位移位和整数提升？
通常，C 要求将二元运算符的操作数提升为较高级别操作数的类型。可以利用这一点来避免用详细的转换填充代码，例如：
if (x-48U<10) ...
y = x+0ULL << 40;
等等。
但是，我发现，至少对于 gcc，这种行为不适用于位移位。即，
int x = 1;
unsigned long long y = x << 32ULL;
我希望右侧操作数的类型会导致左侧操作数提升为unsigned long long，以便移位成功。但相反，gcc 会打印一条警告：
warning: left shift count >= width of type
gcc 是否已损坏，或者标准是否对位移位的类型提升规则做出了一些例外？
Normally, C requires that a binary operator's operands are promoted to the type of the higher-ranking operand. This can be exploited to avoid filling code with verbose casts, for example:
if (x-48U<10) ...
y = x+0ULL << 40;
etc.
However, I've found that, at least with gcc, this behavior does not work for bitshifts. I.e.
int x = 1;
unsigned long long y = x << 32ULL;
I would expect the type of the right-hand operand to cause the left-hand operand to be promoted to unsigned long long so that the shift succeeds. But instead, gcc prints a warning:
warning: left shift count >= width of type
Is gcc broken, or does the standard make some exception to the type promotion rules for bitshifts?
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所谓的通常的算术转换适用于许多二元运算符，但不是全部。例如，它们不适用于位移运算符、&&、||、逗号运算符和赋值运算符。这是移位运算符的规则：
The so-called usual arithmetic conversions apply to many binary operators, but not all of them. For example they do not apply to the bit shift operators, &&, ||, comma operator, and assignment operators. This is the rule for the bit shift operators:
真正的麻烦在于，提升仅适用于您的平台定义为
int的任何内容。正如其他一些答案所述，位移运算符会将左操作数提升为 int。但是，这里int被定义为 32 位值。整数转换不会提升为long long（64 位）。The trouble really is that promotion only works up to whatever your platform defines as an
int. As some other answers have stated, the bit-shift operator will promote the left operand to an int. However, here anintis defined as a 32-bit value. The integer conversion will not promote to along long(64-bit).