Circular shift in c

Question

How does the following code work and what do the variables mean:

y = (x << shift) | (x >> (sizeof(x)*CHAR_BIT - shift));

I foun

Answer 1

This works with unsigned types only. In the case with a signed negative number most left bits will be substituted by the value of most significant bit (with 1-s) by the right-shift operator (">>")

I'd write it like this:

y = (x << shift) | ( (x >> (sizeof(x)*CHAR_BIT - shift)) & (0x7F >> (sizeof(x)*CHAR_BIT - shift) );

In here before "|" operator we do confirm that first n bits ( n = sizeof(x)*CHAR_BIT - shift) are zeroed. We also assume, that x is short (2-bytes long). So, it's also type-dependent.