DCPU-16 Bitwise Shifting

Question

I am trying to learn DCPU.

;Set b to 1
SET B,1 ;00001 (1)
SHL B,1 ;00010 (2)
SHL B,1 ;00100 (4)
SHL B,1 ;01000 (8)
SHL B,1 ;10000 (16)

All I am doing is shifting to the left one so shouldn't it simply double itself each shift. In my code you can see i have the last shift turning the registry B into 16 but when ran B ends up as 10 why is that.

Are you sure you aren't accidentally displaying the registry value in hexadecimal? :) — BlueRaja - Danny Pflughoeft, Jul 25 '12 at 19:04
I wish i could answer that all the code is in the original post and the register window says 0x0010 so i am pretty sure that is hex not sure but i am not sure any other way to view it — Bevilacqua, Jul 25 '12 at 19:45

score 7 · Answer 1 · answered Jul 25 '12 at 20:16

7

the register window says 0x0010...

You're getting the correct answer, you're just viewing it in hexadecimal.

Numbers starting with 0x are the standard way of signifying a hexadecimal number; so the number 0x0010 is is really the number 10₁₆ ("one-zero in base sixteen"), which, in decimal, is the number sixteen.

answered Jul 25 '12 at 20:16

BlueRaja - Danny Pflughoeft

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but then shouldn't it be 0x0016 – Bevilacqua Jul 25 '12 at 20:19
2

@Bevilacqua - Nope, 0x16 would be 0d22 in decimal notation. 0x10 = 0d16 = 0o20 (octal) = 0b10000 (binary); they're all the same number. – Kevin Vermeer Jul 25 '12 at 20:25

DCPU-16 Bitwise Shifting

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