sprintf does not output anything. Maybe you wanted printf? Also, the byte you read from the file needs to be converted to a number before you can output it via %u. Most likely you want ord or unpack for that.

The first warning is simple. sprintf is a function that returns a formatted string. Yet you simple discarded that string. That makes no sense, so you got a warning. Perhaps you wanted printf?

"Numerical context" is a fancy way of saying the argument needs to be a number, and it will be coerced into one if necessary. It's usually very obvious when a number is required. For example, the arguments of addition must be numbers, and so does the argument for %u. Unfortunately, ^R doesn't look anything like a number, so a warning is issued ("isn't a number") and it's treated as zero. You seem to want the character's ordinal value.

printf('%u', ord($foo));
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Perhaps you're from a C background where everything is a number. In Perl, it's more like everything is a string.

                       #  C     Perl
                       #  ----  ----
printf('%u', 4);       #     4     4
printf('%u', '4');     #    52     4
printf('%u', 0x12);    #    18    18
printf('%u', '\x12');  #    18     0*

* - warns "isn't numeric"
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Think about it like this:

A Perl-scalar is not something has only holds a single value, bit rather it consists of a number of "slots" that contain values for the different contexts in which the scalar can be used.

Normally these values are "the same" and are computed by Perl as needed (e.g. a scalar containing a 4 in it's integer-slot will get the value "4" (a string!) in it's string-slot when used in a string-context), but not always.

A good example for this are references. Observe:

my $ref = []; # reference to an array

my $a = $ref + 0; # force numeric context
my $b = "$ref"; # force string-contect

print "$a $b\n";
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