Editing Fixed-point arithmetic (section)

===Binary fixed-point multiplication===
Consider the task of computing the product of 1.2 and 5.6 with binary fixed point using 16 fraction bits.  To represent the two numbers, one multiplies them by 2<sup>16</sup>, obtaining {{thinspace|78|643}}.2 and {{thinspace|367|001}}.6; and round these values the nearest integers, obtaining {{thinspace|78|643}} and {{thinspace|367|002}}.  These numbers will fit comfortably into a 32-bit  word with two's complement signed format.

Multiplying these integers together gives the 35-bit integer {{thinspace|28|862|138|286}} with 32 fraction bits, without any rounding. Note that storing this value directly into a 32-bit integer variable would result in overflow and loss of the most significant bits. In practice, it would probably be stored in a signed 64-bit integer variable or [[register (computing)|register]].

If the result is to be stored in the same format as the data, with 16 fraction bits, that integer should be divided by 2<sup>16</sup>, which gives approximately {{thinspace|440|401}}.28, and then rounded to the nearest integer. This effect can be achieved by adding 2<sup>15</sup> and then shifting the result by 16 bits. The result is {{thinspace|440|401}}, which represents the value 6.{{thinspace|719|985|961|914|062|5}}. Taking into account the precision of the format, that value is better expressed as  6.{{thinspace|719|986}} ± 0.{{thinspace|000|008}} (not counting the error that comes from the operand approximations). The correct result would be 1.2 × 5.6 = 6.72.

For a more complicated example, suppose that the two numbers 1.2 and 5.6 are represented in 32-bit fixed point format with 30 and 20 fraction bits, respectively.  Scaling by 2<sup>30</sup> and 2<sup>20</sup> gives {{thinspace|1|288|490|188}}.8 and {{thinspace|5|872|025}}.6, that round to {{thinspace|1|288|490|189}} and {{thinspace|5|872|026}}, respectively.  Both numbers still fit in a 32-bit signed integer variable, and represent the fractions
: 1.{{thinspace|200|000|000|186|264|514|923|095|703|125}} and
: 5.{{thinspace|600|000|381|469|726|562|50}}
Their product is (exactly) the 53-bit integer {{thinspace|7|566|047|890|552|914}}, which has 30+20 = 50 implied fraction bits and therefore represents the fraction 
: 6.{{thinspace|720|000|458|806|753|229|623|609|513|510|018|587|112|426|757|812|50}}
If we choose to represent this value in signed 16-bit fixed format with 8 fraction bits, we must divide the integer product by 2<sup>50−8</sup> = 2<sup>42</sup> and round the result; which can be achieved by adding 2<sup>41</sup> and shifting by 42 bits. The result is 1720, representing the value 1720/2<sup>8</sup> = 6.{{thinspace|718|75}}, or approximately 6.719 ± 0.002.