无符号长整型和位移位
我有位移位和无符号长整型的问题。这是我的测试代码：
char header[4];
header[0] = 0x80;
header[1] = 0x00;
header[2] = 0x00;
header[3] = 0x00;
unsigned long l1 = 0x80000000UL;
unsigned long l2 = ((unsigned long) header[0] << 24) + ((unsigned long) header[1] << 16) + ((unsigned long) header[2] << 8) + (unsigned long) header[3];
cout << l1 << endl;
cout << l2 << endl;
我希望 l2 也有一个值 2147483648，但它打印的是 18446744071562067968。我认为第一个字节的位移会导致问题？
希望有人可以解释为什么会失败以及我如何修改 l2 的计算以使其返回正确的值。
提前致谢。
I have a problem with bit shifting and unsigned longs. Here's my test code:
char header[4];
header[0] = 0x80;
header[1] = 0x00;
header[2] = 0x00;
header[3] = 0x00;
unsigned long l1 = 0x80000000UL;
unsigned long l2 = ((unsigned long) header[0] << 24) + ((unsigned long) header[1] << 16) + ((unsigned long) header[2] << 8) + (unsigned long) header[3];
cout << l1 << endl;
cout << l2 << endl;
I would expect l2 to also have a value of 2147483648 but instead it prints 18446744071562067968. I assume the bit shifting of the first byte causes problems?
Hopefully somebody can explain why this fails and how I modify the calculation of l2 so that it returns the correct value.
Thanks in advance.
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存储在 char 中的 0x80 值是一个有符号的数量。当您将其转换为更宽的类型时，该值将被符号扩展以保持与更大类型相同的值。
将第一行中的
char类型更改为unsigned char，您将不会发生符号扩展。为了简化您的情况，请运行以下命令：
您将得到以下输出：
64 位整数为 -128（8 位整数为 0x80 为 -128）。
Your value of 0x80 stored in a char is a signed quantity. When you cast this into a wider type, the value is being signed extended to keep the same value as a larger type.
Change the type of
charin the first line tounsigned charand you will not get the sign extension happening.To simplify what is happening in your case, run this:
You get this output:
which is -128 as a 64-bit integer (0x80 is -128 as a 8-bit integer).
这里的结果相同（Linux/x86-64，GCC 4.4.5）。该行为取决于
unsigned long的大小，至少 32 位，但可能更大。如果您想要正好 32 位，请使用
uint32_t（来自标头Same result here (Linux/x86-64, GCC 4.4.5). The behavior depends on the size of
unsigned long, which is at least 32 bits, but may be larger.If you want exactly 32 bits, use a
uint32_tinstead (from the header<stdint.h>; not in C++03 but in the upcoming standard and widely supported).