宽度不兼容的变量
我使用以下代码来简化将大值分配给内存中的特定位置：
int buffer_address = virtual_to_physical(malloc(BUFFER_SIZE));
unsigned long int ring_slot = buffer_address << 32 | BUFFER_SIZE;
但是，编译器抱怨“警告：左移计数>=类型宽度”。但是 C 中的 unsigned long int 是 64 位，因此将 int （32 位）向左移动 32 位应该会产生 64 位值，因此编译器不应该抱怨。但确实如此。
我是否遗漏了一些明显的东西，或者是否有一个简单的解决方法？
I am using the following code to simplify assigning large values to specific locations in memory:
int buffer_address = virtual_to_physical(malloc(BUFFER_SIZE));
unsigned long int ring_slot = buffer_address << 32 | BUFFER_SIZE;
However, the compiler complains "warning: left shift count >= width of type". But an unsigned long int in C is 64 bits, so bit-shifting an int (32 bits) to the left 32 bits should yield a 64 bit value, and hence the compiler shouldn't complain. But it does.
Is there something obvious I'm missing, or otherwise is there a simple workaround?
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unsigned long int不一定是 64 位，但为了简单起见，我们假设它是 64 位。buffer_address的类型为int。buffer_address上没有任何“更高”类型的任何表达式都应返回 int。从而buffer_address << 32 应该返回 int，而不是unsigned long。因此编译器会抱怨。不过，这应该可以解决您的问题：
请注意，
unsigned long不一定一定是 64 位，这取决于实现。使用这个代替：An
unsigned long intis not necessarily 64 bits, but for the simplicity let's assume it is.buffer_addressis of typeint. Any expression without any "higher" types onbuffer_addressshould return int. Therebybuffer_address << 32should return int, and notunsigned long. Thus the compiler complains.This should solve your issue though:
Please note, an
unsigned longis not necessarily 64 bits, this depends on the implementation. Use this instead:buffer_address是一个（32 位）int，因此buffer_size << 32正在将其移动大于或等于其大小的量。请注意，“unsigned long”不必是 64 位（它不在 Windows 上 - 32 位 (ILP32) 或 64 位 (LLP64)；也不在 32 位 Unix 机器 (ILP32) 上）。要获得有保证的（至少）64 位整数，您需要
unsigned long long。很少有机器的
int是 64 位数量 (ILP64)； DEC Alpha 就是其中之一，我相信一些 Cray 机器也使用了它（并且 Cray 也使用“大”char类型 - 每个char超过 8 位）。buffer_addressis a (32-bit)int, sobuffer_size << 32is shifting it by an amount greater than or equal to its size.Note that 'unsigned long' need not be 64-bits (it is not on Windows - 32-bit (ILP32) or 64-bit (LLP64); nor it is on a 32-bit Unix machine (ILP32)). To get a guaranteed (at least) 64-bit integer, you need
unsigned long long.There are few machines where
intis a 64-bit quantity (ILP64); the DEC Alpha was one such, and I believe some Cray machines also used that (and the Cray's also used 'big'chartypes - more than 8 bits perchar).=符号右侧表达式的结果不取决于它被分配的内容。您必须首先强制转换为 unsigned long。The result of the expression on the right side of the
=sign does not depend on what it's assigned to. You must cast to unsigned long first.