awk to compare hex number

Shell Programming and Scripting
1 
$ awk 'BEGIN{ pat111=0x1000000002E3E02; snBegin=0x1000000002E3E01; if (pat111<=snBegin) printf "a\n"}'
a

Result is not correct.
Looks like the number is too big.
Any idea?
Thx!

2

Looks like a bug in awk

awk 'BEGIN{ pat111=sprintf ("%d", 0x1000000002E3E02); snBegin=sprintf("%d",0x1000000002E3E01);print pat111;print snBegin; if (pat111<=snBegin) printf "a\n"}'

72057594040958464
72057594040958464
a

$ awk 'BEGIN{printf "%X", 72057594040958464}'
1000000002E3E00

$ awk 'BEGIN{printf "%X", 72057594040958465}'
1000000002E3E00

$ echo $((0x1000000002E3E02))
72057594040958466

$ echo $((0x1000000002E3E01))
72057594040958465
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3

I believe this is an implementation limit, not a bug, i.e. these numbers are large for awk.

You may use Perl though:

perl -le'
  $pat111  =  0x1000000002E3E02; 
  $snBegin =  0x1000000002E3E01; 
  
  $pat111 <= snBegin
    and print " ... "
    '
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