| Re: IDL calls C calls IDL? [message #62473] |
Fri, 12 September 2008 18:42  |
hotplainrice@gmail.co
Messages: 15
Registered: July 2008
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Junior Member |
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On Sep 11, 10:42 pm, Joost Aan de Brugh <joost...@gmail.com> wrote:
> On Sep 11, 11:53 am, "hotplainr...@gmail.com" <hotplainr...@gmail.com>
> wrote:
>
>
>
>> Hey again,
>
>> I've been thinking, I have code that does
>
>> LINKIMAGE 'func' ,.....
>
>> for t=0,tmax-1
>> func(data)
>> idl_function(data)
>> endfor
>
>> Is it possible to call idl_function from within C ?
>
>> I want it to be like this
>
>> In IDL:
>> func(data)
>
>> Then it passes onto C which does
>> for (t=0; t<tmax;t++)
>> {
>> do whatever it does with data
>> CALL IDL_FUNCTION(data)
>
>> }
>
>> This eliminates me calling func a few million times
>
>> Can it be done?
>
> Hello,
>
> And in the above example, IDL takes the initiative, calls func (which
> is, if I am right, a C function) and does its own IDL function. (IDL
> calls C a million times).
>
> Below, C takes the initiative, creates a forloop, and in each
> iteration it does something with data and then has IDL do something
> with data by calling IDL_FUNCTION. (C calls IDL a million times).
>
> Which of them is less nasty is not easy to say. To figure out how to
> call IDL from C is better known by C-people. So that could better be
> asked in C-groups.
>
> I do not know what your functions do with the data. But if the order
> is not so important, you could try to do somthing like.
>
> In C:
> (t=0; t<tmax;t++)
> {
> C-code(data);
>
> }
>
> and in IDL:
>
> Call_C_Code(data)
>
> ; Now it's IDL time.
> for t=0,tmax-1 do IDL_FUNCTION,data
>
> But here, C does its whole task before IDL starts. This may be a
> problem depending on the actions. It may be possible that C can
> generate the actions that IDL can perform. For example:
>
> In C:
>
> (t=0; t<tmax;t++)
> {
> C-code(data,actions); // Actions is an output by C passed back to
> IDL. Data is left unharmed.
>
> }
>
> in IDL:
>
> Call_C_Code(data,actions)
>
> ; Now it's IDL time.
> for t=0,tmax-1 do begin
> IDL_PERMORM_ACTION,actions,data
> IDL_FUNCTION,data
> End
>
> Best regards,
> Joost
I should have phrased it in a better way.
It is like this
It all starts from within IDL
So IDL does
for t=0,tmax-1 do begin
Part 1 of code written in IDL
Part 2 of code written in IDL
Part 3 of code written in IDL
end
Then I managed to convert part 1 and 3 into C and CUDA which is leaps
ahead faster.
So now it looks like this
LINKIMAGE, 'Part1_C', 'file', 1, 'call_cuda'
LINKIMAGE, 'Part3_C', 'file', 1, 'c_part3'
for t=0,tmax-1 do begin
Part 1 of code written in C that calls CUDA
Part 2 of code written in IDL
Part 3 of code written in C
end
Then I managed to combine the both Part 1 & 3
LINKIMAGE, 'Part1+3_C', 'file', 1, 'call_cuda'
for t=0,tmax-1 do begin
Part (1+3) of code written in C & CUDA
Part 2 of code written in IDL
end
Okay, so this is the nasty bit. Everytime I call Part(1+3), it calls
the C function. There is no memory transfer because they share memory
addresses, which is all great. However, that C function will allocate
memory and transfer data to the graphics card because it is a normal
CUDA procedure, allocate and transfer then execute computation kernel.
The issue is it has to allocate and transfer data to the graphics
card. It takes awhile especially when you have 434 MBs to send every
iteration. That could easily be x4 = 1.5 Gigs of data to the GPU.
Whats worse is that it sends the same data all the time.
So imagine tmax=10^8, that is 10^8 * (the time to transfer to and
back) = 10^8 * 0.01 = 10^6 seconds.
My aim is for
LINKIMAGE, 'fast_C', 'file', 1, 'call_cuda'
Within IDL
<some IDL commands>
fast_C(..., ..., ...)
