如何根据“字节序"以位级别存储数据?

问题描述

我阅读了Endianness并理解了蹲坐...

I read about Endianness and understood squat...

所以我写了这个

main()
{
    int k = 0xA5B9BF9F;

    BYTE *b = (BYTE*)&k;    //value at *b is 9f
    b++;    //value at *b is BF
    b++;    //value at *b is B9
    b++;    //value at *b is A5
}

k等于A5 B9 BF 9F

和(字节)指针"行走"的o/p为9F BF b9 A5

and (byte)pointer "walk" o/p was 9F BF b9 A5

所以我知道它的字节是向后存储的...好吧.

so I get it bytes are stored backwards...ok.

〜

所以现在我想到了如何将其存储在BIT级别...

so now I thought how is it stored at BIT level...

我的意思是将"9f"(1001 1111)存储为"f9"(1111 1001)吗?

I means is "9f"(1001 1111) stored as "f9"(1111 1001)?

所以我写了这个

int _tmain(int argc, _TCHAR* argv[])
{
    int k = 0xA5B9BF9F;
    void *ptr = &k;
    bool temp= TRUE;
    cout<<"ready or not here I come \n"<<endl;

    for(int i=0;i<32;i++)
    {
        temp = *( (bool*)ptr + i );
        if( temp )
            cout<<"1 ";
        if( !temp)
            cout<<"0 ";
        if(i==7||i==15||i==23)
            cout<<" - ";
   }
}

我得到一些随机输出

即使没有.像"32"一样，我什么都不懂.

even for nos. like "32" I dont get anything sensible.

为什么?

推荐答案

Endianness ，您通过实验发现的是指字节在对象中存储的顺序.

Endianness, as you discovered by your experiment refers to the order that bytes are stored in an object.

位没有不同的存储方式，它们始终是8位，并且始终是人类可读的"(高->低).

Bits do not get stored differently, they're always 8 bits, and always "human readable" (high->low).

现在我们已经讨论过，您不需要您的代码...关于您的代码:

Now that we've discussed that you don't need your code... About your code:

for(int i=0;i<32;i++)
{
  temp = *( (bool*)ptr + i );
  ...
}

这没有按照您认为的去做.您要在0-32(一个单词的位数)之间进行迭代-很好.但是您的temp分配都是错误的:)

This isn't doing what you think it's doing. You're iterating over 0-32, the number of bits in a word - good. But your temp assignment is all wrong :)

请务必注意，bool*的大小与int*的大小相同，而与BigStruct*的大小相同.同一台计算机上的所有指针大小都相同-32位计算机上为32位，而64位计算机上为64位.

It's important to note that a bool* is the same size as an int* is the same size as a BigStruct*. All pointers on the same machine are the same size - 32bits on a 32bit machine, 64bits on a 64bit machine.

ptr + i将i字节添加到ptr地址. i>3时，您正在读一个全新单词...这可能会导致段错误.

ptr + i is adding i bytes to the ptr address. When i>3, you're reading a whole new word... this could possibly cause a segfault.

您要使用的是位掩码.这样的事情应该起作用:

What you want to use is bit-masks. Something like this should work:

for (int i = 0; i < 32; i++) {
  unsigned int mask = 1 << i;
  bool bit_is_one = static_cast<unsigned int>(ptr) & mask;
  ...
}

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