Bug #18375: Timeout.timeout(sec, klass: MyExceptionClass) can not retry correctly.

• Status: Rejected
• Assignee: -
• Target version: -
• ruby -v: 3.0.3
• Backport: 2.6: UNKNOWN, 2.7: UNKNOWN, 3.0: UNKNOWN

[zw963 (Wei Zheng)]

Following is a example for describe this issue.

class DelayError < Exception
end

Timeout.timeout(2, DelayError) do |arg|
  puts 'start'
  sleep 10
rescue DelayError
  puts '*'*10
  retry
end

Will output

start
**********
start

What i expected is:

start
**********
start
**********
start
**********
...

Basically, what i expected equivalent is:

begin
  Timeout.timeout(2) do |arg|
    puts 'start'
    sleep 10
  end
rescue TimeoutError
  puts '*'*10
  retry
end

[zw963 (Wei Zheng)]

• Description updated (diff)

[jeremyevans0 (Jeremy Evans)]

• Status changed from Open to Rejected

I don't think this is a bug, and I don't think timeout can work with your proposed code. There's no way the block-level rescue can work the way you want, since what you want requires the timeout method be called again.

This code:

Timeout.timeout(2, DelayError) do |arg|
  puts 'start'
  sleep 10
rescue DelayError
  puts '*'*10
  retry
end

is short for:

Timeout.timeout(2, DelayError) do |arg|
  begin
    puts 'start'
    sleep 10
  rescue DelayError
    puts '*'*10
    retry
  end
end

It's not short for:

begin
  Timeout.timeout(2, DelayError) do |arg|
    puts 'start'
    sleep 10
  end
rescue DelayError
  puts '*'*10
  retry
end

Even if it were short for that, that code doesn't work the way you want either (looks like an infinite loop). You probably want something like:

deadline = Time.now + 10
begin
  Timeout.timeout(2, DelayError) do |arg|
    puts 'start'
    sleep(deadline - Time.now)
  end
rescue DelayError
  puts '*'*10
  retry
end

[zw963 (Wei Zheng)]

jeremyevans0 (Jeremy Evans) wrote in #note-2:

I don't think this is a bug, and I don't think timeout can work with your proposed code. There's no way the block-level rescue can work the way you want, since what you want requires the timeout method be called again.

This code:

Timeout.timeout(2, DelayError) do |arg|
  puts 'start'
  sleep 10
rescue DelayError
  puts '*'*10
  retry
end

is short for:

Timeout.timeout(2, DelayError) do |arg|
  begin
    puts 'start'
    sleep 10
  rescue DelayError
    puts '*'*10
    retry
  end
end

It's not short for:

begin
  Timeout.timeout(2, DelayError) do |arg|
    puts 'start'
    sleep 10
  end
rescue DelayError
  puts '*'*10
  retry
end

Even if it were short for that, that code doesn't work the way you want either (looks like an infinite loop). You probably want something like:

deadline = Time.now + 10
begin
  Timeout.timeout(2, DelayError) do |arg|
    puts 'start'
    sleep(deadline - Time.now)
  end
rescue DelayError
  puts '*'*10
  retry
end

In fact, if consider following code more useful than the actually form.

begin
  Timeout.timeout(2, DelayError) do |arg|
    puts 'start'
    sleep 10
  end
rescue DelayError
  puts '*'*10
  retry
end

Perhaps, we should do some specially process for Timeout.timeout rescue block.