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    copied!<p>Below is some code that I recall seeing a while ago in a compiler construction class. I thought it was particularly clever (if not 'clean') so I held on to it. </p> <p>From <a href="http://www.cs.colorado.edu/~main/a++/tree.h" rel="nofollow noreferrer">http://www.cs.colorado.edu/~main/a++/tree.h</a></p> <pre><code> // If data of type T can be printed with the usual &lt;&lt; operator, then // print&lt;T&gt;(out, p) will interpret *p as a T object and print its // value to out. Otherwise, a message is printed to out, indicating // that objects of type T are not printable. template&lt;typename T&gt; void print(std::ostream&amp; out, const void* p) { // The first part of this code sets an enum value, is_printable, to // be 1 if the data type T can be printed with the usual &lt;&lt; // operator. Otherwise, is_printable is set to zero. The programming // technique is based on a note from Herb Sutter at // http://www.gotw.ca/gotw/071.htm class object { public: object(T convert) { }; }; char operator &lt;&lt; (std::ostream&amp;, const object&amp;); enum { is_printable = sizeof(std::cout &lt;&lt; (*static_cast&lt;T*&gt;(0))) == sizeof(char) ? 0 : 1 }; // Notice that the boolean expression in the if-statement is known at // compile time, so that only one of the two output statements will be // compiled into object code. if (is_printable) out &lt;&lt; *static_cast&lt;const T*&gt;(p); else out &lt;&lt; "(value of type " &lt;&lt; typeid(T).name() &lt;&lt; " cannot be printed)"; } </code></pre> <p>When you construct your container object, hold a pointer to the print function for the variable:</p> <pre><code>void (*printer)(std::ostream&amp;, const void*); printer = print&lt;T&gt;; </code></pre> <p>Then later use the printer() function to display the contained value if possible.</p> <p>Hope this helps.</p>
 

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