Re: [pypy-dev] Shift and typed array

[Tuom Larsen]

Alright, thanks again!

On Mon, Apr 4, 2016 at 3:58 PM, Maciej Fijalkowski  wrote:
> you can write a if in_pypy wrapper that uses __pypy__.intops or
> something, but for exactly the same code is hard to do better. We can
> think about adding some optimizations that do that btw
>
>
> On Mon, Apr 4, 2016 at 3:53 PM, Tuom Larsen  wrote:
>> So other than numpy, do you think:
>>
>> a[0] = (2**63 << 1) & (2**64 - 1)
>>
>> is the best way I can do?
>>
>> On Mon, Apr 4, 2016 at 3:52 PM, Tuom Larsen  wrote:
>>> Yes, I just want to write code which is portable (as for example that
>>> `& ...`) but still favors PyPy.
>>>
>>> On Mon, Apr 4, 2016 at 3:49 PM, Maciej Fijalkowski  wrote:
 right, so there is no way to get wrap-around arithmetics in cpython
 without modifications

 On Mon, Apr 4, 2016 at 3:48 PM, Tuom Larsen  wrote:
> I would like to avoid numpy, if possible. Even if it might be bundled
> with PyPy (is it?) I still would like the code to run in CPython with
> no dependencies.
>
> On Mon, Apr 4, 2016 at 3:45 PM, Maciej Fijalkowski  
> wrote:
>> so numpy64 will give you wrap-around arithmetics. What else are you
>> looking for? :-)
>>
>> On Mon, Apr 4, 2016 at 3:38 PM, Tuom Larsen  
>> wrote:
>>> You mean I should first store the result into numpy's `int64`, and
>>> then to `array.array`? Like:
>>>
>>> x = int64(2**63 << 1)
>>> a[0] = x
>>>
>>> Or:
>>>
>>> x = int64(2**63)
>>> x[0] = x << 1
>>>
>>> What the "real types" goes, is this the only option?
>>>
>>> Thanks in any case!
>>>
>>>
>>> On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  
>>> wrote:
 one option would be to use integers from _numpypy module:

 from numpy import int64 after installing numpy.

 There are obscure ways to get it without installing numpy. Another
 avenue would be to use __pypy__.intop.int_mul etc.

 Feel free to complain "no, I want real types that I can work with" :-)

 Cheers,
 fijal

 On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  
 wrote:
> Hello!
>
> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
> [0])` (item size is 8 bytes). In Python, when an integer does not fit
> machine width it gets promoted to "long" integer of arbitrary size. So
> this will fail:
>
> a[0] = 2**63 << 1
>
> To fix this, one could instead write:
>
> a[0] = (2**63 << 1) & (2**64 - 1)
>
> My question is, when I know that the result will be stored in
> `array.array` anyway, how to prevent the promotion to long integers?
> What is the most performat way to perform such calculations? Is PyPy
> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>
> I mean, in C I wouldn't have to worry about it as everything above the
> 63rd bit will be simply cut off. I would like to help PyPy to generate
> the best possible code, does anyone have some suggestions please?
>
> Thanks!
> ___
> pypy-dev mailing list
> pypy-dev@python.org
> https://mail.python.org/mailman/listinfo/pypy-dev
___
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Re: [pypy-dev] Shift and typed array

[Tuom Larsen]

So other than numpy, do you think:

a[0] = (2**63 << 1) & (2**64 - 1)

is the best way I can do?

On Mon, Apr 4, 2016 at 3:52 PM, Tuom Larsen  wrote:
> Yes, I just want to write code which is portable (as for example that
> `& ...`) but still favors PyPy.
>
> On Mon, Apr 4, 2016 at 3:49 PM, Maciej Fijalkowski  wrote:
>> right, so there is no way to get wrap-around arithmetics in cpython
>> without modifications
>>
>> On Mon, Apr 4, 2016 at 3:48 PM, Tuom Larsen  wrote:
>>> I would like to avoid numpy, if possible. Even if it might be bundled
>>> with PyPy (is it?) I still would like the code to run in CPython with
>>> no dependencies.
>>>
>>> On Mon, Apr 4, 2016 at 3:45 PM, Maciej Fijalkowski  wrote:
 so numpy64 will give you wrap-around arithmetics. What else are you
 looking for? :-)

