What's the difference between tr/// and s/// when using regex in Perl?

Question

I was wondering when one should use s/// over tr/// when working with regular expressions in Perl?

Answer 1

s/// is for substitution:

$string =~ s/abc/123/;

This will replace the first "abc" found in $string with "123".

tr/// is for transliteration:

$string =~ tr/abc/123/;

This will replace all occurrences of "a" within $string with "1", all occurrences of "b" with "2", and all occurrences of "c" with "3".

Answer 2

From perlop: Quote and Quote-like Operators

Note that tr does not do regular expression character classes such as \d or [:lower:]. The tr operator is not equivalent to the tr(1) utility. If you want to map strings between lower/upper cases, see lc and uc, and in general consider using the s operator if you need regular expressions.

Answer 3

tr/// is not a regular expression operator. It is suitable (and faster than s///) for substitutions of one single character with another single character, or (with the d modifier) substituting a single character with zero characters.

s/// should be used for anything more complicated than the narrow use cases of tr.