32位IEEE 754单精度浮点数到十六进制 [英] 32-bit IEEE 754 single precision floating point to hexadecimal

问题描述

我已经学习了如何将数字转换为浮点数(在二进制，八进制和十六进制之上)，并且知道如何将数字转换为浮点数.

I have learnt how to convert numbers to floating point (on top of binary, octal and hexadecimal), and know how to convert numbers to floating point.

但是，在浏览工作表时，遇到了以下问题:

However, while looking through a worksheet I have been given, I have encountered the following question:

使用32位IEEE 754单精度浮点以十六进制表示-12.13.

我尝试查看自己拥有的资源，但仍然不知道如何回答上述问题.给出的答案是 0xc142147b .

I have tried looking at the resources I have and still can't figure out how to answer the above. The answer given is 0xc142147b.

对不起，您不清楚，但我想知道如何手动完成此工作，而不是对其进行编码.

Sorry for not clarifying but I wanted to know how to get this done by hand instead of coding it.

推荐答案

-12.13必须先转换为二进制，然后再转换为十六进制.让我们或多或少像 glibc 库那样做，只使用笔和纸以及Windows计算器.

-12.13 must be converted to binary and then hex. Let's do that more or less like the glibc library does it, using just pen and paper and the Windows calculator.

删除标志，但请记住我们有一个标志:12.13

Remove the sign, but remember we had one: 12.13

整数部分12很简单:C(十六进制)

The integer part, 12 is easy: C (hex)

小数部分0.13有点棘手. 0.13是13/100.我使用Windows计算器(程序员模式，十六进制)，并将13(十六进制D)左移32(*)位:D00000000.用100(十六进制64)除以得到:2147AE14十六进制.

The fractional part, 0.13 is a little trickier. 0.13 is 13/100. I use the Windows calculator (Programmer mode, hex) and shift 13 (hex D) by 32(*) bits to the left: D00000000. Divide that by 100 (hex 64) to get: 2147AE14 hex.

由于我们需要一个小于1的值，因此我们再次右移32位，并得到:0.2147AE14

Since we need a value below 1, we shift right by 32 bits again, and get: 0.2147AE14

现在在左侧添加整数部分:C.2147AE14

Now add the integer part on the left: C.2147AE14

我们只需要24位尾数，所以我们舍入:C.2147B-> C2147B

We only need 24 bits for the mantissa, so we round: C.2147B --> C2147B

现在必须对此进行归一化，因此将二进制点向左移动3位(当然，这些位保持不变).指数(最初为0)相应地提高了3，所以现在是3.

Now this must be normalized, so the binary point is moved 3 bits to the left (but the bits remain the same, of course). The exponent (originally 0) is raised accordingly, by 3, so now it is 3.

现在可以删除隐藏的位:42147B(现在是23个低位)

The hidden bit can now be removed: 42147B (now the 23 low bits)

现在可以将其转换为32位值:0x0042147B

This can be turned into a 32 bit value for now: 0x0042147B

现在让我们接受指数:3 +十六进制7F的偏差=十六进制82或1000 0010二进制.

Now let's take on the exponent: 3 + bias of hex 7F = hex 82, or 1000 0010 binary.

在左侧添加符号位:1 1000 0010.重新组合:1100 0001 0或C10

Add the sign bit on the left: 1 1000 0010. Regrouped: 1100 0001 0 or C10

当然，这些是最高位，所以我们将完整的32位变成0xC1000000

Of course these are top bits, so we turn that into 0xC1000000 for the full 32 bits

0xC100000 | 0x0042147B = 0xC142147B

这就是您想要的值.

(*) 32位，所以以后有足够的位数可以正确取整.

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