 On Mon, Apr 4, 2016 at 3:38 PM, Tuom Larsen  wrote:
> You mean I should first store the result into numpy's `int64`, and
> then to `array.array`? Like:
>
> x = int64(2**63 << 1)
> a[0] = x
>
> Or:
>
> x = int64(2**63)
> x[0] = x << 1
>
> What the "real types" goes, is this the only option?
>
> Thanks in any case!
>
>
> On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  
> wrote:
>> one option would be to use integers from _numpypy module:
>>
>> from numpy import int64 after installing numpy.
>>
>> There are obscure ways to get it without installing numpy. Another
>> avenue would be to use __pypy__.intop.int_mul etc.
>>
>> Feel free to complain "no, I want real types that I can work with" :-)
>>
>> Cheers,
>> fijal
>>
>> On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  
>> wrote:
>>> Hello!
>>>
>>> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
>>> [0])` (item size is 8 bytes). In Python, when an integer does not fit
>>> machine width it gets promoted to "long" integer of arbitrary size. So
>>> this will fail:
>>>
>>> a[0] = 2**63 << 1
>>>
>>> To fix this, one could instead write:
>>>
>>> a[0] = (2**63 << 1) & (2**64 - 1)
>>>
>>> My question is, when I know that the result will be stored in
>>> `array.array` anyway, how to prevent the promotion to long integers?
>>> What is the most performat way to perform such calculations? Is PyPy
>>> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>>>
>>> I mean, in C I wouldn't have to worry about it as everything above the
>>> 63rd bit will be simply cut off. I would like to help PyPy to generate
>>> the best possible code, does anyone have some suggestions please?
>>>
>>> Thanks!
>>> ___
>>> pypy-dev mailing list
>>> pypy-dev@python.org
>>> https://mail.python.org/mailman/listinfo/pypy-dev
___
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Re: [pypy-dev] Shift and typed array

[Tuom Larsen]

Yes, I just want to write code which is portable (as for example that
`& ...`) but still favors PyPy.

On Mon, Apr 4, 2016 at 3:49 PM, Maciej Fijalkowski  wrote:
> right, so there is no way to get wrap-around arithmetics in cpython
> without modifications
>
> On Mon, Apr 4, 2016 at 3:48 PM, Tuom Larsen  wrote:
>> I would like to avoid numpy, if possible. Even if it might be bundled
>> with PyPy (is it?) I still would like the code to run in CPython with
>> no dependencies.
>>
>> On Mon, Apr 4, 2016 at 3:45 PM, Maciej Fijalkowski  wrote:
>>> so numpy64 will give you wrap-around arithmetics. What else are you
>>> looking for? :-)
>>>
>>> On Mon, Apr 4, 2016 at 3:38 PM, Tuom Larsen  wrote:
 You mean I should first store the result into numpy's `int64`, and
 then to `array.array`? Like:

 x = int64(2**63 << 1)
 a[0] = x

 Or:

 x = int64(2**63)
 x[0] = x << 1

 What the "real types" goes, is this the only option?

 Thanks in any case!


 On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  
 wrote:
> one option would be to use integers from _numpypy module:
>
> from numpy import int64 after installing numpy.
>
> There are obscure ways to get it without installing numpy. Another
> avenue would be to use __pypy__.intop.int_mul etc.
>
> Feel free to complain "no, I want real types that I can work with" :-)
>
> Cheers,
> fijal
>
> On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  wrote:
>> Hello!
>>
>> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
>> [0])` (item size is 8 bytes). In Python, when an integer does not fit
>> machine width it gets promoted to "long" integer of arbitrary size. So
>> this will fail:
>>
>> a[0] = 2**63 << 1
>>
>> To fix this, one could instead write:
>>
>> a[0] = (2**63 << 1) & (2**64 - 1)
>>
>> My question is, when I know that the result will be stored in
>> `array.array` anyway, how to prevent the promotion to long integers?
>> What is the most performat way to perform such calculations? Is PyPy
>> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>>
>> I mean, in C I wouldn't have to worry about it as everything above the
>> 63rd bit will be simply cut off. I would like to help PyPy to generate
>> the best possible code, does anyone have some suggestions please?
>>
>> Thanks!
>> ___
>> pypy-dev mailing list
>> pypy-dev@python.org
>> https://mail.python.org/mailman/listinfo/pypy-dev
___
pypy-dev mailing list
pypy-dev@python.org
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Re: [pypy-dev] Shift and typed array

