Whats the reason for this??

[JP]

perl -e '1+1+1'
The answer is 3.

but

perl -e 'print (1+1)+1'

thje answer in this case is 2.
Any explination why I get geting a different value in the second case??

Thanks,
JP

 

[Anno Siegel]

perl -e '1+1+1'
The answer is 3.

but

perl -e 'print (1+1)+1'

thje answer in this case is 2.
Any explination why I get geting a different value in the second case??

Consult the documentation of the function you're using. In "perldoc
-f printf", note the sentence beginning "Also be careful..." near the
end.

Anno

 

[John W. Krahn]

Consult the documentation of the function you're using. In "perldoc
-f printf", note the sentence beginning "Also be careful..." near the
end.

The OP will probably find that sentence easier using "perldoc -f print".



John

 

[Anno Siegel]

The OP will probably find that sentence easier using "perldoc -f print".

Oh my. Where did that "f" come from?

Anno

 

[Eric J. Roode]

perl -e '1+1+1'
The answer is 3.

but

perl -e 'print (1+1)+1'

thje answer in this case is 2.
Any explination why I get geting a different value in the second
case??

For Perl operators like print, the rule is "If it LOOKS like a function
call, it BEHAVES like a function call."

--
Eric
`$=`;$_=\%!;($_)=/(.)/;$==++$|;($.,$/,$,,$\,$",$;,$^,$#,$~,$*,$:,@%)=(
$!=~/(.)(.).(.)(.)(.)(.)..(.)(.)(.)..(.)......(.)/,$"),$=++;$.++;$.++;
$_++;$_++;($_,$\,$,)=($~.$"."$;$/$%[$?]$_$\$,$:$%[$?]",$"&$~,$#,);$,++
;$,++;$^|=$";`$_$\$,$/$:$;$~$*$%[$?]$.$~$*${#}$%[$?]$;$\$"$^$~$*.>&$=`

 

[Tad McClellan]

perl -e '1+1+1'
The answer is 3.

but

perl -e 'print (1+1)+1'

thje answer in this case is 2.

Try it with warnings enabled:

perl -we 'print (1+1)+1'

Any explination why I get geting a different value in the second case??

Message-Id: <[email protected]>

 

[Jürgen Exner]

perl -e '1+1+1'
The answer is 3.

but

perl -e 'print (1+1)+1'

thje answer in this case is 2.
Any explination why I get geting a different value in the second case??

Maybe because (1+1) is 2?

jue

 

[Arndt Jonasson]

perl -e '1+1+1'
The answer is 3.

but

perl -e 'print (1+1)+1'

thje answer in this case is 2.
Any explination why I get geting a different value in the second case??

Adding warnings often helps when 'perl' does something mystifying:

% perl -we 'print (1+1)+1'
print (...) interpreted as function at -e line 1.
Useless use of addition (+) in void context at -e line 1.
2
%

 

[Glenn Jackman]

perl -e '1+1+1'
The answer is 3.

but

perl -e 'print (1+1)+1'

thje answer in this case is 2.
Any explination why I get geting a different value in the second case??

This may shed some light on the destination of the "lost" "+1":
perl -le '$result = print(3+5)+10; print $result'
8
11