Gosh Darn Function Pointers

Re:Gosh Darn Function Pointers by femto113 on Monday January 15, 2018 @04:38PM (#16153679)

I've always found a typedef of the function and doing the pointer in the variable declaration the clearest since it doesn't need funky parentheses or dereferences anywhere:

    typedef int func(int arg);
    int foo(int a) { return a * 10; }
    ...
    func *f = foo;
    ...
    int x = f(10);

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Re:Gosh Darn Function Pointers by B4TMAN on Monday January 15, 2018 @04:44PM (#16153718)

I am a little lost. What effect does `typedef int func(int arg);` have on the code?

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Re:Gosh Darn Function Pointers by inopinatus on Monday January 15, 2018 @05:34PM (#16154063)

It allows the later definition of func * f which is more readable and maintainable than int (* f)(int arg)

This pattern is common in well-structured applications, and is especially useful for clarifying external and internal APIs. e.g. from the Dovecot IMAP server:

    typedef bool event_callback_t(struct event *event,
    			      enum event_callback_type type,
    			      struct failure_context *ctx,
    			      const char *fmt, va_list args);

You can now define a function pointer or declare a parameter using event_callback_t. This is both concise and intention revealing.

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Re:Gosh Darn Function Pointers by majewsky on Monday January 15, 2018 @05:33PM (#16154059)

It says that `func` is a type whose values are all functions with the signature `int foo(into arg)`.

The way to read a typedef (in general) is when you strip off the word "typedef", instead of declaring a type, it declares a value of that type. This applies to all types, not just function types. For example, an array of 5 ints is declared as

  int foo[5];

So the type of all 5-element arrays is defined by

  typedef int FiveLongIntArray[5];

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