[Maciej Fijalkowski]

right, so there is no way to get wrap-around arithmetics in cpython
without modifications

On Mon, Apr 4, 2016 at 3:48 PM, Tuom Larsen  wrote:
> I would like to avoid numpy, if possible. Even if it might be bundled
> with PyPy (is it?) I still would like the code to run in CPython with
> no dependencies.
>
> On Mon, Apr 4, 2016 at 3:45 PM, Maciej Fijalkowski  wrote:
>> so numpy64 will give you wrap-around arithmetics. What else are you
>> looking for? :-)
>>
>> On Mon, Apr 4, 2016 at 3:38 PM, Tuom Larsen  wrote:
>>> You mean I should first store the result into numpy's `int64`, and
>>> then to `array.array`? Like:
>>>
>>> x = int64(2**63 << 1)
>>> a[0] = x
>>>
>>> Or:
>>>
>>> x = int64(2**63)
>>> x[0] = x << 1
>>>
>>> What the "real types" goes, is this the only option?
>>>
>>> Thanks in any case!
>>>
>>>
>>> On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  wrote:
 one option would be to use integers from _numpypy module:

 from numpy import int64 after installing numpy.

 There are obscure ways to get it without installing numpy. Another
 avenue would be to use __pypy__.intop.int_mul etc.

 Feel free to complain "no, I want real types that I can work with" :-)

 Cheers,
 fijal

 On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  wrote:
> Hello!
>
> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
> [0])` (item size is 8 bytes). In Python, when an integer does not fit
> machine width it gets promoted to "long" integer of arbitrary size. So
> this will fail:
>
> a[0] = 2**63 << 1
>
> To fix this, one could instead write:
>
> a[0] = (2**63 << 1) & (2**64 - 1)
>
> My question is, when I know that the result will be stored in
> `array.array` anyway, how to prevent the promotion to long integers?
> What is the most performat way to perform such calculations? Is PyPy
> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>
> I mean, in C I wouldn't have to worry about it as everything above the
> 63rd bit will be simply cut off. I would like to help PyPy to generate
> the best possible code, does anyone have some suggestions please?
>
> Thanks!
> ___
> pypy-dev mailing list
> pypy-dev@python.org
> https://mail.python.org/mailman/listinfo/pypy-dev
___
pypy-dev mailing list
pypy-dev@python.org
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Re: [pypy-dev] Shift and typed array

[Maciej Fijalkowski]

but if you dont actually overflow, the cost is really low btw

On Mon, Apr 4, 2016 at 3:49 PM, Maciej Fijalkowski  wrote:
> right, so there is no way to get wrap-around arithmetics in cpython
> without modifications
>
> On Mon, Apr 4, 2016 at 3:48 PM, Tuom Larsen  wrote:
>> I would like to avoid numpy, if possible. Even if it might be bundled
>> with PyPy (is it?) I still would like the code to run in CPython with
>> no dependencies.
>>
>> On Mon, Apr 4, 2016 at 3:45 PM, Maciej Fijalkowski  wrote:
>>> so numpy64 will give you wrap-around arithmetics. What else are you
>>> looking for? :-)
>>>
>>> On Mon, Apr 4, 2016 at 3:38 PM, Tuom Larsen  wrote:
 You mean I should first store the result into numpy's `int64`, and
 then to `array.array`? Like:

 x = int64(2**63 << 1)
 a[0] = x

 Or:

 x = int64(2**63)
 x[0] = x << 1

 What the "real types" goes, is this the only option?

 Thanks in any case!


 On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  
 wrote:
> one option would be to use integers from _numpypy module:
>
> from numpy import int64 after installing numpy.
>
> There are obscure ways to get it without installing numpy. Another
> avenue would be to use __pypy__.intop.int_mul etc.
>
> Feel free to complain "no, I want real types that I can work with" :-)
>
> Cheers,
> fijal
>
> On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  wrote:
>> Hello!
>>
>> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
>> [0])` (item size is 8 bytes). In Python, when an integer does not fit
>> machine width it gets promoted to "long" integer of arbitrary size. So
>> this will fail:
>>
>> a[0] = 2**63 << 1
>>
>> To fix this, one could instead write:
>>
>> a[0] = (2**63 << 1) & (2**64 - 1)
>>
>> My question is, when I know that the result will be stored in
>> `array.array` anyway, how to prevent the promotion to long integers?
>> What is the most performat way to perform such calculations? Is PyPy
>> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>>
>> I mean, in C I wouldn't have to worry about it as everything above the
>> 63rd bit will be simply cut off. I would like to help PyPy to generate
>> the best possible code, does anyone have some suggestions please?
>>
>> Thanks!
>> ___
>> pypy-dev mailing list
>> pypy-dev@python.org
>> https://mail.python.org/mailman/listinfo/pypy-dev
___
pypy-dev mailing list
pypy-dev@python.org
https://mail.python.org/mailman/listinfo/pypy-dev

