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## Re: [bug-gawk] How does awk implement extremely long integers?

**From**: |
Nelson H. F. Beebe |

**Subject**: |
Re: [bug-gawk] How does awk implement extremely long integers? |

**Date**: |
Tue, 31 Jan 2012 10:53:30 -0700 (MST) |

Peng Yu <address@hidden> asks about the size of an integer
in gawk.
All gawk implementations (gawk, nawk, awk, mawk, tawk, jawk, ...) by
default use a numeric type implemented as a double, which on modern
systems is always the IEEE 754 64-bit format. It has a 53-bit
significand, with a separate sign, so numbers in the range
[0, 2**53 - 1] can be represented exactly.
For the applications for which awk was intended, that is mostly adequate.
I have private versions of mawk and nawk that have been extended for
80-bit and 128-bit long double IEEE 754 formats, and also for 128-bit
IEEE 754-2008 decimal arithmetic. See
http://www.math.utah.edu/pub/mathcw/
if you want to experiment with extended arithmetic in awk.
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