doutuo3899
2015-01-21 11:44
浏览 185
已采纳

Golang:函数的多个返回值的范围

When a function returns more than one variable in Golang, what's the scope of the variables? In the code attached, I can't figure out the scope of b.

package main

import (
    "fmt"
)

func addMulti(x, y int) (int, int) {
    return (x + y), (x * y)
}

func main() {
    //what is the scope of the b variable here?
    a, b := addMulti(1, 2)

    fmt.Printf("%d %d
", a, b)

    //what is the scope of the b variable here?
    c, b := addMulti(3, 4)

    fmt.Printf("%d %d
", c, b)

}   

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当一个函数在Golang中返回多个变量时,这些变量的范围是什么? 在所附的代码中,我无法确定b的范围。

 包main 
 
import(
“ fmt” 
)
 
func addMulti(x,y int)(int,int){
 return(  x + y),(x * y)
} 
 
func main(){
 // b变量的作用范围是什么?
a,b:= addMulti(1,2)
 \  n fmt.Printf(“%d%d 
”,a,b)
 
 // b变量的作用范围是什么?
c,b:= addMulti(3,4)
 
  fmt.Printf(“%d%d 
”,c,b)
 
} 
   
 
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3条回答 默认 最新

  • douken0530 2015-01-21 11:51
    已采纳

    We're not talking about the scope of the return value of a function but rather the scope of the variable you assign the return value to.

    The scope of the variable b in your case is the function body, from the point at which you declare it.

    At first you do it at this line:

    a, b := addMulti(1, 2)
    

    But then you use another Short Variable declaration at this line:

    c, b := addMulti(3, 4)
    

    which - since b is already declared - just assigns a new value to it. b will be in scope until the end of your main() function. Quoting from the Go Language Specification:

    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.

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