new variable that shadows the existing one:
if a, ok := b.(bool); !ok {
fmt.Println("Problem!")
}
you could do this, where a is redeclared:
a, ok := b.(bool)
if !ok {
fmt.Println("Problem!")
}
This works because ok is a new variable, so you can redeclare a and the passage you quoted is in effect.
Similarly, your second code snippet could be written as:
A, err := SomeFunc()
if err != nil {
return err
}
You did redeclare the variable, that is the problem. When you redeclare a variable, it is a different variable. This is known as "shadowing".
package main
import (
"fmt"
)
func main() {
fmt.Println("Hello, playground")
// original a.
var a bool
var b interface{}
b = true
// redeclared a here. This is a a'
if a, ok := b.(bool); !ok {
// a' will last until the end of this block
fmt.Println("Problem!")
}
// a' is gone. So using a here would get the original a.
}
As for your second question. That code looks great. I would probably switch it to if err == nil (and swap the if and else blocks). But that is a style thing.
The error raised by the compiler is saying that both declarations of variable a are not being used.
You're actually declaring a twice, when using short declaration in the if condition you're creating a new scope that shadows the previous declaration of the variable.
The actual problem you're facing is that you're never using the variable value which, in Go, is considered a compile time error.
Regarding your second question, I think that the shortest way to get the value and check the error is to do something similar to this:
func main() {
a, ok := someFunction()
if !ok {
fmt.Println("Problem!")
}
}
This panic recover code works with named return values.
func main() {
result, err := foo()
fmt.Println("result:", result)
if err != nil {
fmt.Println("err:", err)
}
}
func foo() (result int, err error) {
defer func() {
if e := recover(); e != nil {
result = -1
err = errors.New(e.(string))
}
}()
bar()
result = 100
err = nil
return
}
func bar() {
panic("panic happened")
}
Output
result: -1
err: panic happened
But why this code with local variables does not work?
func main() {
result, err := foo()
fmt.Println("result:", result)
if err != nil {
fmt.Println("err:", err)
}
}
func foo() (int, error) {
var result int
var err error
defer func() {
if e := recover(); e != nil {
result = -1
err = errors.New(e.(string))
}
}()
bar()
result = 100
err = nil
return result, err
}
func bar() {
panic("panic happened")
}
Output
result: 0
Any explanation to help me understanding the reason / basic concept of it? In the go tour basics the explanation is as followed.
Named return values
Go's return values may be named. If so, they are treated as variables defined at the top of the function.
So it should be the same, right?
Note that this has nothing to do with panic/recover, it is a feature of the defer statement.
... if the deferred function is a function literal and the surrounding
function has named result parameters that are in scope within the
literal, the deferred function may access and modify the result
parameters before they are returned. If the deferred function has
any return values, they are discarded when the function completes.
Spec: Return statements details this:
There are three ways to return values from a function with a result type:
The return value or values may be explicitly listed in the "return" statement. Each expression must be single-valued and assignable to the corresponding element of the function's result type.
The expression list in the "return" statement may be a single call to a multi-valued function. The effect is as if each value returned from that function were assigned to a temporary variable with the type of the respective value, followed by a "return" statement listing these variables, at which point the rules of the previous case apply.
The expression list may be empty if the function's result type specifies names for its result parameters. The result parameters act as ordinary local variables and the function may assign values to them as necessary. The "return" statement returns the values of these variables.
So basically if you use a return statement that explicitly lists the return values, those will be used, regardless if the result parameters are named or not.
If the result parameters are named, they act as ordinary local variables: you can read and write them. If the result parameters are named, you may use a "naked" return statement, without listing the values to return. If you do so, then the actual return values will be the values of the (named) result parameters. The same thing applies if your function does not reach a return statement due to panicing and recovering: once the deferred functions run, the actual return values will be the values of the named result parameters (which the deferred functions can change and "have a say" in what to return).
If you don't use named result parameters but you declare local variables, they are not special in this way: when the function returns, those are not used "automatically" as the result values (like they would be if they would be named result parameters and not local variables). So if you change them in a deferred function, that will not have any effect on the actual values returned. In fact, if you don't use named result parameters and your function panics and recovers, you can't specify the return values, they will be the zero values of the result types. That's why you see result: 0 (0 is the zero value for int) and no error (because error is an interface type and zero value for interface types is nil and you don't print the error if it's nil).
See related: How to return a value in a Go function that panics?
Might be a brief summary for #icza's anwser:
Named return variables use their final values for returning when the function teminate with no panic(return normally or recover from panic), so you can change them in defer recover func(), and the final values changed, so be the return values.
If use local variables, compiler can not know these local variables will be used as return variables until a normal return. Local variables might be changed in panic recover, but
the return statement has not been executed yet because the panic, so the local variables you defined was not treated as return variables, the return values will be the zero values of the return types.
This code:
package main
import (
"fmt"
)
func main() {
fmt.Println("Hello, playground")
var a bool
var b interface{}
b = true
if a, ok := b.(bool); !ok {
fmt.Println("Problem!")