Re: [pypy-dev] Shift and typed array

[Tuom Larsen]

I would like to avoid numpy, if possible. Even if it might be bundled
with PyPy (is it?) I still would like the code to run in CPython with
no dependencies.

On Mon, Apr 4, 2016 at 3:45 PM, Maciej Fijalkowski  wrote:
> so numpy64 will give you wrap-around arithmetics. What else are you
> looking for? :-)
>
> On Mon, Apr 4, 2016 at 3:38 PM, Tuom Larsen  wrote:
>> You mean I should first store the result into numpy's `int64`, and
>> then to `array.array`? Like:
>>
>> x = int64(2**63 << 1)
>> a[0] = x
>>
>> Or:
>>
>> x = int64(2**63)
>> x[0] = x << 1
>>
>> What the "real types" goes, is this the only option?
>>
>> Thanks in any case!
>>
>>
>> On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  wrote:
>>> one option would be to use integers from _numpypy module:
>>>
>>> from numpy import int64 after installing numpy.
>>>
>>> There are obscure ways to get it without installing numpy. Another
>>> avenue would be to use __pypy__.intop.int_mul etc.
>>>
>>> Feel free to complain "no, I want real types that I can work with" :-)
>>>
>>> Cheers,
>>> fijal
>>>
>>> On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  wrote:
 Hello!

 Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
 [0])` (item size is 8 bytes). In Python, when an integer does not fit
 machine width it gets promoted to "long" integer of arbitrary size. So
 this will fail:

 a[0] = 2**63 << 1

 To fix this, one could instead write:

 a[0] = (2**63 << 1) & (2**64 - 1)

 My question is, when I know that the result will be stored in
 `array.array` anyway, how to prevent the promotion to long integers?
 What is the most performat way to perform such calculations? Is PyPy
 able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?

 I mean, in C I wouldn't have to worry about it as everything above the
 63rd bit will be simply cut off. I would like to help PyPy to generate
 the best possible code, does anyone have some suggestions please?

 Thanks!
 ___
 pypy-dev mailing list
 pypy-dev@python.org
 https://mail.python.org/mailman/listinfo/pypy-dev
___
pypy-dev mailing list
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Re: [pypy-dev] Shift and typed array

[Maciej Fijalkowski]

so numpy64 will give you wrap-around arithmetics. What else are you
looking for? :-)

On Mon, Apr 4, 2016 at 3:38 PM, Tuom Larsen  wrote:
> You mean I should first store the result into numpy's `int64`, and
> then to `array.array`? Like:
>
> x = int64(2**63 << 1)
> a[0] = x
>
> Or:
>
> x = int64(2**63)
> x[0] = x << 1
>
> What the "real types" goes, is this the only option?
>
> Thanks in any case!
>
>
> On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  wrote:
>> one option would be to use integers from _numpypy module:
>>
>> from numpy import int64 after installing numpy.
>>
>> There are obscure ways to get it without installing numpy. Another
>> avenue would be to use __pypy__.intop.int_mul etc.
>>
>> Feel free to complain "no, I want real types that I can work with" :-)
>>
>> Cheers,
>> fijal
>>
>> On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  wrote:
>>> Hello!
>>>
>>> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
>>> [0])` (item size is 8 bytes). In Python, when an integer does not fit
>>> machine width it gets promoted to "long" integer of arbitrary size. So
>>> this will fail:
>>>
>>> a[0] = 2**63 << 1
>>>
>>> To fix this, one could instead write:
>>>
>>> a[0] = (2**63 << 1) & (2**64 - 1)
>>>
>>> My question is, when I know that the result will be stored in
>>> `array.array` anyway, how to prevent the promotion to long integers?
>>> What is the most performat way to perform such calculations? Is PyPy
>>> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>>>
>>> I mean, in C I wouldn't have to worry about it as everything above the
>>> 63rd bit will be simply cut off. I would like to help PyPy to generate
>>> the best possible code, does anyone have some suggestions please?
>>>
>>> Thanks!
>>> ___
>>> pypy-dev mailing list
>>> pypy-dev@python.org
>>> https://mail.python.org/mailman/listinfo/pypy-dev
___
pypy-dev mailing list
pypy-dev@python.org
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Re: [pypy-dev] Shift and typed array

