Question

Guys perl is not as easy i thought its so confusing thing.I just moved to operators and I wrote some codes but I am unable to figure it out how the compiler treating them.

$in = "42" ;

$out = "56"+32+"good";

print $out,;

The output for above code is 88 and where does the good gone? and Now lets see the other one.

$in ="42";

$out="good52"+32;

print $out ;

and for these the output is 32. The question is where does the good gone that we just stored in $out and the value 52 between the ” “why the compiler just printing the value as 32 but not that remaining text.And the other question is

$in=52;

$in="52"; 

both doing the same work “52” not working as a text . becuase when we add “52”+32 it gives as 84. what is happening and

$in = "hello";

$in = hello; 

both do the same work ? or do they differ but if i print then give the same output.Its just eating up my brain.Its so confusing becuase when “52” or 52 and “hello” or hello doing the same job why did they introduce ” “.I just need the explaination why its happening for above codes.

Answer 1

If you put:

use strict;
use warnings;

At the top of your script, you would get warnings such as:

Argument "good" isn't numeric in addition (+) at ...
Argument "good52" isn't numeric in addition (+) at ...

Perl automatically reassigns a string value to numeric, if possible. So “42” + 10 actually becomes 52. But it cannot do that with a proper string value, such as “good”.