Editing Two's complement (section)

===Multiplication===
The product of two {{mvar|N}}-bit numbers requires {{math|2''N''}} bits to contain all possible values.<ref>Bruno Paillard. ''An Introduction To Digital Signal Processors'', Sec. 6.4.2. Génie électrique et informatique Report, Université de Sherbrooke, April 2004.</ref>

If the precision of the two operands using two's complement is doubled before the multiplication, direct multiplication (discarding any excess bits beyond that precision) will provide the correct result.<ref>{{cite web
 |url         = http://pages.cs.wisc.edu/~cs354-1/beyond354/int.mult.html
 |title       = Two's Complement Multiplication
 |date        = August 24, 2007
 |access-date  = April 13, 2015
 |author      = Karen Miller
 |website     = cs.wisc.edu
 |url-status     = dead
 |archive-url  = https://web.archive.org/web/20150213203512/http://pages.cs.wisc.edu/~cs354-1/beyond354/int.mult.html
 |archive-date = February 13, 2015
}}</ref> For example, take {{math|1=6 × (−5) = −30}}. First, the precision is extended from four bits to eight. Then the numbers are multiplied, discarding the bits beyond the eighth bit (as shown by "{{Mono|x}}"):
<pre style="width:25em">
     00000110  (6)
 *   11111011  (−5)
 ============
          110
         1100
        00000
       110000
      1100000
     11000000
    x10000000
 + xx00000000
 ============
   xx11100010
</pre>
This is very inefficient; by doubling the precision ahead of time, all additions must be double-precision and at least twice as many partial products are needed than for the more efficient algorithms actually implemented in computers. Some multiplication algorithms are designed for two's complement, notably [[Booth's multiplication algorithm]]. Methods for multiplying sign-magnitude numbers do not work with two's-complement numbers without adaptation. There is not usually a problem when the multiplicand (the one being repeatedly added to form the product) is negative; the issue is setting the initial bits of the product correctly when the multiplier is negative. Two methods for adapting algorithms to handle two's-complement numbers are common:

* First check to see if the multiplier is negative. If so, negate (i.e., take the two's complement of) both operands before multiplying. The multiplier will then be positive so the algorithm will work. Because both operands are negated, the result will still have the correct sign.
* Subtract the partial product resulting from the MSB (pseudo sign bit) instead of adding it like the other partial products. This method requires the multiplicand's sign bit to be extended by one position, being preserved during the shift right actions.<ref>{{cite book |first=John F. |last=Wakerly |title=Digital Design Principles & Practices |publisher=Prentice Hall |edition=3rd |year=2000 |page=47 |isbn=0-13-769191-2 }}</ref>

As an example of the second method, take the common add-and-shift algorithm for multiplication. Instead of shifting partial products to the left as is done with pencil and paper, the accumulated product is shifted right, into a second register that will eventually hold the least significant half of the product. Since the [[least significant bit]]s are not changed once they are calculated, the additions can be single precision, accumulating in the register that will eventually hold the most significant half of the product. In the following example, again multiplying 6 by −5, the two registers and the extended sign bit are separated by "|":
<pre style="width:90%;overflow:scroll;white-space:pre">
  0 0110  (6)  (multiplicand with extended sign bit)
  × 1011 (−5)  (multiplier)
  =|====|====
  0|0110|0000  (first partial product (rightmost bit is 1))
  0|0011|0000  (shift right, preserving extended sign bit)
  0|1001|0000  (add second partial product (next bit is 1))
  0|0100|1000  (shift right, preserving extended sign bit)
  0|0100|1000  (add third partial product: 0 so no change)
  0|0010|0100  (shift right, preserving extended sign bit)
  1|1100|0100  (subtract last partial product since it's from sign bit)
  1|1110|0010  (shift right, preserving extended sign bit)
   |1110|0010  (discard extended sign bit, giving the final answer, −30)
</pre>