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
[download]

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.
[download]

"\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
[download]

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"
&#9658;                           # not a number, hence 0
[download]

(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
[download]

I hope it's clearer now, since we already had this discussion in one of your last questions!

--> vexed by the hex!