在不使用 long long 的系统中将 23 位数据类型相乘
我正在尝试在微控制器中实现浮点运算，到目前为止我已经取得了巨大的成功。
问题在于我在计算机中进行乘法的方式，并且它工作正常：
unsigned long long gig,mm1,mm2;
unsigned long m,m1,m2;
mm1 = f1.float_parts.mantissa;
mm2 = f2.float_parts.mantissa;
m1 = f1.float_parts.mantissa;
m2 = f2.float_parts.mantissa;
gig = mm1*mm2; //this works fine I get all the bits I need since they are all long long, but won't work in the mcu
gig = m1*m2//this does not work, to be precise it gives only the 32 least significant bits , but works on the mcu
所以你可以看到我的问题是，如果我尝试 gig，微控制器将抛出对 __muldi3 的未定义引用= mm1*mm2 那里。
如果我尝试使用较小的数据类型，它只会保留最低有效位，这是我不希望的。 我需要该产品的 23 msb 位。
有人对我如何做到这一点有任何想法吗？
I am trying to implement floating point operations in a microcontroller and so far I have had ample success.
The problem lies in the way I do multiplication in my computer and it works fine:
unsigned long long gig,mm1,mm2;
unsigned long m,m1,m2;
mm1 = f1.float_parts.mantissa;
mm2 = f2.float_parts.mantissa;
m1 = f1.float_parts.mantissa;
m2 = f2.float_parts.mantissa;
gig = mm1*mm2; //this works fine I get all the bits I need since they are all long long, but won't work in the mcu
gig = m1*m2//this does not work, to be precise it gives only the 32 least significant bits , but works on the mcu
So you can see that my problem is that the microcontroller will throw an undefined refence to __muldi3 if I try the gig = mm1*mm2 there.
And If I try with the smaller data types, it only keeps the least significant bits, which I don't want it to. I need the 23 msb bits of the product.
Does anyone have any ideas as to how I can do this?
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对简短的回答表示歉意，我希望其他人能花时间写一个更完整的解释，但基本上你所做的与在纸上手工乘以两个大数一样！ 只是您不是使用基数 10，而是使用基数 256。也就是说，将您的数字视为字节向量，并对每个字节执行“手动乘法”时对数字执行的操作。
Apologizes for the short answer, I hope that someone else will take the time to write a fuller explanation, but basically you do exactly as when you multiply two big numbers by hand on a paper! It's just that instead of working with base 10, you work in base 256. That is, treat your numbers as a byte vectors, and do with each byte what you do to a digit when you "hand multiply".
FreeBSD 实现 __muldi3() 中的注释对所需过程有很好的解释，请参阅 muldi3.c。 如果您想直接访问源代码（总是一个好主意！），根据评论，此代码基于 Knuth 的 计算机编程的艺术卷。 2（第 2 版），第 4.3.3 节，第 14 页。 278.（注意该链接适用于第三版。）
The comments in the FreeBSD implementation of __muldi3() have a good explanation of the required procedure, see muldi3.c. If you want to go straight to the source (always a good idea!), according to the comments this code was based on an algorithm described in Knuth's The Art of Computer Programming vol. 2 (2nd ed), section 4.3.3, p. 278. (N.B. the link is for the 3rd edition.)
回到 Intel 8088（最初的 PC CPU 和我为其编写汇编代码的最后一个 CPU），当您将两个 16 位数字（32 位？哇哦）相乘时，CPU 将在两个不同的寄存器中返回 2 个 16 位数字 - 一个带有 16 MSB 和LSB 一起。
您应该检查微控制器的硬件功能，也许它有类似的设置（显然，如果有的话，您将需要在汇编中使用此代码）。
否则你将不得不自己实现乘法。
Back on the Intel 8088 (the original PC CPU and the last CPU I wrote assembly code for) when you multiplied two 16 bit numbers (32 bits? whoow) the CPU would return 2 16 bit numbers in two different registers - one with the 16 msb and one with the lsb.
You should check the hardware capabilities of your micro-controller, maybe it has a similar setup (obviously you'll need the code this in assembly if it does).
Otherwise you'll have to implement multiplication on your own.