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copied!<p>If you don't have a 64-bit type, you can do it like the following:</p> <pre><code>uint32_t higher, lower; // your input lower /= 1000; lower += (higher % 1000) * 4294967L; // approximate 2^32 / 1000 higher /= 1000; </code></pre> <p>If the result fitted in <code>lower</code> itself, <code>higher</code> should be <code>0</code>.</p> <p>Just note that as @Mikhail pointed out, this solution is approximate, and has an error of <code>0.296 * higher + 2</code> ms (unless I'm missing something).</p> <hr> <p>If you <em>really</em> want a better precision and don't care about efficiency, you can use a bit of floating-point arithmetic in the middle, and round the results correctly. I doubt if it's worth the effort:</p> <pre><code>uint32_t higher, lower; // your input // simpler without a helper variable if (lower % 1000 >= 500) { lower /= 1000; ++lower; } else lower /= 1000; lower += round((higher % 1000) * 4294967.296); // 2^32 / 1000 higher /= 1000; </code></pre> <p>You'll need to <code>include <cmath></code> for <code>round()</code>.</p> <p>As a note, @Mikhail's solution in this case is probably better and <em>may</em> be faster. Though it's too complex for me.</p> <hr> <p>If you have a 64-bit type, you can convert the split value to it:</p> <pre><code>uint64_t whole_number = higher; whole_number <<= 32; whole_number |= lower; </code></pre> <p>And then you can use <code>whole_number</code> as usual.</p> <hr> <p>Note that if you only need a difference, it will be faster to subtract the values before actually dividing.</p> <p>Assuming that you know which value is bigger:</p> <pre><code>uint32_t higher1, lower1; // smaller value uint32_t higher2, lower2; // bigger value uint32_t del_high = higher2 - higher1; uint32_t del_low = lower2 - lower1; if (lower2 < lower1) --del_high; </code></pre> <p>And now you can convert the result like explained before. Or with a bit luck, <code>del_high</code> will be <code>0</code> (if the difference is smaller than 2^32 μs), and you will have the result in <code>del_low</code> (in μs).</p>

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