help race conditions

Accelerated Computing CUDA CUDA Programming and Performance
1

Please, can someone explain me in a simple manner when race conditions occurs?

if I do this:

if(a[threadIdx.x+off1]<tex1Dfetch(tex,threadIdx.x+off1))
a[threadIdx.x+off1]=tex1Dfetch(tex,threadIdx.x+off1);

if(a[threadIdx.x+off2]<tex1Dfetch(tex,threadIdx.x+off2))
a[threadIdx.x+off2]=tex1Dfetch(tex,threadIdx.x+off2);

are there race conditions?

thanks in advance for any help.

2

Please, can someone explain me in a simple manner when race conditions occurs?

if I do this:

if(a[threadIdx.x+off1]<tex1Dfetch(tex,threadIdx.x+off1))
a[threadIdx.x+off1]=tex1Dfetch(tex,threadIdx.x+off1);

if(a[threadIdx.x+off2]<tex1Dfetch(tex,threadIdx.x+off2))
a[threadIdx.x+off2]=tex1Dfetch(tex,threadIdx.x+off2);

are there race conditions?

thanks in advance for any help.

3

Yes if more than one thread accesses the same element of [font=“Courier New”]a[/font].

4

Yes if more than one thread accesses the same element of [font=“Courier New”]a[/font].

5

Hi, thank you for your reply. Can you explain me how i can remove it in a simple way? where is exactly the race conditions?

thank in advance

6

Hi, thank you for your reply. Can you explain me how i can remove it in a simple way? where is exactly the race conditions?

thank in advance

7

My apologies. In this little snippet there is no race condition because the textures are always read at the same index where [font=“Courier New”]a[/font] is written (and textures are read-only), so if any two threads access the same index they will also write the same value (if they write at all), and order doesn’t matter.

8

My apologies. In this little snippet there is no race condition because the textures are always read at the same index where [font=“Courier New”]a[/font] is written (and textures are read-only), so if any two threads access the same index they will also write the same value (if they write at all), and order doesn’t matter.

9

Hi,

no problem, and thank you very much for your help. I have another question :what happens if I don’t use the texture memory? and I do something like that:

float p=-3;

 if( label[index+1]>=level && p<phi[index+1])

     p= phi[index+1];

  if( &label[index-1]>=level && p<phi[index-1])

      p=phi[index-1];

  if  (  label[index+im_r]>=level && p<phi[index+im_r])

     p=phi[index+im_r];

 if(label[index-im_r]>=level && p<phi[index-im_r]) 

    p=phi[index-im_r];

if (label[index+(im_rim_c)]>=level && p<phi[index+(im_rim_c)])

     p=phi[index+(im_r*im_c)];

if(label[index-(im_rim_c)]>=level && p<phi[index-(im_rim_c)])

    p=phi[index-(im_r*im_c)];

I think that I’m serializing the access to the same memory location, therefor I would not have race conditions. Is it correct?

I really appreciate your help.

Thanks in advance.

10

Hi,

no problem, and thank you very much for your help. I have another question :what happens if I don’t use the texture memory? and I do something like that:

float p=-3;

 if( label[index+1]>=level && p<phi[index+1])

     p= phi[index+1];

  if( &label[index-1]>=level && p<phi[index-1])

      p=phi[index-1];

  if  (  label[index+im_r]>=level && p<phi[index+im_r])

     p=phi[index+im_r];

 if(label[index-im_r]>=level && p<phi[index-im_r]) 

    p=phi[index-im_r];

if (label[index+(im_rim_c)]>=level && p<phi[index+(im_rim_c)])

     p=phi[index+(im_r*im_c)];

if(label[index-(im_rim_c)]>=level && p<phi[index-(im_rim_c)])

    p=phi[index-(im_r*im_c)];

I think that I’m serializing the access to the same memory location, therefor I would not have race conditions. Is it correct?

I really appreciate your help.

Thanks in advance.