Is char c2=i1<<8>>24;
valid C syntax? (Where i1
is and unsigned integer) Additionally, will it yield the result of shifting i1
The syntax is fine (although hard to read) and it will be parsed as c2 = (i1 << 8) >> 24
.
So it will left shift i1
8 positions, thereby shifting the leftmost 8 bits off the lefthand edge, and then right shift the result 24 positions, thereby shifting the rightmost 16 bits off the righthand edge. If that's what you wanted, then you're good. I'd use parentheses to make it more readable.
If you're just going to convert that to a char
, it's not obvious why you feel the need to remove the high-order bits (although it is true that there may be architectures in which int
and char
are the same size.)
Also, as noted by John Bollinger, it is possible for the end result to be larger than can fit in a char
, which is not well defined in the common case that char
is a signed type. (That will be true even if unsigned int
is 32 bits, because you technically cannot assign a value greater than 127 to an 8-bit signed character.)