双位数的位表示 [英] bit representation of a double number

问题描述

我做了这个程序，它的功能如预期，知道浮点的位表示：

I did this program, which functions as expected, to know the bit representation of a float:

float x1=-675.78125;
int *pint1;
pint1=(int *)&x1;


for(int i=0;i<8*sizeof(float);i++)
{

if(*pint1&1)
{
    cout<<1;
    }
else
    cout<<0;
    *pint1>>=1;

}

但它不适用于double： p>

But it doesn't work for a double:

double x=-675.78125;
int *pint;
pint=(int *)&x;

for(int i=0;i<8*sizeof(double);i++)
{

    if(*pint&1)
    {
        cout<<1;
        }
    else
        cout<<0;
        *pint>>=1;

    }

你能解释一下为什么会这样吗？你会怎么做呢？非常感谢您的帮助。

Could you explain me why this is so? how would you do it? Thank you so much for your help.

推荐答案

您的第一个程序似乎工作，而你的第二个不是对于你的特定硬件，一个float的大小和int一样，而一个int没有足够的空间用于 double 。

The reason that your first program seems to work and your second doesn't is that for your particular hardware, the size of a float is the same as int, while an int doesn't have enough room for all the bits in a double.

但是你已经违反了严格的别名规则，所以如果你真的想打印浮点类型的位，正确的方法是转换为 unsigned char * ，然后遍历char的每个位，同时将指针递增到底层浮点类型的每个字节。另请注意，在big-vs-little endian，您的程序的结果可能会有所不同。

But you're already violating the strict aliasing rules, so if you really want to print the bits of a floating point type the right way to do it is to cast to unsigned char* and then iterate over each bit of the char while incrementing the pointer over each byte of the underlying floating point type. Also note that on big-vs-little endian the results of your program may vary.

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