Re^2: Question on simple arithmetic using perl

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Re^3: Question on simple arithmetic using perl by Marshall (Canon) on Sep 28, 2011 at 23:08 UTC
No. A float can represent a wider range of values, but a lesser number of significant digits than an integer. I'm not quite sure what the question is: `#!/usr/bin/perl -w use strict; my $all_bits_are_on = -1; printf ("all_bits => %X\n", $all_bits_are_one); my $x = $all_bits_are_on + 1; print "Adding -1 to +1 is: $x\n"; __END__ all_bits => FFFFFFFF Adding -1 to +1 is: 0` [download]	[reply] [d/l]
Re^4: Question on simple arithmetic using perl by zwon (Abbot) on Sep 29, 2011 at 02:20 UTC
Note that perl is actually using long and double. On 32bit double has more significant digits than long.	[reply]
Re^5: Question on simple arithmetic using perl by Marshall (Canon) on Sep 29, 2011 at 03:29 UTC
Yes, if you mean that a 64 bit floating point number has more significant bits in the mantissa than a 32 bit integer does, I agree. I leave looking up the definition of ANSI C long and double to the interested readers. Update: I looked up what Microsoft did for storage in C. "int" and "long" are both 32 bits (other folks will assign 64 bits to a "long"). A MS "float" is 32 bits and MS "double" is 64 bits with 52 bit mantissa. ANSI C Summary says: that an int will have at least 4 significant decimal digits and a long will have at least 9 significant decimal digits (see link for exact ANSI spec'd ranges). An int can be the same size as a long (bigger than minimum is ok). A float has to have 6 decimal digits of precision and a double has a minimum of 10 decimal digits of precision. So a double is guaranteed to have at least one more decimal digit of precision than a long (10 vs 9).	[reply]