Size of variables in awk

[Dan Nelson]

In the last episode (Mar 03), Wayne Sierke said:
> On Wed, 2004-03-03 at 18:34, Dan Nelson wrote:
> > In the last episode (Mar 03), Wayne Sierke said:
> > > It seems I've run into the 32-bit signed number wall in awk
> > > (5.2-RELEASE).
> > > 
> > > My totals are maxing out at 2147483648.
> > > 
> > > Would anyone happen to know whether that's really the case (that awk is
> > > only implemented with 32-bit number capability - unfortunately I don't
> > > have any other awks nearby to verify nor can I find any reference info
> > > that indicates) and/or can suggest a way around it?
> > 
> > Seems to works fine on -current:
> > 
> > $ jot 8 30 | awk '{ print 2^$1 }'
> 
> Ah, ok. Same for me on 5.2-RELEASE. More info:
> 
> I'm using the printf function in awk but something ain't right:
> 
> # jot 4 30 | awk '{ printf("%u\n", 2^$1-1) }'
> 2147483648
> 
> # jot 4 30 | awk '{ printf("%lu\n", 2^$1-1) }'
> 2147483648
> 
> # jot 4 30 | awk '{ printf("%llu\n", 2^$1-1) }'
> 35186519572480

I see nothing wrong here.  %u is an unsigned int, and on x86 systems,
an int is 32 bits.  %llu is a long long unsigned int, and they are 64
bits.  Since there is no way for C to print a number larger than 64
bits, you won't be able to use the numeric specifiers to print large
numbers.  You can use %s though.  See
/usr/src/contrib/one-true-awk/run.c, the format() function.

-- 
	Dan Nelson
	dnelson at allantgroup.com