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copied!<p>Just to be clear, in your example, because 110 is not after 150, <code>time_after(110,150)</code> should (and does) return false. From the comment:</p> <pre><code>time_after(a,b) returns true if the time a is after time b. </code></pre> <p>Also, note that the code does indeed handle wrapping around to 0. To make the following a bit easier to understand, I'll use unsigned and signed one-byte values, i.e. 8-bit 2's complement. But the argument is general.</p> <p>Suppose b is 253, and five ticks later jiffies has wrapped around to 2. We would therefore expect <code>time_after(2,253)</code> to return true. And it does (using int8_t to denote a signed 8-bit value):</p> <pre><code>(int8_t) 253 - (int8_t) 2 == -3 - 2 == -5 < 0 </code></pre> <p>You can try other values, too. This one is trickier, for <code>time_after(128, 127)</code>, which should be true as well:</p> <pre><code>(int8_t) 127 - (int8_t) 128 == 127 - (-128) == 255 == -1 (for 8-bit 2's complement) < 0 </code></pre> <p>In reality the type of the expression <code>(int8_t) 127 - (int8_t) 128</code> would be an int, and the value really would be 255. But using longs the expression type would be long and the equivalent example would be, for <code>time_after( 2147483648, 2147483647)</code>:</p> <pre><code>(long) 2147483647 - (long) 2147483648 == 2147483647 - (-2147483648) == 4294967295 == -1 < 0 </code></pre> <p>Eventually, after wrapping around, the "after" jiffies value <code>a</code> will begin to catch up with the before value <code>b</code>, and <code>time_after(a,b)</code> will report false. For N-bit 2's complement, this happens when <code>a</code> is 2^(N-1) ticks later than b. For N=8, that happens when <code>a</code> is 128 ticks after b. For N=32, that's 2147483648 ticks, or (with 1 ms ticks) about 25 days.</p> <p>For the mathematically inclined, I believe that in general <code>time_after(a,b)</code> returns true iff the least residue (modulo 2^N) of (a-b) is > 0 and < 2^(N-1).</p>

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