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!

Correct:

unsigned long int

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

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.

For linux kernel articles, visit www.rulingminds.com

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.