如何在没有除法硬件和没有浮点硬件的二进制中实现浮点除法 [英] How to implement floating point division in binary with no division hardware and no floating point hardware

问题描述

我想知道如何在没有除法硬件和浮点硬件的情况下以二进制实现 IEEE-754 32 位单精度浮点除法?

I am wondering how to implement IEEE-754 32-bit single precision floating point division in binary with no division hardware and no floating point hardware?

我有移动硬件、加法、减法和乘法.

I have shifting hardware, add, subtract, and multiply.

我已经使用 16 位字实现了浮点乘法、加法和减法.

I have already implemented floating point multiplication, addition, and subtraction using 16-bit words.

我正在一个专有的多核处理器上实现这些指令，并用汇编语言编写我的代码.之前，我正在使用 matlab 来验证我的算法.

I am implementing these instructions on a proprietary multicore processor and writing my code in assembly. Beforehand, I am using matlab to verify my algorithm.

我知道我需要减去指数，但是如何对尾数执行无符号除法?

I know I need to subtract the exponents, but how do i perform unsigned division on the mantissas?

推荐答案

取决于你想让它变得多复杂.保持相当简单，您可以尝试通过倒数近似除法.

Depends on how complicated you want to make it. Keeping it reasonably simple, you could try division by reciprocal approximation.

而不是计算:(n/d)你会计算:n *(1/d).

Rather than calculating: (n / d) you'd work out: n * (1 / d).

为此，您需要使用某种方法计算倒数，例如 Newton-Raphson，它使用牛顿法连续计算除数倒数的更准确估计，直到它足够"准确地满足您的目的，然后再进行最后的乘法步骤.

To do this you'd need to work out the reciprocal using some method, for example, Newton-Raphson which uses Newton's method to calculate successively more accurate estimates of the reciprocal of the divisor until it's "adequately" accurate for your purpose before doing the final multiplication step.

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