Re: [bug-gawk] Is this a bug

[Manuel Collado]

El 08/03/2014 12:44, Seppo Lehtikangas escribió:
> I have been using some time to find annoying error in my awk-code.
>
> I had to strip almost the whole code to find the error because awk tells me
> that there is syntax-error on else-statement(line 12) while the fact is
> that error is on if-statement(line 9).
>
> Is there some bug or do I have to learn other way interpreted error messages?

Mostly the second. See below.

> My code:
> ===============
> #! /usr/bin/awk -f
> # ~bin/iftest
> #
> {
> #printf "%s %s %s %s \n", ARGV[0], ARGV[1], ARGV[2], ARGV[3]
>
> #if (NF == 0) print "usage iftest [1 | anything]"
>
> If ($1 == 1)
>     {print "number one"
>      printf "%s\n", $0}
> else
>     {if (NF == 0)
>         {print "empty row"}
>      else
>         {print "non one"
>         printf "%s\n", $0}}
> }
> ===============
> running code gives:
> ========
> iftest test
> awk: /mnt/L6EVM/seppo/bin/iftest.error:12: else
> awk: /mnt/L6EVM/seppo/bin/iftest.error:12: ^ syntax error
> ========

This diagnostic just tells that the compiler cannot accept the 'else' token 
as part of the current syntax construction been analyzed. This doesn't 
imply that the 'else' itself (in line 12) is the culprit.

As you correctly assume, it seems that the fault is to write 'If' 
(capitalized, in line 9) instead of 'if' (lowercase).

And, as posted elsewhere, the capitalized 'If' is accepted as the name of 
an uninitialized variable. So the code fragment:

If ($1 == 1)
   {print "number one"
    printf "%s\n", $0}

is valid up to this point:

If ($1 == 1)           ---> arithmetic expression statement (concatenation)
   {print "number one" ---> block statement
    printf "%s\n", $0}

Valid continuations would be '}' (to close the current pattern-action rule) 
or another statement.

Hope tis helps.
--
Manuel Collado - http://lml.ls.fi.upm.es/~mcollado