}
}
Yields this error in the golang playground:
tmp/sandbox791966413/main.go:11:10: a declared and not used
tmp/sandbox791966413/main.go:14:21: a declared and not used
This is confusing because of what we read in the short variable declaration golang docs:
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.
So, my questions:
Why can't I redeclare the variable in the code snippet above?
Supposing I really can't, what I'd really like to do is find
a way to populate variables with the output of functions while
checking the error in a more concise way. So is there any way
to improve on the following form for getting a value out of an
error-able function?
var A RealType
if newA, err := SomeFunc(); err != nil {
return err
} else {
A = newA
}
This is happening because you are using the short variable declaration in an if initialization statement, which introduces a new scope.
Rather than this, where a is a new variable that shadows the existing one:
if a, ok := b.(bool); !ok {
fmt.Println("Problem!")
}
you could do this, where a is redeclared:
a, ok := b.(bool)
if !ok {
fmt.Println("Problem!")
}
This works because ok is a new variable, so you can redeclare a and the passage you quoted is in effect.
Similarly, your second code snippet could be written as:
A, err := SomeFunc()
if err != nil {
return err
}
You did redeclare the variable, that is the problem. When you redeclare a variable, it is a different variable. This is known as "shadowing".
package main
import (
"fmt"
)
func main() {
fmt.Println("Hello, playground")
// original a.
var a bool
var b interface{}
b = true
// redeclared a here. This is a a'
if a, ok := b.(bool); !ok {
// a' will last until the end of this block
fmt.Println("Problem!")
}
// a' is gone. So using a here would get the original a.
}
As for your second question. That code looks great. I would probably switch it to if err == nil (and swap the if and else blocks). But that is a style thing.
The error raised by the compiler is saying that both declarations of variable a are not being used.
You're actually declaring a twice, when using short declaration in the if condition you're creating a new scope that shadows the previous declaration of the variable.
The actual problem you're facing is that you're never using the variable value which, in Go, is considered a compile time error.
Regarding your second question, I think that the shortest way to get the value and check the error is to do something similar to this:
func main() {
a, ok := someFunction()
if !ok {
fmt.Println("Problem!")
}
}
How do you assign a default value if an environment variable isn't set in Go?
In Python I could do mongo_password = os.getenv('MONGO_PASS', 'pass') where pass is the default value if MONGO_PASS env var isn't set.
I tried an if statement based on os.Getenv being empty, but that doesn't seem to work due to the scope of variable assignment within an if statement. And I'm checking for multiple env var's, so I can't act on this information within the if statement.
There's no built-in to fall back to a default value,
so you have to do a good old-fashioned if-else.
But you can always create a helper function to make that easier:
func getenv(key, fallback string) string {
value := os.Getenv(key)
if len(value) == 0 {
return fallback
}
return value
}
Note that as #michael-hausenblas pointed out in a comment,
keep in mind that if the value of the environment variable is really empty, you will get the fallback value instead.
Even better as #ŁukaszWojciechowski pointed out, using os.LookupEnv:
func getEnv(key, fallback string) string {
if value, ok := os.LookupEnv(key); ok {
return value
}
return fallback
}
What you're looking for is os.LookupEnv combined with an if statement.
Here is janos's answer updated to use LookupEnv:
func getEnv(key, fallback string) string {
value, exists := os.LookupEnv(key)
if !exists {
value = fallback
}
return value
}
Go doesn't have the exact same functionality as Python here; the most idiomatic way to do it though, I can think of, is:
mongo_password := "pass"
if mp := os.Getenv("MONGO_PASS"); mp != "" {
mongo_password = mp
}
To have a clean code I do this:
myVar := getEnv("MONGO_PASS", "default-pass")
I defined a function that is used in the whole app
// getEnv get key environment variable if exist otherwise return defalutValue
func getEnv(key, defaultValue string) string {
value := os.Getenv(key)
if len(value) == 0 {
return defaultValue
}
return value
}
Had the same question as the OP and found someone encapsulated the answers from this thread into a nifty library that is fairly simple to use, hope this help others!
https://github.com/caarlos0/env
For more complex application you can use tooling such as viper, which allows you to set global custom default values, parse configuration files, set a prefix for your app's env var keys (to ensure consistency and name spacing of env var configurations) and many other cool features.
Sample code:
package main
import (
"fmt"
"github.com/spf13/viper"
)
func main() {
viper.AutomaticEnv() // read value ENV variable
// Set default value
viper.SetEnvPrefix("app")
viper.SetDefault("linetoken", "DefaultLineTokenValue")
// Declare var
linetoken := viper.GetString("linetoken")
fmt.Println("---------- Example ----------")
fmt.Println("linetoken :", linetoken)
}
I also had the same problem and I just created a small package called getenvs exactly to answer this problem.
