重新用C浮动presentation [英] Representation of float in C

问题描述

我试图用这个code，了解在C浮点再presentation（均浮动和 INT 4个字节我的机器上）：

I was trying to understand the floating point representation in C using this code (both float and int are 4 bytes on my machine):

int x = 3;
float y = *(float*) &x;
printf("%d %e \n", x, y);

我们知道X的二进制重新presentation将以下

We know that the binary representation of x will be the following

00000000000000000000000000000011

所以我本来期望y将被重新psented如下$ P $

Therefore I would have expected y to be represented as follows

• 符号位= 0

指数（位2-9左起）= 0

Exponent (bits 2-9 from left) = 0

尾数（位10-32）： 1 + 2 ^（ - 22）+2 ^（ - 23）

Mantissa (bits 10-32): 1 + 2^(-22)+2^(-23)

通往 Y =（-1）^ 0 * 2 ^（0-127）*（1 + 2 ^（ - 22）+ 2 ^（ - 23））= 5.87747E-39

不过我的程序打印出

3 4.203895e-45

也就是说，Y的值的4.203895e-45 而不是 5.87747E-39 如我所料。为什么会发生这种情况。我在做什么错了？

That is, y has the value 4.203895e-45 instead of 5.87747E-39 as I expected. Why does this happen. What am I doing wrong?

P.S。我也直接从GDB打印的值，因此它不与printf命令有问题。

P.S. I have also printed the values directly from gdb so it is not a problem with the printf command.

推荐答案

与所有的0指数领域的IEEE浮点数被规格化。这意味着，在尾数前面的隐式1不再是活动的。这使得真正的小号码重新presented。请参见本作更多的解释维基百科的文章。在你的榜样，其结果将是3 * 2 ^ -149

IEEE floating point numbers with exponent fields of all 0 are 'denormalized'. This means that the implicit 1 in front of the mantissa no longer is active. This allows really small numbers to be represented. See This wikipedia article for more explanation. In your example the result would be 3 * 2^-149

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