I want to enter file names to be worked with dynamically i.e I dont want to provide file names at the begnning of the program. I have written following code, but its not working properly. 

print "file1 ="; 
$file1=<stdin>;
print "file2 =" ;
$file2=<stdin>;
@ARGV = {"$file1", "$file2"} ;
chomp($remove = <stdin>);
chomp($replace = <stdin>);
while(defined($line=<>)){
 $line =~ s/$remove/$replace/g;
 print "$line"; }

[download]

Pls let me know whats wrong.

In reply to providing filenames dynamically by manish.rathi

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