find and replace text in a perl script - perl

Question

so I know that there are many ways of doing what I am asking for, but all that I have found is not helping me for what I am trying to do.
It is supposed to be a simple find and replace script using stdin and stdout.
I have a script called replace.pl and this is what i have in it:
#!/usr/bin/perl
use strict;
use warnings;
while(<STDIN>){
$_ = s/$ARGV[0]/$ARGV[1]/g;
print STDOUT $_;
}
When I run echo "replace a with b please" | replace.pl 'a' 'b'
all I get back is a "1". My desire output is "replace b with b please" but what ever I try to do, it is not changing it. Could any one tell me what i am doing wrong here?

Answer 1

Try
s/$ARGV[0]/$ARGV[1]/g;
instead of
$_ = s/$ARGV[0]/$ARGV[1]/g;
as s/// returns 1 when substitution was successful.
You can also quote search pattern if it should be literal string (not a regex),
s/\Q$ARGV[0]\E/$ARGV[1]/g;

Answer 2

From perlop:
s/PATTERN/REPLACEMENT/msixpodualgcer
Searches a string for a pattern, and if found, replaces that pattern with the replacement text and returns the number of substitutions made. Otherwise it returns false (specifically, the empty string).
That's why the code sets $_ = 1. You just want to do s/$ARGV[0]/$ARGV[1]/g; for its substitution side-effect, without assigning its return value to $_.
while (<STDIN>) {
s/$ARGV[0]/$ARGV[1]/g;
print;
}