Getenvs supports string, bool, int and float and it can be used like below:
package main
import (
"fmt"
"gitlab.com/avarf/getenvs"
)
func main() {
value := getenvs.GetEnvString("STRING_GETENV", "default-string-value")
bvalue, _ := getenvs.GetEnvBool("BOOL_GETENV", false)
ivalue, _ := getenvs.GetEnvInt("INT_GETENV", 10)
fmt.Println(value)
fmt.Println(bvalue)
fmt.Println(ivalue)
}
In case you are OK with adding little dependency you can use something like https://github.com/urfave/cli
package main
import (
"os"
"github.com/urfave/cli"
)
func main() {
app := cli.NewApp()
app.Flags = []cli.Flag {
cli.StringFlag{
Name: "lang, l",
Value: "english",
Usage: "language for the greeting",
EnvVar: "APP_LANG",
},
}
app.Run(os.Args)
}
In go there are functions which return two values or more values, commonly one is an error. Suppose that I want to store the first return value into an already initialized variable, but I would like to initialize the variable to contain the error inline. Is there a way to do this?
For example, say I had this code
var a int
//This code doesn't compile because err doesn't exist
a, err = SomeFuncWithTwoReturnValues()
//This code doesn't compile either
a, err := SomeFuncWithTwoReturnValues()
I know you could do this, but I was hoping there was a way to do it all inline
var a int
var err error
a, err = SomeFuncWithTwoReturnValues()
or
a, err := SomeFuncWithTwoReturnValues()
EDIT: The code above actually compiles, so I looked back at my code to drill down more and have created a quick sample that actually replicates the problem (not just in my mind...).
package main
func myfunc() (int, int) {
return 1, 1
}
func main() {
a := make([]int, 1)
a[0], b := myfunc()
a[0] = b
}
Compiler says main.go|9| non-name a[0] on left side of :=. If I make it = instead of := though then b is never created. I get the feeling that there is not shorthand way to do it though.
As you've mentioned in the comments, you'll need to use the = operator in order to assign to a variable you've already declared. The := operator is used to simultaneously declare and assign a variable. The two are the same:
var x int
x = 5
//is the same as
x := 5
This solution will at least compile:
package main
func myfunc() (int, int) {
return 1, 1
}
func main() {
var b int
a := make([]int, 1)
a[0], b = myfunc()
a[0] = b
}
To answer your question, I don't think there is a way to simultaneously use an undeclared and a declared variable when returning multiple values. That would be trying to use two different operators simultaneously.
Edit: just saw your example from the code that compiles, so it appears you're already familiar with go's assignment operators. I'll leave the example up anyway.
Golang is not a very consistent language. This is a good example. At the beginning I was confused and it would be much simpler if they would always allow the := operator. The compiler is smart enough to detect already declared variables:
package main
import "fmt"
func testFunc() (int,error) {
return 42,fmt.Errorf("Test Error")
}
func main() {
number1,err := testFunc() // OK
number2,err := testFunc() // OK, even if err is already defined
number1,err = testFunc() // OK
// number1,err := testFunc() // ERROR: no new variables on left side of :=
fmt.Println(number1,number2,err)
}
Playground Link: https://play.golang.org/p/eZVB-kG6RtX
It's not consistent, because golang allows you to use := for already declared variables if you assign to them while also introducing a new variable. So the compiler can detect that variables already exists and skip their declaration. But the golang developers decided to allow that only if you introduce at least one new value. The last example shows that.
I ran into this situation like this:
package main
import "os"
func main() {
var cache struct { dir string }
// undefined: err
cache.dir, err = os.UserCacheDir()
// non-name cache.dir on left side of :=
cache.dir, err := os.UserCacheDir()
if err != nil {
panic(err)
}
println(cache.dir)
}
as you discovered, this issue does not have a clean solution. You can declare
an extra variable:
dir, err := os.UserCacheDir()
if err != nil {
panic(err)
}
cache := userCache{dir}
Or, while more verbose, you can declare the error beforehand. This can save
memory, as Go does not use a Rust ownership model:
var (
cache struct { dir string }
err error
)
cache.dir, err = os.UserCacheDir()
As mention in the spec, while using:=, if one of the variables is new, then the old one will just be assigned with the new data.
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.
field1, offset := nextField(str, 0)
field2, offset := nextField(str, offset) // redeclares offset
As mentioned by the other answers you cannot use assignment and declaration in the same return statement. You have to use either.
However I guess the main reason for your question is cleaning up the code so you don't have to declare an extra err variable above the method or function statement.
You can solve this in two ways:
Declare a global var err error variable and use it in the assignment:
var err error
func MyFunc(someInput string) {
var a int
a, err = someOtherFunction()
}
If your method or function returns an error you can use the declared return variable
func MyFunc(someInput string) (err error) {
var a int
a, err = someOtherFunction()
return
}
I mainly have the problem in methods when I want to assign something to a struct member, e.g.:
type MyStruct struct {
so string
}
func (m *MyStruct) SomeMethod() (err error) {
m.so, err = SomeFunction()
// handle error and continue or return it
return
}