Unexepected behavior from multiple bitwise shifts on the same line [duplicate]

Question

This question already has an answer here:

• Why does combining two shifts of a uint8_t produce a different result? 2 answers

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?

Answer 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 int can represent all values of the original type ... the value is converted to an int...

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.