Use means that a module-function import() is called at BEGIN-time.

Normally this function is inherited by Exporter module and does only an aliasing of imported functions.

And aliasing costs only some bytes, not enough to be discussed.

But a module author can chose to define his own import and do whatever he wants... so no general answer here.

A great counterexample is POSIX:

Q:\>perl -MPOSIX -wle "print for grep {$_ =~ /$$/} `tasklist`"
perl.exe                    5508 Console                 0      3.936 
+K


Q:\>perl -MPOSIX=strftime -wle "print for grep {$_ =~ /$$/} `tasklist`
+"
perl.exe                    4904 Console                 0      3.672 
+K
[download]

POSIX.pm exports all its functions and constants by default, which costs 300K memory, while importing just the few needed functions reduces the amount of memory spent.

tl; dr - just like you said, no general answer here.

If you use use something::or::other qw(funt) this method, it will export only funt symbol into the caller namespace.

If you use use something::or::other method, it will export list of symbols (subroutines and variables) of the module to be exported into the caller namespace from @EXPORT array defined in something::or::other

The question was "will it use less memory" not "what is the difference between ..."

The question was "will it use less memory" not "what is the difference between ..."

And then?

/\ thanks to all above

¹ If the used module is built around Autoloader/AutoSplit/SelfLoader, this doesn't hold, as they allow the module to only load and compile subs on-demand instead of compiling the whole thing at program start-up. In my experience, though, this isn't done all that often, so it's probably safe to assume that any random module will be loaded in full if used.