Time environment variable not working as expected

Question

I trying to measure time using Windows command line.

echo %TIME% & timeout 5 > NUL & echo %TIME%

But this command prints two equal timestamps. Why? What I'am doing wrong? timeout 5 is just for example.

Add `setlocal ENABLEDELAYEDEXPANSION` – Hans Passant Dec 18 '18 at 14:25 — Hans Passant, Dec 18 '18 at 14:25
`setlocal` has no effect outside of batchfiles. – Stephan Dec 18 '18 at 14:27 — Stephan, Dec 18 '18 at 14:27

score 4 · Accepted Answer · answered Dec 18 '18 at 14:26

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The interpreter first parses the whole line (replacing both %time% with the same value (because same time of parsing)) before executing anything. You need delayed expansion to make it work.

If you need a command line solution:

cmd /v /c "echo !time! & timeout 5 & echo !time!"

answered Dec 18 '18 at 14:26

Stephan

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Please, you are a 34.4k reputation, you flag a duplicate and still answer the question? – Matthieu Brucher Dec 18 '18 at 20:35
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@MatthieuBrucher I answered the question, and then someone notified me of the duplicate, so I flagged it as such. (it's a quite wordy dupe handling batchfiiles (the command line solution is hidden somewhere in the middle), so it isn't very obvious and I guess most newbies wouldn't catch the point, so I left the answer) – Stephan Dec 18 '18 at 20:45

score 1 · Answer 2 · answered Dec 18 '18 at 14:26

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This article https://blogs.msdn.microsoft.com/oldnewthing/20060823-00/?p=29993 says the ENV variables are expanded at the time the command is read. Your results would seem to indicate that the entire command line is read/parsed/expanded at once before anything is executed.

answered Dec 18 '18 at 14:26

JJF

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LOL - the very same second `:D` – Stephan Dec 18 '18 at 14:34

Time environment variable not working as expected

2 Answers2