十六进制转换为十进制当没有数据类型可以保存完整号码 [英] Convert Hex to Decimal when no datatype can hold the full number

问题描述

好了，我正与一个PIC单片机，在C.这是16F，所以它不能持有整数大于32位的（无符号INT32提供最大的数据大小）

Ok, so I am working with a PIC microprocessor, in C. It's a 16F, so it can't hold integers larger than 32bits (unsigned int32 is the largest datasize available)

这是一个读者，我收到了5个字节的ID code。进行传递，我一定要带连接coded到BCD码，数字一个数字。我不能冲刺它为一个字符串，因为它是大的数据大小，并且不能处理它。我不能把它，因为没有操作定义它。

From a reader, I receive a 5 byte ID code. To transmit it, I have to encoded to BCD, digit by digit. I can't sprint it to a string, as it is larger that the data size, and can't process it. I can't divide it because no operation is defined for it.

我想不出任何可能的解决方案，没有任何人有处理过？

I can't figure out any solution possible, does anyone have dealt with this before?

编辑：

我接收在一系列的5个字节的数目：FF-FF-FF-FF-FF。我需要将其转换为十进制0123456789012（13位，256 ^ 5十进制长度），它通过RS232发送。第二个函数（以ASCII码，并将其发送）我已经有工作，但我需要重新串的全部数presentation之前，我可以用它做什么。

I receive the number in a series of 5 bytes: "FF-FF-FF-FF-FF". I need to convert it to decimal "0123456789012" (13 digits, length of 256^5 in decimal) to send it through RS232. The second function (Take the ASCII, and send it) I already have it working, but I need the string representation of the full number before I can do anything with it.

推荐答案

假设你有32位运算：2 ** 24 = 16777216，因此服用x作为最显著2个字节和y为至少显著3：

Assuming you have 32 bit arithmetic: 2**24 = 16777216, so taking x as the most significant 2 bytes and y as the least significant 3:

  (16777216 * x + y) / 1000 
= (16777000 * x + 216 * x + y) / 1000
= 16777 * x + (216 * x + y) / 1000

第一项可以在不溢出来计算在32位（因为 X 2 ** 16 ）。 第二项也可以不溢出（因为计算 X 2 ** 16 和 Y'2 ** 24 ）。

The first term can be calculated in 32 bits without overflow (since x < 2**16). The second term can also be calculated without overflow (since x < 2**16 and y < 2**24).

这基本上是长除法的基础 2 ** 24 2位数的价值，但条件pre-计算明知除数为1000。一干就是选择，因为它是10更大的最少的功率比 2 ** 8 。

This is basically long division in base 2**24 of a 2-digit value, but with terms pre-calculated knowing that the divisor is 1000. One thousand is chosen because it's the least power of 10 greater than 2**8.

因此​​，首先计算最低的三个数字，用事实（2 ** 32）％1000 == 296 。所以这一次，我们将采取x作为最高字节和y作为低4个字节

So, first compute the lowest three digits, use the fact that (2**32) % 1000 == 296. So this time we'll take x as the highest byte and y as the low 4 bytes

((2**32) * x + y) % 1000 = ((2**32) * x) % 1000 + y % 1000 (modulo 1000)
                         = (296 * x) % 1000 + y % 1000     (modulo 1000)
((2**32) * x + y) % 1000 = ((296 * x) % 1000 + y % 1000) % 1000

然后1000使用上面的公式划分原来的号码。然后你安全地进入32位的领土，并且可以使用正常的循环生产出剩余的数字。

Then divide the original number by 1000 using the formula above. Then you're safely into 32 bit territory and can churn out the remaining digits using the normal loop.

顺便说一句，我会检查的结果，如果我是你，我没有测试过这一点，这是可能的，我什么地方犯了一个错误。应该很容易进行比较的的BCD转换结果使用通常的装置在PC上的64位整数进行。

Btw, I'd check the results if I were you: I haven't tested this and it's possible I've made an error somewhere. It should be easy to compare against the results of bcd conversions done using the usual means in a 64-bit integer on a PC.

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