位相乘
Mulitplying bits
我的教科书是这样说的：
我在乘法部分没有得到这句话：
The first observation is that the number of digits in the product is considerably larger than the number in either the multiplicand or the multiplier. In fact, if we ignore the sign bits, the length of the multiplication of an n-bit multiplicand and an m-bit multiplier is a product that is n + m bits long. That is, n + m bits are required to represent all possible products.
我在产品中看到 7 位数字，但此引用暗示应该有 8 位
报价是在谈论存储产品所需的space（粗体强调我的）：
That is, n + m bits are required to represent all possible products.
您只查看了一个产品。您需要查看 n 位数字和 m 位数字的所有可能乘积。例如，对于 m = n = 4 和基数 10：
9999×9999 = 99980001
很明显，这是最大尺寸，而不是每个产品的尺寸，否则任何n-digit number and 0 should have n+1 digits 当它显然只有 1 数字.
我的教科书是这样说的：
我在乘法部分没有得到这句话：
The first observation is that the number of digits in the product is considerably larger than the number in either the multiplicand or the multiplier. In fact, if we ignore the sign bits, the length of the multiplication of an n-bit multiplicand and an m-bit multiplier is a product that is n + m bits long. That is, n + m bits are required to represent all possible products.
我在产品中看到 7 位数字，但此引用暗示应该有 8 位
报价是在谈论存储产品所需的space（粗体强调我的）：
That is, n + m bits are required to represent all possible products.
您只查看了一个产品。您需要查看 n 位数字和 m 位数字的所有可能乘积。例如，对于 m = n = 4 和基数 10：
9999×9999 = 99980001
很明显，这是最大尺寸，而不是每个产品的尺寸，否则任何n-digit number and 0 should have n+1 digits 当它显然只有 1 数字.