正确的转变 [英] right shifting
问题描述
如果我有愚蠢的代码，
main（）
{
unsigned int i = 34;
i>> = 32;
}
为什么我的价值仍为34？我认为它应该是零。
（g ++编译器给我一个关于移动> =变量大小的警告。）
if I have the foolowing code,
main()
{
unsigned int i=34;
i>>=32;
}
why does the value of i remain 34? I thought it should be a zero.
(the g++ compiler gives me a warning about shifting >= size of variable.)
推荐答案
crookie写道：
crookie wrote:
如果我有愚蠢的代码，
main（）
{
unsigned int i = 34;
i>> = 32;
}
为什么我的价值仍为34？我认为它应该是零。
（g ++编译器给我一个关于移位> =变量大小的警告。）
if I have the foolowing code,
main()
{
unsigned int i=34;
i>>=32;
}
why does the value of i remain 34? I thought it should be a zero.
(the g++ compiler gives me a warning about shifting >= size of variable.)
关于班次操作员的标准[5.8 / 1]说：
[...]如果右操作数为负数，则行为未定义，或者
大于或等于提升的左操作数的位长度。
看来你的平台上的sizeof（int）< = 32，你有没有
未定义的行为：任何事情都可能发生。
Best
Kai-Uwe Bux
The standard [5.8/1] says about the shift operators:
[...] The behavior is undefined if the right operand is negative, or
greater than or equal to the length in bits of the promoted left operand.
It would appear that on your platform sizeof(int)<=32, whence you have
undefined behavior: anything could happen.
Best
Kai-Uwe Bux
crookie写道：
crookie wrote:
如果我有愚蠢的代码，
main（）
{
unsigned int i = 34;
i>> = 32;
}
为什么我的价值仍然是34？我认为它应该是零。
（g ++编译器给我一个关于移位> =变量大小的警告。）
if I have the foolowing code,
main()
{
unsigned int i=34;
i>>=32;
}
why does the value of i remain 34? I thought it should be a zero.
(the g++ compiler gives me a warning about shifting >= size of variable.)
因为在你的系统上，unsigned int有32位或更少的位，并且（最近学到的是
）移位任何大于或等于
位数的数字是未定义的行为。这意味着i
的任何价值同样有效。
欲了解更多细节，请在夏天看到这个帖子：
http://tinyurl.com/y382v3
祝你好运，
Tom
Because on your system, an unsigned int has 32 or fewer bits, and (as I
recently learned) shifting by any number greater than or equal to the
number of bits is Undefined Behavior. Which means that any value of i
is equally valid.
For more detail, see this thread from over the summer:
http://tinyurl.com/y382v3
Best regards,
Tom
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