The warning is as follows:
Last Edited: Fri. May 24, 2019 - 03:07 PM
can I ignore the following warning
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This reply has been marked as the solution. #2
Wouldn't make more sense to make temp an 8 element array, and place the bytes as needed without all the shifts?
Then cast temp to long long in the return?
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What is the endianism of the buffer data (same or different to CPU)? If the same just cast a unit64_t pointer onto byte 0 and read through the pointer.
The warning seems to imply a long long is only
32 bits on your platform.
--Mike
long long has to be at least 64 bits,
endian[n] is type char, which is then promoted to int for the shift.
Presumably int is 32 bits on this platform, which means a shift is valid over range 0 - 31.
You could do
temp += (unsigned long long)endian[n] << whatever
That's what I actually done.
Maybe
unsigned long long data_to_cpu64(char endian[8]) {
/* little endian assumed */
return *(unsigned long long *)endian;
}works. See also https://onlinegdb.com/BJVM5D4C4.
Last Edited: Tue. Jun 4, 2019 - 10:02 PM
avr-mike wrote:
The warning seems to imply a long long is only
32 bits on your platform.
No, it doesn't. `long long` is not involved in that shift expression at all. That warning implies that `int` is 32 bits on that platform. This `int` is a result of integral promotion applied to the original `char` value.
skotti wrote:
works.
Formally, that's a violation of strict aliasing semantics. Today it "works", tomorrow it might stop working.
Last Edited: Fri. May 24, 2019 - 11:05 PM
AndreyT wrote:
avr-mike wrote:The warning seems to imply a long long is only
32 bits on your platform.
No, it doesn't. `long long` is not involved in that shift expression at all.
Yeah, MrKendo already corrected my erroneous thinking.
--Mike
AndreyT wrote:
skotti wrote:works.
Formally, that's a violation of strict aliasing semantics. Today it "works", tomorrow it might stop working.
I read recently about strict aliasing, and the only valid approach is to use memcpy to copy data from one type to another. So maybe this should work:
unsigned long long data_to_cpu64(char endian[8]) {
unsigned long long ll;
/* little endian assumed */
memcpy(&ll, endian, sizeof ll);
return ll;
}
The compiler should be able to optimize that to basically the same machine code as skotti's.
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I think you want to cast your endian[4], otherwise it shifts the char 32 bits in an 8 bit char.
You CAN ignore warnings -- at your peril.
If you don't know my whole story, keep your mouth shut.
If you know my whole story, you're an accomplice. Keep your mouth shut. 
Last Edited: Sat. May 25, 2019 - 07:32 PM
My recollection is that unsigned char may alias any object.
If char is unsigned, I expect the rule applies to char.
That said, char should be reserved for text.
The only exception I can think of at the moment
is dealing with legacy code that does it wrong.
Some day your code might be legacy code.