Maybe there's something obvious here - you say that count goes from 0 ..
7, and the shift amount is (8+count)*8, or from 0
to 120?? You're far beyond a reasonable shift count, and perhaps the
compiler is seeing that. And with volatile, it actually does
the shifts, and your cpu does some masking with the shift amount.
What DO you want dout to be after shifting right 120 bits? What are you
calling the "right" answer?
Manish Baphna wrote:
A quick update, Though I couldn't really found real issue , I found an interesting solution which worked. magic word is 'volatile' :)
Original two lines were
#Line1# dout.data[15 - count] = (uint8_t)(((long long unsigned int)(*(soc_cpr_dbusdataout_->sig_value_upper)) & (mask << ((8+count)*8)))>>((8+count)*8));
#Line2# cout << "Dout data " << (uint64_t)((dout.data)[15-count]) << endl;
What I did was inserting this line just above Line#1
volatile uint32_t shift_val = ((8+count)*8) ;
and used this in above Line#1 , and then I removed Line#2 and this time it worked.
( First I tried without using 'volatile' and it didn't help )
Another workaround that did the same thing was making
count as volatile.
These two lines are in for loop which goes for count = 0 to 7 , I don't know why compiler is being lazy here to read it everytime unless I make it volatile.
Should it be logged as bug in gcc somewhere ?
-Manish
