The rules haven't changed. It's still technically undefined.
Quoting from the C standard (Section 6.5.7, paragraph 4, of n1548):
The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.
It clearly says if E1 is not unsigned or signed with nonnegative value, the behavior is undefined.
C++03 standard tells us that the result of applying the bitwise shift operators to signed types can be UB and Impl. defined for negative values. My question is following: why for operator << it has undefined behaviour, while for operator >> it is just implementation defined ? Is there a strict reason why the result of << couldn't be implementation defined also ?
Thanks in advance.
Signed integers on right-hand side are undefined behavior in the C language.
ISO 9899:1999 6.5.7 Bit-wise shift operators §3
The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand. If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined.
The right operand to a shift operator may be negative (behavior undefined).
https://www.securecoding.cert.org/confluence/display/cplusplus/INT34-CPP.+Do+not+shift+a+negative+number+of+bits+or+more+bits+than+exist+in+the+operand