<some IDL commands>
Within the C code,
Call_CUDA
end of file
Within CUDA code,
Allocate memory and transfer data
for (t=0; t<tmax;t++)
{
Part 1
Part 3 // They don't necessarily have to be in order
Transfer data from GPU to host
CALL_IDL ( Part 2)
}
So I essentially have eliminated transfering to the GPU 10^8 times and
thus a significant fraction of 10^6 seconds.
Why don't I convert Part 2?
I think its too hard for me to re-invent this wheel, its image writing
code and some special code that is written by someone else
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| Re: IDL calls C calls IDL? [message #62515 is a reply to message #62473] |
Thu, 11 September 2008 05:42  |
Joost Aan de Brugh
Messages: 16
Registered: July 2008
|
Junior Member |
|
|
On Sep 11, 11:53 am, "hotplainr...@gmail.com" <hotplainr...@gmail.com>
wrote:
> Hey again,
>
> I've been thinking, I have code that does
>
> LINKIMAGE 'func' ,.....
>
> for t=0,tmax-1
> func(data)
> idl_function(data)
> endfor
>
> Is it possible to call idl_function from within C ?
>
> I want it to be like this
>
> In IDL:
> func(data)
>
> Then it passes onto C which does
> for (t=0; t<tmax;t++)
> {
> do whatever it does with data
> CALL IDL_FUNCTION(data)
>
> }
>
> This eliminates me calling func a few million times
>
> Can it be done?
Hello,
And in the above example, IDL takes the initiative, calls func (which
is, if I am right, a C function) and does its own IDL function. (IDL
calls C a million times).
Below, C takes the initiative, creates a forloop, and in each
iteration it does something with data and then has IDL do something
with data by calling IDL_FUNCTION. (C calls IDL a million times).
Which of them is less nasty is not easy to say. To figure out how to
call IDL from C is better known by C-people. So that could better be
asked in C-groups.
I do not know what your functions do with the data. But if the order
is not so important, you could try to do somthing like.
In C:
(t=0; t<tmax;t++)
{
C-code(data);
}
and in IDL:
Call_C_Code(data)
; Now it's IDL time.
for t=0,tmax-1 do IDL_FUNCTION,data
But here, C does its whole task before IDL starts. This may be a
problem depending on the actions. It may be possible that C can
generate the actions that IDL can perform. For example:
In C:
(t=0; t<tmax;t++)
{
C-code(data,actions); // Actions is an output by C passed back to
IDL. Data is left unharmed.
}
in IDL:
Call_C_Code(data,actions)
; Now it's IDL time.
for t=0,tmax-1 do begin
IDL_PERMORM_ACTION,actions,data
IDL_FUNCTION,data
End
Best regards,
Joost
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| Re: IDL calls C calls IDL? [message #62555 is a reply to message #62473] |
Mon, 15 September 2008 09:39 |
Karl[1]
Messages: 79
Registered: October 2005
|
Member |
|
|
On Sep 12, 7:42 pm, "hotplainr...@gmail.com" <hotplainr...@gmail.com>
wrote:
> On Sep 11, 10:42 pm, Joost Aan de Brugh <joost...@gmail.com> wrote:
>
>
>
>> On Sep 11, 11:53 am, "hotplainr...@gmail.com" <hotplainr...@gmail.com>
>> wrote:
>
>>> Hey again,
>
>>> I've been thinking, I have code that does
>
>>> LINKIMAGE 'func' ,.....
>
>>> for t=0,tmax-1
>>> func(data)
>>> idl_function(data)
>>> endfor
>
>>> Is it possible to call idl_function from within C ?
>
>>> I want it to be like this
>
>>> In IDL:
>>> func(data)
>
>>> Then it passes onto C which does
>>> for (t=0; t<tmax;t++)
>>> {
>>> do whatever it does with data
>>> CALL IDL_FUNCTION(data)
>
>>> }
>
>>> This eliminates me calling func a few million times
>
>>> Can it be done?
>
>> Hello,
>
>> And in the above example, IDL takes the initiative, calls func (which
>> is, if I am right, a C function) and does its own IDL function. (IDL
>> calls C a million times).
