Why Does an Infinite Loop in One Go Routine Block Others from Sending Data to a Timeout Channel?

Go Routine Blocking Others in Go

In Go, a common issue arises when one go routine blocks the execution of other routines. This question explores why a go routine with an infinite loop might prevent other routines from sending data to a timeout channel.

The problem lies in the use of cooperative scheduling in Go. In this system, goroutines must yield execution back to the scheduler to allow other routines to run. There are several situations where goroutines yield, including:

• Sending or receiving on an unbuffered channel
• Performing system calls (e.g., file reads/writes)
• Allocating memory
• Calling time.Sleep()
• Invoking runtime.Gosched()

In the given code, the go routine with the infinite loop is constantly running without yielding, thus preventing the other routine from writing to the timeout channel. This infinite loop effectively blocks the entire program.

The solution to this issue is to use preemptive scheduling, which would force goroutines to yield periodically. However, Go currently employs cooperative scheduling.

One potential workaround is to increase the number of processors through GOMAXPROCS. This allows multiple goroutines to run simultaneously, but it's important to note that the garbage collector can still stop the world and halt all goroutines if the CPU-bound routine is not yielding.

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