Sams Teach Yourself Perl in 24 Hours

Workshop

Quiz

Examine the following block of code:

sub bar {
    ($a,$b)=@_;
    $b=100;
    $a=$a+1;
}
sub foo {
    my($a)=67;
    local($b)=@_;
    bar($a, $b);
}
foo(5,10)

Q1:	After you run bar($a, $b), what is the value in $b? 5 100 68
Q2:	What is the return value from foo()? 67 68 undef
Q3:	Inside foo(), what is $b? Lexically scoped Dynamically scoped Globally scoped

Answers

A1:	b. $b is declared with local in foo() so that every called subroutine shares the same value for $b (unless they later declare $b again with local or my). After calling bar(), where $b is modified, $b is set to 100.
A2:	b. Surprised? The last statement in foo() is bar($a, $b). bar() returns 68 because the value of $a is passed to bar(), and it's incremented. foo() returns the value of the last expression, ...