Re^2: Why does localtime() return 1900-$YEAR?

Further, you need 11 bits to specify 1999, while only 7 are necessary for 99. Believe it or not, being able to to save 4 bits a record (or structure) was a big deal back in the '70s...

Yes, I'm a dinosaur, as I've been programming since 1972...

Comment on Re^2: Why does localtime() return 1900-$YEAR?

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Re^3: Why does localtime() return 1900-$YEAR? by szabgab (Priest) on Jan 31, 2011 at 13:51 UTC
Perl was first released in 1987. The memory issue was much less important by that time and I am quite sure people were already aware of the bug 2000. Though that's an interesting other question. When did people start to talk about "bug 2000"?	[reply]
Re^4: Why does localtime() return 1900-$YEAR? by ikegami (Patriarch) on Jan 31, 2011 at 16:15 UTC
Dunno, but Perl has already addressed the y2038 bug. That's when the years would overflow even when subtracting 1900.	[reply]
Re^4: Why does localtime() return 1900-$YEAR? by Utilitarian (Vicar) on Feb 01, 2011 at 09:28 UTC
Hmm, I worked on a COBOL system with 2 digit dates in the late 80's / early 90's and there was no pressure to modify the date structures. So I'm not sure people were as prescient as you assume. The first time I heard mention of the Y2K issue was '97 when a friend let it be known to a recruiter that I could work in COBOL `>>>Shudder<<<` `print "Good ",qw(night morning afternoon evening)[(localtime)[2]/6]," fellow monks."`	[reply] [d/l] [select]
Re^4: Why does localtime() return 1900-$YEAR? by apl (Monsignor) on Feb 01, 2011 at 11:11 UTC
Perl was first released in 1987. Quite true, but the Unix time structure (which defined the year as an offset from 1900) has been a de facto standard since around 1975.	[reply]