bash script

[Patrick Drechsler]

Kaushal Shriyan wrote:
> Hi,
> 
> while read epoch mem; do d=$(date -d "@$epoch");echo "$d    $mem";done
> < enterprise-core-01.memusage >> ksks.txt
> 
> what does @$epoch means ?

See info date:

===========================================================
28.8 Seconds since the Epoch
============================

If you precede a number with `@', it represents an internal time stamp
as a count of seconds.  The number can contain an internal decimal
point (either `.' or `,'); any excess precision not supported by the
internal representation is truncated toward minus infinity.  Such a
number cannot be combined with any other date item, as it specifies a
complete time stamp.

    Internally, computer times are represented as a count of seconds
since an epoch--a well-defined point of time.  On GNU and POSIX
systems, the epoch is 1970-01-01 00:00:00 UTC, so `@0' represents this
time, `@1' represents 1970-01-01 00:00:01 UTC, and so forth.  GNU and
most other POSIX-compliant systems support such times as an extension
to POSIX, using negative counts, so that `@-1' represents 1969-12-31
23:59:59 UTC.

    Traditional Unix systems count seconds with 32-bit two's-complement
integers and can represent times from 1901-12-13 20:45:52 through
2038-01-19 03:14:07 UTC.  More modern systems use 64-bit counts of
seconds with nanosecond subcounts, and can represent all the times in
the known lifetime of the universe to a resolution of 1 nanosecond.

    On most hosts, these counts ignore the presence of leap seconds.
For example, on most hosts `@915148799' represents 1998-12-31 23:59:59
UTC, `@915148800' represents 1999-01-01 00:00:00 UTC, and there is no
way to represent the intervening leap second 1998-12-31 23:59:60 UTC.
===========================================================

HTH, Patrick