Re: Re: time() function

So, I understand this is true. I'm O.K. with overflows. I can handle overflows. But why does

$var = time * 1000;
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overflow and

 
$var = time; $var *= 1000;
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not overflow?

I believe that on my system (Win2k, perl 5.6.1) integers are 32 bit, just like returned by time() I verified this with print ($var&(~0)) which gives me (2**32)-1.

Does perl automatically decide (not) to use BigInts? Any ideas?

Ira,

"So... What do all these little arrows mean?"
~unknown

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Re: Re: Re: time() function by tachyon (Chancellor) on Sep 28, 2001 at 22:48 UTC
time() is a Perl function, implemented in C with 32 bit (long int) math relevant to the unix epoch. A perl scalar is an interesting beast (also implemented in C) that can hold a string or a number or a reference so is not directly comparable to a short, int, long, float, double, char in C. It is in fact a struct. Have a look at this code and perlman:perlguts. On the Win95 box in front of me Perl will represent up to 249 (562,949,953,421,312) as an integer before resorting to scientific notation. `for ( 0..64 ) { $var = 2$_; print "Perl can do 2**$_ : $var\n"; }` [download] cheers tachyon s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print	[reply] [d/l]

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Re: Re: Re: time() function
by tachyon (Chancellor) on Sep 28, 2001 at 22:48 UTC

time() is a Perl function, implemented in C with 32 bit (long int) math relevant to the unix epoch.

A perl scalar is an interesting beast (also implemented in C) that can hold a string or a number or a reference so is not directly comparable to a short, int, long, float, double, char in C. It is in fact a struct. Have a look at this code and perlman:perlguts. On the Win95 box in front of me Perl will represent up to 2**49 (562,949,953,421,312) as an integer before resorting to scientific notation.

for ( 0..64 ) {
    $var = 2**$_;
    print "Perl can do 2**$_ : $var\n";
}
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cheers

tachyon

s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

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