$ perl -e"@f=4..5;print scalar( 1,2,3,@f) "
2
$ perl -e"@f=4..5;print scalar( 1,2,3,) "
3
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Nevermind the warnings warn and not using numbers is clear

$ perl -we"@f=( qw/ r p n /);print scalar( q/a/, q/b/, q/c/ ,@f) "
Useless use of a constant ("a") in void context at -e line 1.
Useless use of a constant ("b") in void context at -e line 1.
Useless use of a constant ("c") in void context at -e line 1.
3
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actually you have proven how problematic the wording "list in scalar context" is:

the scalar comma operator is more like a "weak" semicolon, i.e. separating evaluated expressions without terminating a statement.

you'll sometimes see people writing stuff like

  DB<119> $a=0
 => 0

  DB<120> while ($a++,$a<5) {print "$a\n"}
1
2
3
4
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while only excepts one statement, so no semicolon possible.

But every comma separated part is executed, but only the last result is returned

so in your case, you'll get a warning, cause "a" is executed but never returned.

  DB<108> use warnings; ("a","b")
 => ("a", "b")

  DB<109> use warnings; scalar ("a","b")
Useless use of a constant (a) in void context at (eval 34)[multi_perl5
+db.pl:2279] line 1.
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Cheers Rolf

PS: Je suis Charlie!

edit

maybe it's best to say that scalar comma list is kind of a poor man's do block, but called in scalar context.

update

  DB<135> sub ctx {print wantarray ? 'list' : defined wantarray ? 'sca
+lar' : 'void'; return}

  DB<136> @a= do {ctx();ctx()}   # list-do
voidlist
  DB<137> do {ctx();ctx()};1     # void-do
voidvoid
  DB<138> $a=do {ctx();ctx()}    # scalar-do
voidscalar

  DB<140> scalar (ctx(),ctx())   # scalar comma
voidscalar
  DB<141> 1 while (ctx(),ctx())  # scalar (boolean) comma
voidscalar
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