data = ((hwirq / 32) << 5) | ((hwirq % 32) & 0x1F)
Which doesn't seem to actually do anything?
It's not a no-op, because hwirq is signed. It probably should be
unsigned, like most things.
You'll have to draw me a picture.
In C, signed division is round-towards-zero, while unsigned division
is round-towards-negative-infinity. Suppose hwirq is -1, then
hwirq/32 is 0 and hwirq%32 is -1, so that the full expression above
will be 0x1f, not -1. There is no such problem if hwirq would be
unsigned; the compiler can generate better code in that case.
Segher
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