Question

Is the following undefined and why?

int i = 0xFF;
unsigned int r = i << 24;

Answer 1

The behaviour is technically undefined unless the int type has more than 32 bits.

From C++11, 5.8/2 (describing an expression E1 << E2):

if E1 has a signed type and non-negative value, and E1×2^E2 is representable
in the result type, then that is the resulting value; otherwise, the behavior is undefined.

The result type of i << 24 is (signed) int; if that has 32 bits or less, then 0xff * 2^24 == 0xff000000 is not representable (the maximum representable 32-bit signed value being 0x7fffffff), so behaviour is undefined as specified in that clause.