两个16位数字的乘法-为什么结果为32位长? [英] Multiplication of two 16-bit numbers - Why is the result 32-bit long?

问题描述

如果我将两个16位数字相乘，结果将是32位长.但是为什么会这样呢? 对此的明确解释是什么?

据我正确的理解: 计算方法是:将n位数字乘以m位数字得出(n + m)位数字?

解决方案

(2 ⁿ -1)*(2 ^m -1)= 2 ^{n + m} -2 ⁿ -2 ^m + 1

-(2 ⁿ + 2 ^m )就像清除索引 n 和 m ，与2 ^{n + m} 相比，对结果的影响不大，因此您需要n + m位来表示结果.

例如 1111 ₂ * 1111 ₂ = 11100001 ₂ (15 * 15 = 225)

通常(b ⁿ -1)*(b ^m -1)= b ^{n + m} -b ⁿ -b ^m + 1 ，因此在任意基数 b 中将n位数字乘以m位数字最多n + m个数字

您可以轻松地以10为底: 9 * 9 = 81 (1位* 1位= 2位)或 99 * 99 = 9801 (2位) * 2位= 4位)

If I multiplie two 16-bit numbers, the result will be 32-bit long. But why is this so? What is the clear explanation for this?

And for my right understanding: The calculation for this is: n-bit number multiplied with a m-bit number gives a (n+m) bit number?

解决方案

(2ⁿ - 1)*(2^m - 1) = 2^n+m - 2ⁿ - 2^m + 1

-(2ⁿ + 2^m) is like clearing the bits at index n and m, which does not affect much the result compared to 2^n+m, so you need n+m bits to represent the result.

For example 1111₂*1111₂ = 11100001₂ (15*15 = 225)

In general, (bⁿ - 1)*(b^m - 1) = b^n+m - bⁿ - b^m + 1, so multiply an n-digit by an m-digit number in an arbitrary base b results in a number at most n+m digits

You can see that easily in base 10: 9*9 = 81 (1 digit * 1 digit = 2 digit) or 99*99 = 9801 (2 digit * 2 digit = 4 digit)

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