From: bc@freeuk.com   
      
   On 02/01/2026 07:24, Michael Sanders wrote:   
   > i have:   
   >   
   > void moo(char HISTORY[][64], int hst_len, int invalid, const char *gme_msg)   
   >   
   > void mastermind(char HISTORY[][64], int hst_len, int invalid, const char   
   *gme_msg)   
   >   
   > to use either i have:   
   >   
   > void (*render)(char [][64], int, int, const char *) = MOO ? moo : mastermind;   
   >   
   > my multi-part question:   
   >   
   > why is void required for the function pointer?   
   >   
   > A: because both moo() & mastermind return void?   
      
   Neither return anything. But in C there is no special syntax to denote   
   routines that don't return values. So a dummy 'void' return type is used.   
      
   >   
   > B: because every function must have a return type   
   >     *including function pointers*?   
      
   If you have any function that returns type T (including when T is void   
   like your examples), then its type is 'function(...)returning T'.   
      
   A pointer to such a function will have a type:   
      
      'pointer to function(...)returning T'.   
      
   In C syntax, that return type always goes at the start of the   
   declaration. In your examples,  that will be 'void'.   
      
   > C: what about tyedef?   
      
   Typedefs create convenient aliases for types. So if you created an alias   
   U for that pointer type in my last example, then you can subsequently   
   just use U:   
      
       U p, q, r;   
      
   This declares three variable all with type 'pointer to ... void'. So in   
   this case the details are hidden, including that void.   
      
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    * Origin: you cannot sedate... all the things you hate (1:229/2)