Answering my own question after much help from Josko Plazonic-
local mem_to_use = 0
-- This is the logic -
-- either min_mem_per_node or min_mem_per_cpu will be set
-- Both can't be set, so only act in those two cases
if job_desc.min_mem_per_node ~= nil then
mem_to_use = job_desc.min_mem_per_node
end
if job_desc.min_mem_per_cpu ~= nil then
mem_to_use = job_desc.min_mem_per_cpu * job_desc.min_cpus
end
log_info("slurm_job_submit: Got total memory: %d", mem_to_use)
Bill
On 1/24/20 8:52 AM, William G. Wichser wrote:
> Resurrecting an older thread where I need to obtain the value for memory
> in a submitted job. Turns out this is not an easy case with the method
> I'm trying to use so hope that there is just some variable I am overlooking.
>
> The trivial case was simply to look at job_desc.pn_min_memory. And this
> works fine as long as jobs are submitted with a --mem= flag. But there
> are two other ways that jobs get submitted which make this value
> something like 2^63.
>
> The first is when no memory is specified and users rely on the default.
> The second is with --mem-per-cpu=X
>
>
> For that second case I can detect using
> (job_desc.pn_min_memory - slurm.MEM_PER_CPU) * job_desc.min_cpus
>
> But I find that when users are using the default memory allocation, it
> isn't so easy to detect since it appears that both of the memory values
> are set to 2^63 or close to that number. Maybe it's 2^64 -1. Whatever.
>
> I just feel that there has to be a better way! Is there soemthing that
> I'm missing? Perhaps a tres.memory or something which has the right
> value when in job_submit.lua?
>
> Thanks,
> Bill
>
