What is the default memory alignment used by Memory_alloc()

Mark Goldberg

According to the xdc.runtime.Memory module documentation, calling Memory_alloc() with the align argument set to 0 denotes "maximum default type alignment".

My question is, what is the maximum default type alignment? Can I rely on at least 8-byte alignment if the align argument is 0?

I am using CCS v5.2 and a C6474 multicore DSP.

Many thanks,

Mark

over 12 years ago

0 Sasha Slijepcevic over 12 years ago

TI__Mastermind 37985 points

If alignment is not specified, the value that will be used is the default alignment for a primitive type that requires the largest alignment. For 6747, double type requires 8-byte alignment, so you'll certainly get at least that.

0 Mark Goldberg over 12 years ago in reply to Sasha Slijepcevic

Prodigy 100 points

Thank you Sasha for your response. However I have a follow-up question. My device is not a 6747 floating point as you referenced but rather it is a C6474 fixed point device which does not natively support double precision floating point types.

For C6474 what is the default memory alignment generated by Memory_Alloc()?

Thanks,

Mark

0 Sasha Slijepcevic over 12 years ago in reply to Mark Goldberg

TI__Mastermind 37985 points

Mark,
I misstyped, I was looking at the target ti.targets.C64P, which corresponds to C6474. Anyway, for both ti.targets.C64P and ti.targets.C674, which corresponds to C6747, the default memory alignment is 8.

0 Mark Goldberg over 12 years ago in reply to Sasha Slijepcevic

Prodigy 100 points

Sasha,

Thank you for your timely response. You've answered my question.

Mark

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What is the default memory alignment used by Memory_alloc()