CODEWITHSUNDEEP
go
G4143
G4143

Reputation: 2829

Golang: Scope of multi return values from function

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\n", a, b)

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

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

}

Upvotes: 1

Views: 1911

Answers (4)

user3067510
user3067510

Reputation: 37

If that's the case, try the following code:

func makeRequest(uri string, payload string) {
    var resp http.Response
    var err error

    if payload == "" {
        resp, err := http.Get(uri)
    }
    if err != nil {
        return
    }

    bytes, err := io.ReadAll(resp.Body)
    fmt.Println(bytes)
}

Upvotes: -1

Uvelichitel
Uvelichitel

Reputation: 8490

The scope of b variable is main.main(). In second assignment c, b := addMulti(3, 4) you introduce new variable c, and assign variable b introduced in first assignment. If you change second assignment to be a, b := addMulti(3, 4) same as first it want not to compile.

Upvotes: 0

Toni Cárdenas
Toni Cárdenas

Reputation: 6820

It's a normal variable inside a block. From the spec:

The scope of a constant or variable identifier declared inside a function begins at the end of the ConstSpec or VarSpec (ShortVarDecl for short variable declarations) and ends at the end of the innermost containing block.

In the second call, you're just reassigning the value of the same b variable. Its scope is the same.

Upvotes: 0

icza
icza

Reputation: 417452

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.

Upvotes: 10

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