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FamilyGuy
Journeyman III

Another LDS question

Let's assume I have a thread group of size 64, one wavefront. That group uses the entire LDS by means of

dcl_lds_id(0) 32768

Is it correct that I cannot issue a transaction with external memory such as a read without stalling execution on that SIMD unit until this particular read is returned? Meaning, the wavefront cannot be descheduled because it occupies the entire LDS?

If so, is there a way my kernel can inform the chip that it doesn't need the LDS anymore, so that in the above scenario other threads can run?

thx.

 

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FamilyGuy
Journeyman III

Another LDS question

Really nobody with an answer / opinion / comment on this?

Would still be interested...

 

thx.

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Jawed
Adept II

Another LDS question

Memory access latency is hidden by finding another hardware thread for the ALUs to work on. You have declared that you only have a single hardware thread for the ALUs to work on. So you will get no latency hiding.

It's not possible to define a kernel so that it "releases" its LDS allocation. You need to start a new kernel (with no LDS allocation), which would mean saving data from the first kernel so that the second kernel could see it.

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