[string] String compare in Perl with "eq" vs "=="

I am (a complete Perl newbie) doing string compare in an if statement:

If I do following:

if ($str1 == "taste" && $str2 == "waste") { }

I see the correct result (i.e. if the condition matches, it evaluates the "then" block). But I see these warnings:

Argument "taste" isn't numeric in numeric eq (==) at line number x.
Argument "waste" isn't numeric in numeric eq (==) at line number x.

But if I do:

if ($str1 eq "taste" && $str2 eq "waste") { }

Even if the if condition is satisfied, it doesn't evaluate the "then" block.

Here, $str1 is taste and $str2 is waste.

How should I fix this?

== does a numeric comparison: it converts both arguments to a number and then compares them. As long as $str1 and $str2 both evaluate to 0 as numbers, the condition will be satisfied.

eq does a string comparison: the two arguments must match lexically (case-sensitive) for the condition to be satisfied.

"foo" == "bar";   # True, both strings evaluate to 0.
"foo" eq "bar";   # False, the strings are not equivalent.
"Foo" eq "foo";   # False, the F characters are different cases.
"foo" eq "foo";   # True, both strings match exactly.

Maybe the condition you are using is incorrect:

$str1 == "taste" && $str2 == "waste"

The program will enter into THEN part only when both of the stated conditions are true.

You can try with $str1 == "taste" || $str2 == "waste". This will execute the THEN part if anyone of the above conditions are true.

Did you try to chomp the $str1 and $str2?

I found a similar issue with using (another) $str1 eq 'Y' and it only went away when I first did:

chomp($str1);
if ($str1 eq 'Y') {
....
}

works after that.

Hope that helps.