Updated Examples and Recipes (markdown)

Examples-and-Recipes.md

@@ -5,6 +5,7 @@ This page contains examples and recipes contributed by community members. Feel f
 - [Obtaining `y/m/d h:m:s` components from a `time_point`](#components_to_time_point)
 - [Normalizing `y/m/d` when it is `!ok()`](#normalize)
 - [Converting from {year, microseconds} to CCSDS](#ccsds)
+- [Difference in months between two dates](#deltamonths)
 
 ***
 
@@ -192,6 +193,56 @@ The next step is to find when the day started that is associated with `utc`.  To
 
 Now the number of microseconds since the start of the day needs to be computed.  The start of the day, `dp`, is converted back into the leap-second aware system, and subtracted from the microsecond time point: `utc`.  The variable `us` is reused to hold “microseconds since midnight”.  Now it is a simple computation to split this into milliseconds since midnight, and microseconds since the last millisecond.
 
+<a name="delta months"></a>
+### Difference in months between two dates
+(by [Howard Hinnant](https://github.com/HowardHinnant))
+
+I recently stumbled across this Stack Overflow question:
+
+http://stackoverflow.com/q/2536379/576911
+
+And I decided to see if I could answer it (here) using this library.  The question asks:  How can I get the number of months between two dates?  And it gives two example dates:
+
+    auto d1 = 1_d/oct/2013;
+    auto d2 = 30_d/oct/2016;
+
+(I've converted the syntax to that of this library).
+
+The question isn't perfectly clear since "months" is not a very precise unit.  Do we want the number of "full months"?  Or perhaps we should round to the nearest number of integral months?  Or do we want a floating-point representation of months which can show fractional months?
+
+These are all reasonable possibilities, and you can compute all of these things with this library.  Sometimes the hardest part of a question is sufficiently refining it until you know what is really being asked.
+
+If we want to just ignore the day-field in these dates, and then compute the number of months, that is easily done like so:
+
+    std::cout << (d2.year()/d2.month() - d1.year()/d1.month()).count() << '\n';
+
+This creates two `year_month` objects and subtracts them.  This gives a `std::chrono::duration` that represents a signed-integral number of months.  The output is:
+
+    36
+
+To include the influence of the day-fields, it is best to convert `d1` and `d2` to `day_point`s:
+
+    auto dp1 = day_point(d1);
+    auto dp2 = day_point(d2);
+
+Now we could (for example) subtract the two `day_point`s, and round the result to the nearest integral month:
+
+    std::cout << round<months>(dp2-dp1).count() << '\n';
+
+This outputs:
+
+    37
+
+Or we could create a new `std::chrono::duration` type based on `float`, but with a period of months, and convert the difference to that:
+
+    std::cout << duration<float, months::period>(dp2-dp1).count() << '\n';
+
+This outputs:
+
+    36.9617
+
+These are all reasonable answers to the question, and all easily computable with this library.
+
 ***
 
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