IEEE 754 Float 调试 - 从内存小端到实际浮点数

发布于 2024-11-28 21:30:08 字数 435 浏览 2 评论 0原文

我正在使用 VS2008 测试 IEEE 754 浮动格式，使用下面的示例：

int main(int argc, char *argv[])
{
    float i = 0.15625;
}

我将 &i 放入 VS2008 手表，我看到地址是 0x0012FF60，我可以从内存调试窗口看到地址的内容是 00 00 20 3e，如下所示：

0x0012FF60 00 00 20 3e cc cc cc cc

顺便说一句，我有 IEEE754 浮点格式的基本知识，并且我知道 IEEE 754 浮点格式由三个字段组成：符号位、指数和分数。小数是没有最高有效位的有效数。

但是我如何精确计算从小端 00 00 20 3e 到 0.15625 ？

非常感谢

原文

I am testing IEEE 754 floating format with VS2008 using the example below:

int main(int argc, char *argv[])
{
    float i = 0.15625;
}

I put &i to the VS2008 watch and I see the address is 0x0012FF60 and I can see address's content is 00 00 20 3e from Memory debug window, see below:

0x0012FF60 00 00 20 3e cc cc cc cc

BTW I have the basic knowledge of IEEE754 floating format and I know IEEE 754 floating format consist of three fields: sign bit, exponent, and fraction. The fraction is the significand without its most significant bit.

But how did I calcuate exactly from little endian 00 00 20 3e to 0.15625 ?

Many thanks

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美胚控场 2024-12-05 21:30:09

你正在打印出一些损坏的东西。我们只需要 32 位，它们是：

00 00 20 3E

二进制变量：

00000000 00000000 00100000 00111110

小尾数的逻辑值：

00111110 00100000 00000000 00000000

根据 IEEE：

0 01111100 01000000000000000000000
S E - 127  M - 1

所以现在很清楚：

符号是 +1 (S = 0)
指数是124 - 127 = -3
尾数为 1.01b，即 5/4，

因此该值为 5/4 / 8 = 5/32 = 0.15625。

You are printing out something broken. We only need 32 bits, which are:

00 00 20 3E

Your variable in binary:

00000000 00000000 00100000 00111110

Logical value accounting for little endian:

00111110 00100000 00000000 00000000

According to IEEE:

0 01111100 01000000000000000000000
S E - 127  M - 1

So now it's clear:

the sign is +1 (S = 0)
the exponent is 124 - 127 = -3
the mantissaa is 1.01b, which is 5/4

So the value is 5/4 / 8 = 5/32 = 0.15625.

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彩扇题诗 2024-12-05 21:30:09

您的值为十六进制 0x3E200000 或

0011 1110 0010 0000 0000 0000 0000 0000

或重新排列：

s ----e--- ----------m------------
0 01111100 01000000000000000000000

sign_bit    = 0 (i.e. positive)
exponent    = 0x7C = 124 ---> subtract 127 to get -3
significand = 1 + 0.0100... = 1.0100... = 1*2^0 + 0*2^-1 + 1*2^-2 = 1.25

significand * 2^exponent = 1.25 * 2^-3 = 1.25 * 0.125 = 0.15625

Your value is hex 0x3E200000 or

0011 1110 0010 0000 0000 0000 0000 0000

or rearranged:

s ----e--- ----------m------------
0 01111100 01000000000000000000000

sign_bit    = 0 (i.e. positive)
exponent    = 0x7C = 124 ---> subtract 127 to get -3
significand = 1 + 0.0100... = 1.0100... = 1*2^0 + 0*2^-1 + 1*2^-2 = 1.25

significand * 2^exponent = 1.25 * 2^-3 = 1.25 * 0.125 = 0.15625

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再浓的妆也掩不了殇 2024-12-05 21:30:09

IEEE 浮点的基本格式基于四字节
value，这样显示的话更容易分析。在那
在这种情况下，最高位是指数，接下来的 8 位是指数（超过
127)，其余为尾数。最简单的解释方法是
可能是为了显示将访问单独字段的 C++ 代码：

double d;
// ...
uint32_t const* p = reinterpret_cast<uint32_t const*>( &d );
bool isNegative = (*p & 0x80000000) != 0;
int exp = ((*p & 0x78000000) >> 23) - 127;
int mantissa = (*p & 0x07FFFFFF) | 0x08000000 ;

尾数应该在 24 上方有一个隐式小数位
位（但我不知道如何将其表示为整数:-)）。

如果你拥有的只是一个字节序列，那么你必须组装它们，
根据字节顺序，然后应用上面的。

编辑：继 Rudy Velthuis 指出我的错误之后，常量值已得到纠正。

The basic format of an IEEE floating point is based on a four byte
value, and it is simpler to analyse if you display it as such. In that
case, the top bit is the exponent, the next 8 the exponent (in excess
127), and the rest the mantissa. The simplest way to explain it is
probably to show the C++ code which would access the separate fields:

double d;
// ...
uint32_t const* p = reinterpret_cast<uint32_t const*>( &d );
bool isNegative = (*p & 0x80000000) != 0;
int exp = ((*p & 0x78000000) >> 23) - 127;
int mantissa = (*p & 0x07FFFFFF) | 0x08000000 ;

The mantissa should have an implicit decimal place just above the 24
bits (But I don't know how to represent this as an integer:-)).

If all you have is a sequence of bytes, you have to assemble them,
according to the byte order, and then apply the above.

Edited: the constant values have been corrected, following up on Rudy Velthuis' pointing out my error.

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