Re: bin/48017: awk(1) fails to treat var as integer (may be related to #47840)

[David Holland <dholland-bugs%netbsd.org@localhost>]

The following reply was made to PR bin/48017; it has been noted by GNATS.

From: David Holland <dholland-bugs%netbsd.org@localhost>
To: gnats-bugs%NetBSD.org@localhost
Cc: gnats-admin%netbsd.org@localhost, netbsd-bugs%netbsd.org@localhost, 
sdaoden%gmail.com@localhost
Subject: Re: bin/48017: awk(1) fails to treat var as integer (may be related
 to #47840)
Date: Wed, 10 Jul 2013 17:06:44 +0000

 On Wed, Jul 10, 2013 at 08:55:01AM +0000, Steffen Daode Nurpmeso wrote:
  >  David Holland <dholland-bugs%netbsd.org@localhost> wrote:
  >   | sprintf witih %d doesn't produce an number value; it produces a
  >   | string value, which you have to coerce to a number by adding zero to
  >   | it to get it to behave like a number.
  >  
  >  (Adding +0 was my final solution too, because GNU awk(1) didn't
  >  make it by the (presumably more expensive, too) sprintf("%X")
  >  call just as all other tested awk(1)s did.)
  >  
  >  So there is a problem with the implicit type conversion, since
  >  
  >    echo f001 f00d |\
  >    awk '{ a=sprintf("%d", "0x" $1); b=sprintf("%d", "0x" $2); while (a < b) 
{ print a; a++; }}'
  >  
  >  works just fine?!?
 
 Compare
 
 % echo f001 | awk '{ a="0x" $1; print a, (a > 700); }'  
 0xf001 0
 
 % echo f001 | awk '{ a="0x" $1; a=sprintf("%d", a); print a, (a > 700); }'
 61441 0
 
 It's not a number; however, the string 61141 is less than the string
 61453.
 
 There isn't, as far as I can see, a problem; it's just that while most
 of the time this stuff is fully transparent and it always does what
 you want, if you start poking at the corners the exact behavior is
 sometimes a bit odd.
 
  >  And because the `Table: Expressions in Decreasing Precedence in awk'
  >  contains the line
  >  
  >    expr < expr   Less than   Numeric   None
 
 What about < on strings?
 
 In particular, if you compare a number and a string, the comparison is
 done as a string.
 
 % echo boo | awk '{ a = 1234; a+=0; print (a > "!") }'
 1
 
 I was arguing elsewhere recently (PR 47840) that in some cases
 pertaining to numeric constants this produces unexpected/unwanted
 results and should produce a warning; but in general it's better than
 doing such comparisons as numbers.
 
 -- 
 David A. Holland
 dholland%netbsd.org@localhost