>
>> Below, C takes the initiative, creates a forloop, and in each
>> iteration it does something with data and then has IDL do something
>> with data by calling IDL_FUNCTION. (C calls IDL a million times).
>
>> Which of them is less nasty is not easy to say. To figure out how to
>> call IDL from C is better known by C-people. So that could better be
>> asked in C-groups.
>
>> I do not know what your functions do with the data. But if the order
>> is not so important, you could try to do somthing like.
>
>> In C:
>> (t=0; t<tmax;t++)
>> {
>> C-code(data);
>
>> }
>
>> and in IDL:
>
>> Call_C_Code(data)
>
>> ; Now it's IDL time.
>> for t=0,tmax-1 do IDL_FUNCTION,data
>
>> But here, C does its whole task before IDL starts. This may be a
>> problem depending on the actions. It may be possible that C can
>> generate the actions that IDL can perform. For example:
>
>> In C:
>
>> (t=0; t<tmax;t++)
>> {
>> C-code(data,actions); // Actions is an output by C passed back to
>> IDL. Data is left unharmed.
>
>> }
>
>> in IDL:
>
>> Call_C_Code(data,actions)
>
>> ; Now it's IDL time.
>> for t=0,tmax-1 do begin
>> IDL_PERMORM_ACTION,actions,data
>> IDL_FUNCTION,data
>> End
>
>> Best regards,
>> Joost
>
> I should have phrased it in a better way.
>
> It is like this
>
> It all starts from within IDL
>
> So IDL does
>
> for t=0,tmax-1 do begin
> Part 1 of code written in IDL
> Part 2 of code written in IDL
> Part 3 of code written in IDL
> end
>
> Then I managed to convert part 1 and 3 into C and CUDA which is leaps
> ahead faster.
>
> So now it looks like this
> LINKIMAGE, 'Part1_C', 'file', 1, 'call_cuda'
> LINKIMAGE, 'Part3_C', 'file', 1, 'c_part3'
>
> for t=0,tmax-1 do begin
> Part 1 of code written in C that calls CUDA
> Part 2 of code written in IDL
> Part 3 of code written in C
> end
>
> Then I managed to combine the both Part 1 & 3
>
> LINKIMAGE, 'Part1+3_C', 'file', 1, 'call_cuda'
>
> for t=0,tmax-1 do begin
> Part (1+3) of code written in C & CUDA
> Part 2 of code written in IDL
> end
>
> Okay, so this is the nasty bit. Everytime I call Part(1+3), it calls
> the C function. There is no memory transfer because they share memory
> addresses, which is all great. However, that C function will allocate
> memory and transfer data to the graphics card because it is a normal
> CUDA procedure, allocate and transfer then execute computation kernel.
>
> The issue is it has to allocate and transfer data to the graphics
> card. It takes awhile especially when you have 434 MBs to send every
> iteration. That could easily be x4 = 1.5 Gigs of data to the GPU.
> Whats worse is that it sends the same data all the time.
>
> So imagine tmax=10^8, that is 10^8 * (the time to transfer to and
> back) = 10^8 * 0.01 = 10^6 seconds.
>
> My aim is for
>
> LINKIMAGE, 'fast_C', 'file', 1, 'call_cuda'
>
> Within IDL
>
> <some IDL commands>
> fast_C(..., ..., ...)
> <some IDL commands>
>
> Within the C code,
>
> Call_CUDA
> end of file
>
> Within CUDA code,
> Allocate memory and transfer data
>
> for (t=0; t<tmax;t++)
> {
> Part 1
> Part 3 // They don't necessarily have to be in order
> Transfer data from GPU to host
> CALL_IDL ( Part 2)
>
> }
>
> So I essentially have eliminated transfering to the GPU 10^8 times and
> thus a significant fraction of 10^6 seconds.
>
> Why don't I convert Part 2?
> I think its too hard for me to re-invent this wheel, its image writing
> code and some special code that is written by someone else
I think that there is a C function called IDL_Execute that you can
call from your C program that will let you run IDL statements.
See
http://idlastro.gsfc.nasa.gov/idl_html_help/Executing_IDL_St atements.html
or google for IDL_Execute.
Or read the IDL External Programming Guide
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