CODEWITHSUNDEEP
functiongoclosuresreturn-value
Bob
Bob

Reputation: 1075

Golang closure not storing captured variables?

In this block of code below, I expect the output to be 0, 1, 2, 3,..., 9. However, only the output 10, 10,...,10 is produced.

package main

import "fmt"

func adder() []func() {
    out := []func(){}
    for i := 0; i < 10; i++ {
        out = append(out, func() { fmt.Println(i) })
    }

    return out
}

func main() {
    out := adder()
    for i := 0; i < 10; i++ {
        out[i]()
    }
}

My mental model suggests that a new function is generated and stored every time in the for-loop, but it doesn't seem to be the case. Is there a solution to generating new functions without modifying the signature (no parameters or global variables) but still keeping the current value of i?

Upvotes: 6

Views: 1372

Answers (2)

Mohammad Reza Karimi
Mohammad Reza Karimi

Reputation: 2057

This behavior has changed in the Go version 1.22, have a look at the release note, which says:

Previously, the variables declared by a "for" loop were created once and updated by each iteration. In Go 1.22, each iteration of the loop creates new variables, to avoid accidental sharing bugs.

And this example on the playground: https://go.dev/play/p/djpXI9ZqunG

Upvotes: 1

Adrian
Adrian

Reputation: 46423

A new function is created every iteration, but they all close over the same variable, i. That variable is updated with every iteration until the last, so when you go through and execute all those functions, they all refer to the same i, with the same value, 10.

If you don't want that behavior, you must not close over the loop variable:

for i := 0; i < 10; i++ {
    temp := i
    out = append(out, func() { fmt.Println(temp) })
}

Upvotes: 9

Related Questions