问题
I get a different result depending on if I combine multiple bitwise shifts on one line or put them on separate lines.
unsigned char a = 73;
a = (a << 6) >> 2;
printf("%d\n", a);
prints 144 when I expect 16
unsigned char a = 73;
a = a << 6;
a = a >> 2;
printf("%d\n", a);
prints the expected result of 16
when I shift left 6, then right 1 on one line, it prints the expected result of 32, so it looks like the 2nd shift right is shifting in a 1 instead of a 0, but why?
Also
a = (unsigned char) (a << 6 ) >> 2;
produces the expected result (16).
Is the first left shift returning a signed char instead of an unsigned char, and if so why?
回答1:
Yes, the result of the first shift is signed. signed int specifically. Check out the rules for the "integer promotions" in the C spec:
If an
intcan represent all values of the original type ... the value is converted to anint...
So, since your int - usually a range of [-231, 231) - can represent all the values of unsigned char - usually [0, 255] - it's converted to int, and the << and >> take place on the resulting values.
来源：https://stackoverflow.com/questions/25393058/unexepected-behavior-from-multiple-bitwise-shifts-on-the-same-line