Table of Contents
Is This a Result of Compiler Optimization?
The Go Memory Model
Omitting Synchronization
Optimization by the Compiler
Example with Synchronization
Home Backend Development Golang Why Does My Goroutine Incrementing a Variable Produce Unexpected Results?

Why Does My Goroutine Incrementing a Variable Produce Unexpected Results?

Oct 29, 2024 pm 08:38 PM

Why Does My Goroutine Incrementing a Variable Produce Unexpected Results?

Is This a Result of Compiler Optimization?

In this code snippet, a goroutine is launched and repeatedly increments the variable i:

<code class="go">package main

import "time"

func main() {
    i := 1
    go func() {
        for {
            i++
        }
    }()
    <-time.After(1 * time.Second)
    println(i)
}</code>
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However, the output is always 1. This behavior can be attributed to the Go memory model and the specific implementation of this code.

The Go Memory Model

The Go memory model defines the conditions under which reads of a variable in one goroutine can be guaranteed to observe values produced by writes to the same variable in a different goroutine. It emphasizes the importance of synchronization for concurrent access to shared data.

Omitting Synchronization

In the given code:

  • The assignment to i (i.e., i ) is not followed by any synchronization event, indicating that changes may not be immediately visible to other goroutines.
  • The compiler may optimize this loop optimization by simplifying it to a no-op.

Optimization by the Compiler

An aggressive compiler might delete the i statement, effectively reducing the goroutine to:

<code class="go">for {}</code>
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Example with Synchronization

To demonstrate that the issue stems from the lack of synchronization, consider the following code:

<code class="go">package main

import (
    "sync"
    "time"
)

func main() {
    mx := new(sync.Mutex)
    i := 1
    go func() {
        for {
            mx.Lock()
            i++
            mx.Unlock()
        }
    }()
    <-time.After(1 * time.Second)
    mx.Lock()
    println(i)
    mx.Unlock()
}</code>
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In this case, the output is no longer 1, but a large number, as expected. The sync.Mutex provides synchronization and ensures that both goroutines access i in a controlled manner, allowing the goroutine to increment i and making the changes visible to the main routine.

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