位移运算符似乎换行？
bitwise shift operator seems to wrap?
这是 Windows7 32 位，Visual Studio 2017，在 C 文件中。
int i = 65536;
不出所料，
i >> 0 = 65536
i >> 1 = 32768
:
:
i >> 16 = 1
i >> 17 to 31 = 0.
i >> 32 又神奇地变成了 65536。这怎么合法？
ISO/IEC 9899:TC2 表示以下内容，我假设 C++ 规范相同？
The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has
an unsigned type or if E1 has a signed type and a nonnegative value,
the value of the result is the integral part of the quotient of E1 /
2^E2. If E1 has a signed type and a negative value, the resulting value
is implementation-defined.
... 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.
使用 clang、gcc 和 icc，如果您尝试移动一个大于或等于移动值宽度的常量，您将 get a warning，并且您甚至不需要任何额外的command-line 旗帜。
这是 Windows7 32 位，Visual Studio 2017，在 C 文件中。
int i = 65536;
不出所料，
i >> 0 = 65536
i >> 1 = 32768
:
:
i >> 16 = 1
i >> 17 to 31 = 0.
i >> 32 又神奇地变成了 65536。这怎么合法？
ISO/IEC 9899:TC2 表示以下内容，我假设 C++ 规范相同？
The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2^E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined.
... 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.
使用 clang、gcc 和 icc，如果您尝试移动一个大于或等于移动值宽度的常量，您将 get a warning，并且您甚至不需要任何额外的command-line 旗帜。