[Tuom Larsen]

You mean I should first store the result into numpy's `int64`, and
then to `array.array`? Like:

x = int64(2**63 << 1)
a[0] = x

Or:

x = int64(2**63)
x[0] = x << 1

What the "real types" goes, is this the only option?

Thanks in any case!


On Mon, Apr 4, 2016 at 3:32 PM, Maciej Fijalkowski  wrote:
> one option would be to use integers from _numpypy module:
>
> from numpy import int64 after installing numpy.
>
> There are obscure ways to get it without installing numpy. Another
> avenue would be to use __pypy__.intop.int_mul etc.
>
> Feel free to complain "no, I want real types that I can work with" :-)
>
> Cheers,
> fijal
>
> On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  wrote:
>> Hello!
>>
>> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
>> [0])` (item size is 8 bytes). In Python, when an integer does not fit
>> machine width it gets promoted to "long" integer of arbitrary size. So
>> this will fail:
>>
>> a[0] = 2**63 << 1
>>
>> To fix this, one could instead write:
>>
>> a[0] = (2**63 << 1) & (2**64 - 1)
>>
>> My question is, when I know that the result will be stored in
>> `array.array` anyway, how to prevent the promotion to long integers?
>> What is the most performat way to perform such calculations? Is PyPy
>> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>>
>> I mean, in C I wouldn't have to worry about it as everything above the
>> 63rd bit will be simply cut off. I would like to help PyPy to generate
>> the best possible code, does anyone have some suggestions please?
>>
>> Thanks!
>> ___
>> pypy-dev mailing list
>> pypy-dev@python.org
>> https://mail.python.org/mailman/listinfo/pypy-dev
___
pypy-dev mailing list
pypy-dev@python.org
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Re: [pypy-dev] Shift and typed array

[Maciej Fijalkowski]

one option would be to use integers from _numpypy module:

from numpy import int64 after installing numpy.

There are obscure ways to get it without installing numpy. Another
avenue would be to use __pypy__.intop.int_mul etc.

Feel free to complain "no, I want real types that I can work with" :-)

Cheers,
fijal

On Mon, Apr 4, 2016 at 3:10 PM, Tuom Larsen  wrote:
> Hello!
>
> Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
> [0])` (item size is 8 bytes). In Python, when an integer does not fit
> machine width it gets promoted to "long" integer of arbitrary size. So
> this will fail:
>
> a[0] = 2**63 << 1
>
> To fix this, one could instead write:
>
> a[0] = (2**63 << 1) & (2**64 - 1)
>
> My question is, when I know that the result will be stored in
> `array.array` anyway, how to prevent the promotion to long integers?
> What is the most performat way to perform such calculations? Is PyPy
> able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?
>
> I mean, in C I wouldn't have to worry about it as everything above the
> 63rd bit will be simply cut off. I would like to help PyPy to generate
> the best possible code, does anyone have some suggestions please?
>
> Thanks!
> ___
> pypy-dev mailing list
> pypy-dev@python.org
> https://mail.python.org/mailman/listinfo/pypy-dev
___
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[pypy-dev] Shift and typed array

[Tuom Larsen]

Hello!

Suppose I'm on 64-bit machine and there is an `a = arrar.array('L',
[0])` (item size is 8 bytes). In Python, when an integer does not fit
machine width it gets promoted to "long" integer of arbitrary size. So
this will fail:

a[0] = 2**63 << 1

To fix this, one could instead write:

a[0] = (2**63 << 1) & (2**64 - 1)

My question is, when I know that the result will be stored in
`array.array` anyway, how to prevent the promotion to long integers?
What is the most performat way to perform such calculations? Is PyPy
able to optimize away that `& (2**64 - 1)` when I use `'L'` typecode?

I mean, in C I wouldn't have to worry about it as everything above the
63rd bit will be simply cut off. I would like to help PyPy to generate
the best possible code, does anyone have some suggestions please?

Thanks!
___
pypy-dev mailing list
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