In Perl, why does variable interpolation fail for a hexadecimal escape sequence?

Question

In Perl, if I run the code:

print "Literal Hex: \x{50} \n";

I get this: "Literal Hex: P"

However, if I run the code:

my $hex_num = 50; 
print "Interpolated Hex: \x{$hex_num}";

The variable does not interpolate properly and I get this: "Interpolated Hex:"

Similar failure results when I attempt to use variable interpolation in unicode and octal escape sequences.

Is it possible to use escape sequences (e.g. \x, \N) with interpolated string variables? I was under the impression that a $variable contained within double quotes is always interpolated, but is this the exception?

Note: Thanks to this question, I am aware of the workaround: chr(hex($hex_num)), but my above questions regarding variable interpolation for escape sequences still stand.

Consider: my $n = 'n'; print "\$n"; It prints "$n", not a newline. The string is only going to be interpolated once. — rutter
– rutter, Commented Aug 2, 2013 at 6:45
Had to read the whole thing just to get to the workaround of... print(chr(hex($hex_num))); — HoldOffHunger
– HoldOffHunger, Commented Apr 12, 2016 at 21:41

HoldOffHunger · Accepted Answer · 2018-08-27 18:52:07Z

7

Interpolation is not recursive, everything is interpolated just once, from left to right. Therefore, when \x{$hex} is being processed, the following applies (cited from perlop):

If there are no valid digits between the braces, the generated character is the NULL character ("\x{00}").

Zero is really there:

perl -MO=Deparse -e '$h=50;print "<\x{$h}>"'
$h = 50;
print "<\000>";
-e syntax OK

edited Aug 27, 2018 at 18:52

HoldOffHunger

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answered Aug 2, 2013 at 6:40

choroba

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1 Comment

schulwitz Over a year ago

Wow, thank you, answered perfectly! I don't know if I ever would have found this on my own.

Miguel Prz · Accepted Answer · 2013-08-02 06:19:21Z

1

You should put in your variable the complete scape sequence:

my $hex_num = "\x50"; 
print "Interpolated Hex: $hex_num", "\n";

answered Aug 2, 2013 at 6:19

Miguel Prz

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3 Comments

schulwitz Over a year ago

I appreciate your answer but it doesn't get at the fundamental question I'm trying to answer here, which is why does "\x{$hex_num}" fail? Your answer relies on using a string literal, which works in one instance, but is not very practical for cycling $hex_num through every hex character from "\x001" to "\x501".

Miguel Prz Over a year ago

short answer: because there is not a double (2-phases) interpolation. There is only a single step

HoldOffHunger Over a year ago

Interesting alternative, did not think of this. Wouldn't fit my case, as I wanted a for-loop of integers and their sequences, but may be useful for someone else.

Taryn · Accepted Answer · 2014-04-01 11:36:11Z

1

The issue I had was adding an escaped var into another variable such as:

$MYVAR = "20";
$myQuery = "\x02\x12\x10\x$MYVAR\x10";

Tried a number of \\x, \Q\x and various other escape sequences to no avail!!!

My workaround was not a direct escape but converting the var prior to adding to the string.

$MYVAR = chr(hex(20));

Did quite a bit of searching for a direct regex solution but had to run with this in the end.

edited Apr 1, 2014 at 11:36

Taryn

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answered Mar 27, 2014 at 19:31

PodTech.io

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