I have this macro which, obviously assumes that the provided type is the correct and expected type.
I'm currently required to get rid of this macro in the code and I can't help thinking whether I could in a way make it "type safe". Could someone have any ideas ?Code:#define ENCODE_EVENT(devid, info, addr, rshift) do { \ *(uint16_t *)&evt[0] = devid; \ *(uint8_t *)&evt[3] = info; \ *(uint64_t *)&evt[4] = rshift ? cpu_to_le64(addr) :\ cpu_to_le64(addr) >> rshift; \ } while (0)
> *(uint64_t *)&evt[4] = rshift ? cpu_to_le64(addr) :\
> cpu_to_le64(addr) >> rshift; \
What does this do?
If rshift is 0, you do cpu_to_le64(addr) >> 0; otherwise you do nothing.
What is evt? Is it always correctly aligned to permit the type punning to be valid?
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Huh, You're right.Originally Posted by Salem
It should be something like
Is there a cleaner way ? Thanks for spotting that!Code:*(uint64_t *)&evt[4] = rshift != 0 ? cpu_to_le64(addr) :\ cpu_to_le64(addr) >> rshift;
This macro assumes that there's a local 64-bit integer array of size 5 of named 'evt'.
Turn it into a function!
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> This macro assumes that there's a local 64-bit integer array of size 5 of named 'evt'.
> *(uint16_t *)&evt[0] = devid;
OK, so how does this differ from
evt[0] = devid;
Is it important that some of the bits in your evt are not assigned by simple assignment?
Another thought, does the type punning imply that there are some endian issues which a simple assignment-like operator couldn't achieve?
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
That's an option!
No, There's no difference between the two I just wrote that because I was blindly following what's specified ;-)
I might be doing this wrong but here's the idea. 'evt' is just a chunk of 256 bits. I want to assign say bits 16:23 some value that's why I do things like.
Okay, the question should be am I addressing the right bits in this case ?Code:*(uint8_t *)&evt[3] = info;
I would start by building a simple function to put a set of bits into your bit-buffer at a given position.
Then write an ENCODE_EVENT function which calls setBits three times.Code:#include <stdio.h> // This doesn't account for values which have start+length spanning word boundaries. void setBits ( unsigned long *buffer, unsigned long value, int start, int length ) { int index = start / 32; int bitpos= start % 32; unsigned long mask = ( ( 1 << length ) - 1 ) << bitpos; buffer[index] &= ~mask; buffer[index] |= (value << bitpos) & mask; } int main() { unsigned long buffer[5] = { 0 }; setBits(buffer,0x12,0,5); setBits(buffer,0x34,8,8); setBits(buffer,0xFFFF,36,16); for ( int i = 0 ; i < 5 ; i++ ) { printf("%08lx\n",buffer[i]); } return 0; }
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Thanks you. This is much more flexible than I was hoping or could ever come up with :-)
