Interrupts

ravi.chakram

while understanding the interrupts, I saw the below line in hardirq.h.

#define SOFTIRQ_OFFSET (1UL << SOFTIRQ_SHIFT)</p>

Can anyone please tell me what is 1UL means in the above line.

Thanks in advance

Regards

Kumar

marc

ravi.chakram wrote:

while understanding the interrupts, I saw the below line in hardirq.h.

#define SOFTIRQ_OFFSET (1UL << SOFTIRQ_SHIFT)

Can anyone please tell me what is 1UL means in the above line.


Thanks in advance

Regards
Kumar

1 Unsigned Long = an "unsigned long" full of ones

Regards

PS: I'm no expert, I might be wrong!

asedt

Correct:

unsigned long int

http://www.cs.umbc.edu/portal/help/C++/summary.shtml#Expressions

rulingminds

UL means long unsigned int. 1 is left shifted by SOFTIRQ_SHIFT(8) times, that means value will be

0x00000100.

This macro is used to increment the softirq count in the thread_info->preempt_count at 0x00000100 offset.

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Goineasy9

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chekkizhar

rulingminds wrote:

This macro is used to increment the softirq count in the thread_info->preempt_count at 0x00000100 offset.

Great . But, is it incrementing the count at 0x00000100 or assigning/updating the soft-interrupts value as 0x00000100

My opinion is, it is doing the second one.

rulingminds

No, This is incrementing or decrementing the softirq count. The preempt_count is divided as follows.

For softirq count there are 8 bits(from 8 - 15) in preempt_count variable. So, the variable can be incremented upto 255.
This count says number of levels the softirqs are disabled(number of tymes local_bh_disable() is called) . decremented using local_bh_enable(). See the following code in softirq.c and preempt.h

# define add_preempt_count(val) do { preempt_count() += (val); } while (0)

static inline void __local_bh_disable(unsigned long ip, unsigned int cnt)
{
add_preempt_count(cnt);
barrier();
}

Here, count is 0x100.


-Rulingminds