Re: attribution of ++ to the wrong variable in "a ++b"

[Pascal Maugis]

Thanks Ed,

I didn't know that awk cared little for spaces as separators. I will 
switch to parenthesis, then.

May I suggest this aspect be stressed somewhere in the documentation, 
maybe after §1.6, or §6.2 Operators: Doing Something with Values, or 
specifically for ++ in §6.2.4. ?

Cheers,

Pascal.

Le 13/09/2022 à 13:40, Ed Morton a écrit :
> On 9/13/2022 5:42 AM, Pascal Maugis wrote:
> > Hi, as promised here is the strange, unexpected behavior of ++ that 
> > looks like an error to me, because in "a ++b", a is incremented 
> > instead of b. "a b++" and "a++ b" produce the correct results however :
> >
> >    touch a.awk
> >    gawk -D -f ./a.awk
> >    eval "a=1 ; b=2"
> >    eval "print a ++b"   ; output = 12 : (variables will be incremented
> >    after the operation, ok with me as documented)
> >    eval "print a b"     ; output = 22 : a has been incremented 
> > instead of b
> >    eval "print a++b"    ; output = 22 :
> >    eval "print a b"     ; output = 32 : a has been incremented
> >
> > The operation seems to be interpreted as "a++b", the space been 
> > uninterpreted as a separator
> Pascal - with a few exceptions, white space generally doesn't matter 
> to awk. If you want to ensure that `a ++b` gets interpreted as `a 
> (++b)` then that latter is the code you need to write since once you 
> remove/ignore white space `a++ b` and `a ++b` are both `a++b` which 
> visually could be interpreted either way.
>
>     Ed.
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*Pascal MAUGIS*
Modélisation hydrologique et changement climatique, PhD

LSCE

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