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POFastest way to determine if an integer is between two integers (inclusive) with known sets of values
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copied!<p>Is there a faster way than <code>x >= start && x <= end</code> in C or C++ to test if an integer is between two integers?</p> <p><em>UPDATE</em>: My specific platform is iOS. This is part of a box blur function that restricts pixels to a circle in a given square.</p> <p><em>UPDATE</em>: After trying the <a href="https://stackoverflow.com/a/17095534/1165522">accepted answer</a>, I got an order of magnitude speedup on the one line of code over doing it the normal <code>x >= start && x <= end</code> way.</p> <p><em>UPDATE</em>: Here is the after and before code with assembler from XCode:</p> <p><strong>NEW WAY</strong></p> <pre><code>// diff = (end - start) + 1 #define POINT_IN_RANGE_AND_INCREMENT(p, range) ((p++ - range.start) < range.diff) Ltmp1313: ldr r0, [sp, #176] @ 4-byte Reload ldr r1, [sp, #164] @ 4-byte Reload ldr r0, [r0] ldr r1, [r1] sub.w r0, r9, r0 cmp r0, r1 blo LBB44_30 </code></pre> <p><strong>OLD WAY</strong></p> <pre><code>#define POINT_IN_RANGE_AND_INCREMENT(p, range) (p <= range.end && p++ >= range.start) Ltmp1301: ldr r1, [sp, #172] @ 4-byte Reload ldr r1, [r1] cmp r0, r1 bls LBB44_32 mov r6, r0 b LBB44_33 LBB44_32: ldr r1, [sp, #188] @ 4-byte Reload adds r6, r0, #1 Ltmp1302: ldr r1, [r1] cmp r0, r1 bhs LBB44_36 </code></pre> <p>Pretty amazing how reducing or eliminating branching can provide such a dramatic speed up.</p>

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