Why does two's-complement multiplication need to do sign extension?

Question

In the book Computer Systems A Programmer's Perspective (2.3.5), the method to calculate two's-complement multiplication is described as follows:

Signed multiplication in C generally is performed by truncating the 2w-bit product to w bits. Truncating a two’s-complement number to w bits is equivalent to first computing its value modulo 2^w and then converting from unsigned to two’s-complement.

Thus, for similar bit-level operands, why is unsigned multiplication different from two’s-complement multiplication? Why does two's-complement multiplication need to do sign extension?

To calculate same bit-level representation of unsigned and two’s-complement addition, we can convert the arguments of two’s-complement, then perform unsigned addition, and finally convert back to two’s-complement.

Since multiplication consists of multiple additions, why are the full representations of unsigned and two’s-complement multiplication different?

Answer 1

Figure 2.27 demonstrates the example below:

+------------------+----------+---------+-------------+-----------------+
|       Mode       |    x     |    y    |    x · y    | Truncated x · y |
+------------------+----------+---------+-------------+-----------------+
| Unsigned         |  5 [101] | 3 [011] | 15 [001111] |  7 [111]        |
| Two's complement | –3 [101] | 3 [011] | –9 [110111] | –1 [111]        |
+------------------+----------+---------+-------------+-----------------+

If you multiply 101 by 011, you will get 1111 (which is equal to 001111). How did they get 110111 for two's complement case then?

The catch here is that to get a correct 6-bit two's complement product you need to multiply 6-bit two's complement numbers. Thus, you need first to convert -3 and 3 to 6-bit two's complement representation: -3 = 111101, 3 = 000011 and only then multiply them 111101 * 000011 = 10110111. You also need to truncate the result to 6 bits to eventually get 110111 from the table above.