Editing Strength reduction (section)

== Other strength reduction operations ==
{{hatnote|This material is disputed. It is better described as [[peephole optimization]] and instruction assignment.}}

Operator strength reduction uses [[Identity (mathematics)|mathematical identities]] to replace slow math operations with faster operations. The benefits depend on the target CPU and sometimes on the surrounding code (which can affect the availability of other functional units within the CPU).

* replacing integer division or multiplication by a power of 2 with an [[arithmetic shift]] or [[logical shift]]<ref>In languages such as C and Java, integer division has round-towards-zero semantics, whereas a bit-shift always rounds down, requiring special treatment for negative numbers. For example, in Java, <code>-3 / 2</code> evaluates to <code>-1</code>, whereas <code>-3 >> 1</code> evaluates to <code>-2</code>. So in this case, the compiler ''cannot'' optimize division by two by replacing it by a bit shift.</ref>
* replacing integer multiplication by a constant with a combination of shifts, adds or subtracts
* replacing integer division by a constant with a multiplication, taking advantage of the limited range of machine integers.<ref>{{cite web|last=Granlund|first=Torbjörn|title=Division by Invariant Integers Using Multiplication|url=http://gmplib.org/~tege/divcnst-pldi94.pdf|author2=Peter L. Montgomery }}</ref> This method also works if divisor is a non-integer sufficiently greater than 1, e.g. √2 or π.<ref>{{cite web |last=Jones |first=Nigel |title=Division of integers by constants |url=https://embeddedgurus.com/stack-overflow/2009/06/division-of-integers-by-constants/ |archive-url=https://web.archive.org/web/20240326221708/https://embeddedgurus.com/stack-overflow/2009/06/division-of-integers-by-constants/ |archive-date=26 March 2024}}</ref>

{| class="wikitable"
|-
! Original calculation
! Replacement calculation
|-
| y+= 1
| y++
|-
| y%2 != 0
| y & 1
|-
| y = x * 2
| y = x << 1
|-
| y = x / 2
| y = x >> 1
|-
| y = x % 2
| y = x & 1
|-
| y = x * 15
| y = (x << 4) - x
|-
| y = (uint16_t)x / 10
| y = ((uint32_t)x * (uint32_t)0xCCCD) >> 19)
|-
| y = (uint16_t)x / π
| y = (((uint32_t)x * (uint32_t)0x45F3) >> 16) + x) >> 2)
|}