Why Does the Go Playground Timeout While Running the Same Code on a Local Machine?

Discrepancies Between Go Playground and Go on Your Machine: Goroutine Execution

To resolve confusion regarding Goroutine execution, a code snippet was executed on both the Go Playground and a local machine. The Playground produced the error "Process took too long," suggesting the creation of a perpetually running Goroutine, while the local machine resulted in almost instantaneous output. This raises the question of whether the Goroutine within the other() function exits when the main Goroutine finishes or continues running in the background.

Explanation

On the Go Playground, GOMAXPROCS is set to 1. This indicates that only one Goroutine is executed at a time. Without blocking, the scheduler will not switch to other Goroutines.

Within the code, the main Goroutine starts and blocks on the done channel. The scheduler must switch to the other() Goroutine, which sends a value on the done channel, making both Goroutines runnable. The scheduler opts to continue executing other(). Since GOMAXPROCS is 1, the main Goroutine remains blocked.

other() then launches an infinite loop Goroutine. This Goroutine continues running, preventing the main Goroutine from resuming. Consequently, the Playground's timeout occurs.

Locally, GOMAXPROCS typically exceeds 1. As a result, while the infinite loop Goroutine is running, another Goroutine (likely the main Goroutine) is simultaneously executed. When the main Goroutine returns, the program terminates, as it does not await the completion of other non-main Goroutines.

Go Playground Modification

By explicitly setting GOMAXPROCS to 1 on the Go Playground, it is possible to reproduce the "Process took too long" behavior. Conversely, setting GOMAXPROCS to a higher value, such as 2, will often result in a different execution order and successful termination without a timeout.

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