When you incorporate @array into a double-quoted construct it is interpolated as a list of elements, rather than with scalar context. The list separator in double-quoted constructs is usually " " (the space character), but can be changed to just about anything by setting $".

So you might think, "Well, I'll just do this: print @array, "\n";.", which would be wrong again, because once again @array is being evaluated in list context, which just returns the elements (this time separated by whatever is set in "$,".

You have to give @array scalar context. Here are a couple of ways:

my $size = @array;
print "$size\n";
# or...
print scalar(@array), "\n";
# or ..
print @array . "\n"; # Because '.' invokes scalar context on its opera
+nds.
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You have (at least) a couple of options:

Your pick. You need to force scalar context, which I explicitly did by using the scalar function.

Context is everything. Used in a quoted string, "@array" will print the elements. You must use it in a numeric context (i.e. if ( @array > 0 )) to have it behave that way.

You may have noticed that Perl uses sigils in front of variable names? They aren't part of the variable name, they show that type of thing the variable (or expression) returns. So a $ sigil returns a scaler 'value' - regardless of how that 'value' is being used. An @ returns a list. A % returns a hash (which is a list of key value pairs). Sigils are more like conversion operators than part of variable identifiers.

This might be the most easiest question in perl. You can get array size by any of the following

my $size = @array;
print $size;
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or

I prefer the following

my @Array = (1,2,3,4);
say 0+@Array; #@Array as a number)
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Its clearer than

If you just want to print the size, this is the simplest way.

Try this !!

Try this trick to evaluate inside quotation string:

print "Array has @{[0+@Array]} elements";
print "Hash has @{[0+keys %Hash]} keys";
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Sorry for necrobumping ^))