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epsilon

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Joined: 30/07/2020
Location: Belgium
Posts: 255

Posted: 12:27pm 04 Jan 2021

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Hi,

I was doing some fixed point arithmetic experiments and I noticed that right shift appears to be implemented as an unsigned shift even though the document says it's signed:

Documentation:

Quote
These operate in a special way. << means that the value returned
will be the value of x shifted by y bits to the left while >> means the
same only right shifted. They are integer functions and any bits
shifted off are discarded. For a right shift any bits introduced are set
to the value of the top bit (bit 63). For a left shift any bits introduced
are set to zero.

On 5.06 FW:

> PRINT HEX$(&H8000000000000000>>1)
4000000000000000

Is there way to make it signed?

Thanks,
Ruben/Epsilon.

Epsilon CMM2 projects

Volhout
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Joined: 05/03/2018
Location: Netherlands
Posts: 5091

Posted: 12:46pm 04 Jan 2021

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Hi Epsilon,

Normally all shifts are unsigned since they are binary operations. They don't care if the content of a memory location is a character, or an integer number, or a float.

If you want to shift to the right and treat it as a numeric, then you have to do a numeric operation on it. So a divide by 2 replacing the right shift.

PicomiteVGA PETSCII ROBOTS

jirsoft

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Joined: 18/09/2020
Location: Czech Republic
Posts: 533

Posted: 12:55pm 04 Jan 2021

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You can by-self protect the highest bit:

DIM INTEGER a
a = 1 << 63

and then when you need shift right:

DIM INTEGER b = a AND (1 << 63)
a = (a >> 1) OR b

Jiri
Napoleon Commander and SimplEd for CMM2 (GitHub), CMM2.fun

matherp
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Joined: 11/12/2012
Location: United Kingdom
Posts: 10315

Posted: 01:36pm 04 Jan 2021

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Sorry didn't see the original post. This is either a manual error or a bug depending on what is preferred. The origin of this is that on the PIC32 all shift right operations on 64-bit numbers are signed - there is no choice even though Geoff's C source asks for unsigned

Quote void op_shiftright(void) {
iret = (long long int)((unsigned long long int)iarg1 >> (long long int)iarg2);
}

The CMM2 uses the same C source but the STM32 implements it properly so it gives an unsigned answer. I can either change the source to match the documentation or visa-versa

Thoughts?

epsilon

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Joined: 30/07/2020
Location: Belgium
Posts: 255

Posted: 01:46pm 04 Jan 2021

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matherp said I can either change the source to match the documentation or visa-versa
Thoughts?

Although I would prefer a signed shift, since MMBasic integers are signed, correcting the documentation is probably the best option at this point. Changing the source would break compatibility.

In Java >>> is unsigned and >> is signed. Maybe you can do it the other way around in MMBasic? Or maybe add an ASR function?

Epsilon CMM2 projects

matherp
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Posts: 10315

Posted: 02:12pm 04 Jan 2021

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Quote In Java >>> is unsigned and >> is signed. Maybe you can do it the other way around in MMBasic?

I can do that - watch for 5.07.00b2

epsilon

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Joined: 30/07/2020
Location: Belgium
Posts: 255

Posted: 02:25pm 04 Jan 2021

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That's great! Thank you!

Epsilon CMM2 projects

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