Perl's conditional operator in string context -

this question has answer here:

let's take following minimalistic script:

#!/usr/bin/perl  # # conditional_operator.pl #  use strict;  print ( 1 ? "true" : "false" )." statement\n";  exit;

i expect output "true statement". when execute snippet, see ...

deviolog@home:~/test$ perl conditional_operator.pl true

the " statement\n" concatenation seems ignored.

my perl version v5.14.2. read perlop manual conditional operator , think, string concatenation should possible.

can explain behaviour?

always include use warnings; @ top of every script.

to desired behavior, add parenthesis print called entire argument instead of first part:

print(( 1 ? "true" : "false" )." statement\n");

if you'd had warnings turned on, would've gotten alert:

useless use of concatenation (.) or string in void context

you can avoid undesired behavior leading blank concatenation, or put plus sign before parenthesis:

print +( 1 ? "true" : "false" )." statement\n"; print ''.( 1 ? "true" : "false" )." statement\n";

EIght