英文:
How to differentiate between assigning and declaring values from a function with multiple returns?
问题
当从函数中检索多个返回值时，你可以使用:=在使用中声明变量来获取返回值，或者使用=将返回值赋给已经存在的变量。但是，当我想将一个返回值赋给已经存在的变量，并为另一个返回值声明一个新变量时，我遇到了问题。
目前，我通过先声明所需的变量（在这种情况下是bar），然后只赋值来解决了这个问题，就像这个代码片段中所示：
package main
import (
"fmt"
)
func getFooAndBar() (foo string, bar string) {
return "Foo", "Bar"
}
func main() {
var foo = "default"
var condition = true
if condition {
var bar string // 如果可能的话，我想避免这一步
foo, bar = getFooAndBar()
fmt.Println(bar)
}
fmt.Println(foo)
}
如果我使用:=，它无法构建，因为会出现以下错误：
./app.go:16: foo declared and not used
所以，有没有办法避免单独声明bar这一步呢？
英文:
When retrieving multiple returns from a function, I get that you can declare variables for the values on the fly by using := or assign the values to already existing variables by simply using =. My issue occurs when I want to assign one of the return values to an already existing variable while declaring a new variable for the other.
I have currently solved it by only assigning the values and declaring the required variables (bar in this case) beforehand, as in this snippet:
package main
import (
"fmt"
)
func getFooAndBar() (foo string, bar string) {
return "Foo", "Bar"
}
func main() {
var foo = "default"
var condition = true
if condition {
var bar string // Would like to avoid this step if possible
foo, bar = getFooAndBar()
fmt.Println(bar)
}
fmt.Println(foo)
}
If I use := it fails to build due to:
>./app.go:16: foo declared and not used
So, is it possible to somehow avoid the step declaring bar separately?
答案1
得分: 5
在这种情况下，根据规范，你不能使用短变量声明:=来重新声明foo变量：
与常规变量声明不同，短变量声明可以重新声明变量，前提是它们在同一块中以相同的类型最初被声明，并且至少有一个非空白变量是新的。因此，重新声明只能出现在多变量的短声明中。重新声明不会引入新变量；它只是给原始变量赋予新值。
通过消除./app.go:16: foo declared and not used错误。
func main() {
var foo = "default"
var condition = true
if condition {
foo, bar := getFooAndBar()
fmt.Println(bar) // 输出：Bar
fmt.Println(foo) // 输出：Foo
// _ = foo
}
fmt.Println(foo) // 输出：default
}
在这种情况下，foo在if块中被声明，这个声明将创建一个新变量，遮蔽了外部块中的原始foo变量，只有在同一块中使用多变量短声明重新声明foo时，才会发生重新声明。
func main() {
var foo = "default"
foo, bar := getFooAndBar()
fmt.Println(bar) // 输出：Bar
fmt.Println(foo) // 输出：Foo
}
英文:
In this case you can't use the short variable declarations ":=" for redeclaring the foo variable, according to the spec:
> Unlike regular variable declarations, a short variable declaration may
> redeclare variables provided they were originally declared earlier in
> the same block 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.
by eliminating ./app.go:16: foo declared and not used.
func main() {
var foo = "default"
var condition = true
if condition {
foo, bar := getFooAndBar()
fmt.Println(bar) // prints: Bar
fmt.Println(foo) // prints: Foo
// _ = foo
}
fmt.Println(foo) // prints: default
}
in this case foo is declared in the if block, this declaration will create a new variable shadowing the original foo variable in the outer block, the redeclaration of foo will happen only if you have declared foo and redeclared it with multi-variable short declaration within the same block.
func main() {
var foo = "default"
foo, bar := getFooAndBar()
fmt.Println(bar) //prints: Bar
fmt.Println(foo) //prints: Foo
}
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