C++ 对齐/移位问题

Question

我有一个字符串:

"the quic"

它被传递给一个函数，其中构成字符串的位存储在一个无符号的 __int64 中。产生以下输出:

0111010001101000011001010010000001110001011101010110100101100011

但是当我传递一个包含这些值的字符串时:

0xDE, 0x10, 0x9C, 0x58, 0xE8, 0xA4, 0xA6, 0x30, '\0'

输出不像我预期的那样正确:

1111111111111111111111111111111111111111111111111010011000110000

我使用的代码与第一个字符串中的相同，内容为:

(((unsigned __int64)Message[0]) << 56) | (((unsigned __int64)Message[1]) << 48) |
(((unsigned __int64)Message[2]) << 40) | (((unsigned __int64)Message[3]) << 32) |
(((unsigned __int64)Message[4]) << 24) | (((unsigned __int64)Message[5]) << 16) |
(((unsigned __int64)Message[6]) << 8) | (((unsigned __int64)Message[7]));

Answer 1

我猜你做过这样的事情，

char a[] = {0xDE, 0x10, 0x9C, 0x58, 0xE8, 0xA4, 0xA6, 0x30};

将其更改为 unsigned char 将解决您的问题，

unsigned char a[] = {0xDE, 0x10, 0x9C, 0x58, 0xE8, 0xA4, 0xA6, 0x30};

我都试过了，char 版本在 VC++ 中是错误的，但 unsigned 版本是正确的。

如果你想知道原因，看一个更简单的版本，

char a = 0xDE;
unsigned char b = 0xDE;

有什么区别？ 0xDE 是一个 int 类型。对于第一个，您将 int 转换为 signed char，对于第二个，您将 int 转换为 unsigned char。

来自标准 4.7/2、4.7/3

If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined.

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2n where n is the number of bits used to represent the unsigned type). [ Note: In a two’s complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). —end note ]