https://gcc.gnu.org/bugzilla/show_bug.cgi?id=103194
--- Comment #22 from Jakub Jelinek <jakub at gcc dot gnu.org> --- (In reply to Hongtao.liu from comment #15) > > Is the behavior well defined for n >= 64? I got > > > > foo.c:11:19: warning: left shift count >= width of type > > [-Wshift-count-overflow] > > 11 | long mask = 1ll << 65; > > | ^~ > According to C99 > 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 > × 2E2, reduced modulo one more than the maximum value representable in the > result type. If E1 has a signed type and nonnegative value, and E1 × 2E2 is > representable in the result type, then that is the resulting value; > otherwise, the behavior is undefined. > > So yes, it's well defined, and the result is zero. No, that isn't well defined in any C or C++ versions. E.g. in C++: https://eel.is/c++draft/expr.shift#1 The behavior is undefined if the right operand is negative, or greater than or equal to the width of the promoted left operand. C99 6.5.7: 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. What varies is whether shifting negative values left is well defined or not, or whether shifting bits into the sign bit of signed type is well defined or not.
