## digitalmars.D.learn - signed nibble

• Michelle Long (3/3) Jan 07 Is there any direct way to convert a signed nibble in to a signed
Michelle Long <HappyDance321 gmail.com> writes:
Is there any direct way to convert a signed nibble in to a signed
byte with the same absolute value? Obviously I can do some bit
comparisons but just curious if there is a very quick way.
Jan 07
Patrick Schluter <Patrick.Schluter bbox.fr> writes:
On Monday, 7 January 2019 at 17:23:19 UTC, Michelle Long wrote:
Is there any direct way to convert a signed nibble in to a
signed byte with the same absolute value? Obviously I can do
some bit comparisons but just curious if there is a very quick
way.
byte b = nibble | ((nibble & 0x40)?0xF0:0);
Jan 07
Adam D. Ruppe <destructionator gmail.com> writes:
On Monday, 7 January 2019 at 18:42:13 UTC, Patrick Schluter wrote:
byte b = nibble | ((nibble & 0x40)?0xF0:0);
don't you mean & 0x80 ?
Jan 07
Patrick Schluter <Patrick.Schluter bbox.fr> writes:
On Monday, 7 January 2019 at 18:47:04 UTC, Adam D. Ruppe wrote:
On Monday, 7 January 2019 at 18:42:13 UTC, Patrick Schluter
wrote:
byte b = nibble | ((nibble & 0x40)?0xF0:0);
don't you mean & 0x80 ?
He asked for signed nybble. So mine is wrong and yours also :-) It's obviously 0x08 for the highest bit of the low nybble. byte b = nibble | ((nibble & 0x08)?0xF0:0);
Jan 07
"H. S. Teoh" <hsteoh quickfur.ath.cx> writes:
On Mon, Jan 07, 2019 at 06:42:13PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
On Monday, 7 January 2019 at 17:23:19 UTC, Michelle Long wrote:
Is there any direct way to convert a signed nibble in to a signed
byte with the same absolute value? Obviously I can do some bit
comparisons but just curious if there is a very quick way.
byte b = nibble | ((nibble & 0x40)?0xF0:0);
This is equivalent to doing a bit comparison (implied by the ? operator). You can do it without a branch: cast(byte)(nibble << 4) >> 4 will use the natural sign extension of a (signed) byte to "stretch" the upper bit. It just takes 2-3 CPU instructions. T -- Written on the window of a clothing store: No shirt, no shoes, no service.
Jan 07
Patrick Schluter <Patrick.Schluter bbox.fr> writes:
On Monday, 7 January 2019 at 18:56:17 UTC, H. S. Teoh wrote:
On Mon, Jan 07, 2019 at 06:42:13PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
On Monday, 7 January 2019 at 17:23:19 UTC, Michelle Long wrote:
Is there any direct way to convert a signed nibble in to a
signed byte with the same absolute value? Obviously I can do
some bit comparisons but just curious if there is a very
quick way.
byte b = nibble | ((nibble & 0x40)?0xF0:0);
This is equivalent to doing a bit comparison (implied by the ? operator). You can do it without a branch: cast(byte)(nibble << 4) >> 4 will use the natural sign extension of a (signed) byte to "stretch" the upper bit. It just takes 2-3 CPU instructions.
Yeah, my bit-fiddle-fu goes back to pre-barrel-shifter days. Up to 32 bit processors, shifting was more expensive than branching.
Jan 07
"H. S. Teoh" <hsteoh quickfur.ath.cx> writes:
On Mon, Jan 07, 2019 at 08:06:17PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
On Monday, 7 January 2019 at 18:56:17 UTC, H. S. Teoh wrote:
On Mon, Jan 07, 2019 at 06:42:13PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
[...]
byte b = nibble | ((nibble & 0x40)?0xF0:0);
This is equivalent to doing a bit comparison (implied by the ? operator). You can do it without a branch: cast(byte)(nibble << 4) >> 4 will use the natural sign extension of a (signed) byte to "stretch" the upper bit. It just takes 2-3 CPU instructions.
Yeah, my bit-fiddle-fu goes back to pre-barrel-shifter days. Up to 32 bit processors, shifting was more expensive than branching.
Really? Haha, never knew that, even though I date all the way back to writing assembly on 8-bit processors. :-D T -- One reason that few people are aware there are programs running the internet is that they never crash in any significant way: the free software underlying the internet is reliable to the point of invisibility. -- Glyn Moody, from the article "Giving it all away"
Jan 07
Patrick Schluter <Patrick.Schluter bbox.fr> writes:
On Monday, 7 January 2019 at 20:28:21 UTC, H. S. Teoh wrote:
On Mon, Jan 07, 2019 at 08:06:17PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
On Monday, 7 January 2019 at 18:56:17 UTC, H. S. Teoh wrote:
On Mon, Jan 07, 2019 at 06:42:13PM +0000, Patrick Schluter
via Digitalmars-d-learn wrote:
[...]
byte b = nibble | ((nibble & 0x40)?0xF0:0);
This is equivalent to doing a bit comparison (implied by the ? operator). You can do it without a branch: cast(byte)(nibble << 4) >> 4 will use the natural sign extension of a (signed) byte to "stretch" the upper bit. It just takes 2-3 CPU instructions.
Yeah, my bit-fiddle-fu goes back to pre-barrel-shifter days. Up to 32 bit processors, shifting was more expensive than branching.
Really? Haha, never knew that, even though I date all the way back to writing assembly on 8-bit processors. :-D
Most of my career was programming for 80186. Shifting by one was 2 cycles in register and 15 in memory. Shifting by 4, 9 cycles for regs/21 for mem. And 80186 was a fast shifter compared to 8088/86 or 68000 (8+2n cycles).
