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    <p><code>typedef</code> is a language construct that associates a name to a type.<br/> You use it the same way you would use the original type, for instance</p> <pre><code> typedef int myinteger; typedef char *mystring; typedef void (*myfunc)(); </code></pre> <p>using them like</p> <pre><code> myinteger i; // is equivalent to int i; mystring s; // is the same as char *s; myfunc f; // compile equally as void (*f)(); </code></pre> <p>As you can see, you could just replace the <em>typedefed</em> name with its definition given above.<br/></p> <p>The difficulty lies in the pointer to functions syntax and readability in C and C++, and the <code>typedef</code> can improve the readability of such declarations. However, the syntax is appropriate, since functions - unlike other simpler types - may have a return value and parameters, thus the sometimes lengthy and complex declaration of a pointer to function.</p> <p>The readability may start to be really tricky with pointers to functions arrays, and some other even more indirect flavors.</p> <p>To answer your three questions</p> <ul> <li><p><strong>Why is typedef used?</strong> To ease the reading of the code - especially for pointers to functions, or structure names. </p></li> <li><p><strong>The syntax looks odd (in the pointer to function declaration)</strong> That syntax is not obvious to read, at least when beginning. Using a <code>typedef</code> declaration instead eases the reading</p></li> <li><p><strong>Is a function pointer created to store the memory address of a function?</strong> Yes, a function pointer stores the address of a function. This has nothing to do with the <code>typedef</code> construct which only ease the writing/reading of a program ; the compiler just expands the typedef definition before compiling the actual code.</p></li> </ul> <p>Example:</p> <pre><code>typedef int (*t_somefunc)(int,int); int product(int u, int v) { return u*v; } t_somefunc afunc = &amp;product; ... int x2 = (*afunc)(123, 456); // call product() to calculate 123*456 </code></pre>
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    1. COin the last example, wouldn't just 'square' refer to the same thing i.e pointer to the function instead of using &square.
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    2. COQuestion, in your first typedef example you have of the form `typedef type alias` but with function pointers there only seems to be 2 arguments, `typedef type`. Is alias defaulted to the name specified in type argument?
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    3. CO@pranavk: Yes, `square` and `&square` (and, indeed, `*square` and `**square`) all refer to the same function pointer.
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      1. USJonathan Leffler
 

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