将 0 向左移动是否被认为是 C 中的未定义行为？

Question

是否是未定义行为（UB）：

1. 0<< 32
2. 0LL<< 64

？

如果答案包含对 C 标准的引用，我将不胜感激。

Answer 1

请参阅 ISO C 标准 (ISO/IEC 9899)，6.5.7 位移运算符，第 4 段：

E1结果<< 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.

0

<<

32 和 0LL

<<

64 都会导致 C 中的未定义行为 (UB)。