Jan 07
"H. S. Teoh" <hsteoh quickfur.ath.cx> writes:
On Mon, Jan 07, 2019 at 08:41:32PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
On Monday, 7 January 2019 at 20:28:21 UTC, H. S. Teoh wrote:
On Mon, Jan 07, 2019 at 08:06:17PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
[...]
Up to 32 bit processors, shifting was more expensive than
branching.
Really? Haha, never knew that, even though I date all the way back to writing assembly on 8-bit processors. :-D
Most of my career was programming for 80186. Shifting by one was 2 cycles in register and 15 in memory. Shifting by 4, 9 cycles for regs/21 for mem. And 80186 was a fast shifter compared to 8088/86 or 68000 (8+2n cycles).
I used to hack 6502 assembly code. During the PC revolution I wrote an entire application in 8088 assembly. Used to know many of the opcodes and cycle counts by heart like you do, but it's all but a faint memory now. T -- Right now I'm having amnesia and deja vu at the same time. I think I've forgotten this before.
Jan 07
Patrick Schluter <Patrick.Schluter bbox.fr> writes:
On Monday, 7 January 2019 at 21:46:21 UTC, H. S. Teoh wrote:
On Mon, Jan 07, 2019 at 08:41:32PM +0000, Patrick Schluter via
Digitalmars-d-learn wrote:
On Monday, 7 January 2019 at 20:28:21 UTC, H. S. Teoh wrote:
On Mon, Jan 07, 2019 at 08:06:17PM +0000, Patrick Schluter
via Digitalmars-d-learn wrote:
[...]
Up to 32 bit processors, shifting was more expensive than
branching.
Really? Haha, never knew that, even though I date all the way back to writing assembly on 8-bit processors. :-D
Most of my career was programming for 80186. Shifting by one was 2 cycles in register and 15 in memory. Shifting by 4, 9 cycles for regs/21 for mem. And 80186 was a fast shifter compared to 8088/86 or 68000 (8+2n cycles).
I used to hack 6502 assembly code.
Yeah, that's also what I started with, on the Apple II in the early 80s. I was quite surprized that my 6502 knowledge came in very handy when we worked on dial-in modems in the late 90s as the Rockwell modems all used 6502 derived micro-controllers for them.
During the PC revolution I wrote an entire application in 8088
assembly.  Used to know many of the opcodes and cycle counts by
heart like you do, but it's all but a faint memory now.
I had to lookup the exact cycle counts ;-) . I remember the relative costs, more or less, but not the details anymore.
Jan 08
Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= <ola.fosheim.grostad gmail.com> writes:
On Tuesday, 8 January 2019 at 09:30:14 UTC, Patrick Schluter
wrote:
During the PC revolution I wrote an entire application in 8088
assembly.  Used to know many of the opcodes and cycle counts
by heart like you do, but it's all but a faint memory now.
I had to lookup the exact cycle counts ;-) . I remember the relative costs, more or less, but not the details anymore.
Heh, I remember they had a friday-night trivia contest at the mid-90s students pub (for natural sciences) where one of the questions was the opcode for 6502 LDA (or was it NOP?), and I believe I got it right. The opcode for NOP is burned into my memory as \$EA was used for erasing code during debugging in a monitor. And it was also the letters for the big game company Electronic Arts... The cycle counts for 6502 are pretty easy though as they tend to be related to the addressing mode and most of them are in the range 1-5... No instruction for multiplication or division... Oh the fun...
Jan 08
Patrick Schluter <Patrick.Schluter bbox.fr> writes:
On Tuesday, 8 January 2019 at 10:32:25 UTC, Ola Fosheim GrĂ¸stad
wrote:
On Tuesday, 8 January 2019 at 09:30:14 UTC, Patrick Schluter
wrote:
[...]
Heh, I remember they had a friday-night trivia contest at the mid-90s students pub (for natural sciences) where one of the questions was the opcode for 6502 LDA (or was it NOP?), and I believe I got it right. The opcode for NOP is burned into my memory as \$EA was used for erasing code during debugging in a monitor. And it was also the letters for the big game company Electronic Arts... The cycle counts for 6502 are pretty easy though as they tend to be related to the addressing mode and most of them are in the range 1-5... No instruction for multiplication or division... Oh the fun...
2-7 cycles ;-)
Jan 08
Ola Fosheim =?UTF-8?B?R3LDuHN0YWQ=?= <ola.fosheim.grostad gmail.com> writes:
On Tuesday, 8 January 2019 at 10:55:59 UTC, Patrick Schluter
wrote:
The cycle counts for 6502 are pretty easy though as they tend
to be related to the addressing mode and most of them are in
the range 1-5... No instruction for multiplication or
division... Oh the fun...
2-7 cycles ;-)
There you go, the last one in that series I touched was 6800 in 1990. But it kinda make sense, the instructions were not pipelined, so even NOP takes 2 cycles... Another world really.
Jan 08
"H. S. Teoh" <hsteoh quickfur.ath.cx> writes:
On Mon, Jan 07, 2019 at 05:23:19PM +0000, Michelle Long via Digitalmars-d-learn
wrote:
Is there any direct way to convert a signed nibble in to a signed byte
with the same absolute value? Obviously I can do some bit comparisons
but just curious if there is a very quick way.
Assuming you have the nibble stored in the lower bits of a ubyte: import std.stdio; byte nibSgnExt(ubyte nib) { return cast(byte)(nib << 4) >> 4; } void main() { writefln("%02X", nibSgnExt(0x0F)); writefln("%02X", nibSgnExt(0x07)); } T -- If it breaks, you get to keep both pieces. -- Software disclaimer notice
Jan 07