First there is nothing like a "hexadecimal division" in Perl.
Numbers are represented internally as integers or floats in binary format.
You are only free to use different base systems for the notation, see perlnumber .
But the division is always the same!
> the division or 0x05 by 0x10 that failed as the result was 0x30 not 0x00
really?
DB<17> p 0x05/0x10
0.3125
DB<18> p int(0x05/0x10)
0
But "\x10" is not a number it's a one-character string with the ASCII code 16 (= 0x10)
DB<25> p int(0x05/"\x10")
Illegal division by zero at (eval 33)[C:/Perl_64/lib/perl5db.pl:646] l
+ine 2.
"\x{Hex}" is just a way to use chr inside interpolated strings!
(See "escape sequences" in perlop#Quote-and-Quote-like-Operators for details.)
Stop using ASCII codes in strings for hex-numbers!
Perl is not C, it will ONLY try to treat a string which looks like a decimal number as it's look-a-like number, all others as number zero in numerical context.
DB<31> p "05"/"16" # strings with decimal(!) numbers
0.3125
Again Perl is not C and will NOT try to take the byte-code of a character as a number, (unless you use ord explicitly ) !
"\x10" doesn't look like a decimal number, it doesn't even look like any number at all!
It's just one character which has position 16 (=0x10) in the ASCII table.
DB<27> p ord("\x10")
16
DB<28> p "\x10"
► # not a number, hence 0
(NB: ► shows as ► on my display this is not a number, not even a character.
It's the Data_Link_Escape character in original ASCII, some OS try to overload it with something more meaningful.)
This can only succeed (somehow) if you use the ASCII-code of a number character.
Only the 10 characters "0".."9" look like a digit in ASCII.
DB<1> p ord("1")
49
DB<2> p ord("\x31")
49
DB<3> p 0 + "\x31" # string = "1" looks like number
1
DB<4> p 5 / "\x31\x36" # string = "16" looks like number
0.3125
I hope it's clearer now, since we already had this discussion in one of your last questions!
--> vexed by the hex!
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