mjhjmtu

I need to understand the code in "Real time clock" function rtc_interrupt. Code is

rtc_irq_data += 0x100; 
rtc_irq_data &= ~0xff; 
rtc_irq_data |= (CMOS_READ(RTC_INTR_FLAGS) & 0xF0);

I am unable to understand why it is += 0x100 and the rest of code.

From the Book "Linux kernel development", from Robert Love, that snippet of code has the following comment(s):

/*
 * Can be an alarm interrupt, update complete interrupt,
 * or a periodic interrupt. We store the status in the
 * low byte and the number of interrupts received since
 * the last read in the remainder of rtc_irq_data.
 */

As for rtc_irq_data += 0x100; So, we know there is a counter for the interrupts received in the high byte. Hence the 0x100. If a 16 bit hexadecimal number representation, where the highest byte is being added +1 (more one interrupt on the counter).

As for the second line, rtc_irq_data &= ~0xff; rtc_irq_data is being logically ANDED with the negation of 0xff, eg, with possibly 0xff00. The high part of the integer is being kept, and the low part being discarded. So supposing this was the first time being called, the value would now guaranteed to be 0x0100.

The last part rtc_irq_data |= (CMOS_READ(RTC_INTR_FLAGS) & 0xF0); is doing a logical OR |= of the low byte (that is now 0 / 0x00) with as the RTC current status. Hence the comment "We store the status in the low byte".

As for doing a logical AND with 0xF0 in (CMOS_READ(RTC_INTR_FLAGS) & 0xF0) , consulting the original AT compatible RTC datasheet, INTR_FLAGS is REGISTER C, a register byte where only the 4 upwards bits are used. b7 = IRQF, b6 = FP, b5 = AF, b4 = UF,

b3 to b0

The unused bits of Status Register 1 are read as "0s". They cannot be
writen.

From RTC datasheet

Hence then as a good standard coding practice, making sure with the AND logical 0xF0 that the lower 4 bits are ignored.