The repetition operator x, being in list context, creates a list of as many question marks as there are items in @array. join joins them with commas.

This line of code creates a placeholder string for a SQL statement sent to the DBI. It has to have one question mark in the statement for each parameter -- and @array contains them.

So...

@array = qw(one two three);
$p_str=join ",",@array;
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But, your sample code has an extra little something... the list of items is ('?')x@array.

The x after a string ('?' in this case) is an operator that says, repeat this string x number of times, where x is the number following it.

So what is x? It is @array, and since this is in scalar context (Perl knows that it is supposed to supply a number not a list), @array returns the number of elements within the array.

so...

@array = qw(one two three);
$p_str = '?'x@array;
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Finally,

@array = qw(one two three);
$p_str = join ",",'?'x@array;
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@array = qw(one two three);
$p_str = join ",","???";
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Since there is only one item, there is no comma, and the final result is:

???
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A minor correction...

@array = qw(one two three);
$p_str = join ',',('?')x@array; ## parens added here puts the '?'s int
+o list format
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The join will return the '?' seperated by commas. "?,?,?"
In the post that you took this from, the '?'s are used as placeholders

$p_str = join ",",'?'x@array;

Sandy's explanation is almost correct. The problem is that in the OP, there are parentheses around the '?' string. Those parentheses matter, because they cause the x operator to act on a list instead of a scalar. So actually, it's more like

my @array = qw(a b c);
my @tmp = ('?') x @array; # @tmp now contains ('?', '?', '?')
my $p_str = join ",", @tmp; # $p_str now contains '?,?,?'
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It does the same thing my_str = ( [ '?' ] * array.size ).join( "," ) does in Ruby.

See also perldoc -f join, and perldoc perlop for the x operator in list context.