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Suppose I have a function pointer, What's the more conventional way of assigning a function to it?

void (*myFunc)(void*)

1) Use & to get function's address, and dereference to invoke.

myFunc = &function;

(*myFunc)(input);

Or

2) Use function's name directly, and invoke without dereferencing.

myFunc = function;

myFunc(input);

Steven Shang
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  • This might answer the question: [Why do function pointer definitions work with any number of ampersands '&' or asterisks '*'?](https://stackoverflow.com/questions/6893285/why-do-function-pointer-definitions-work-with-any-number-of-ampersands-or-as). – Lundin Sep 13 '17 at 06:40

1 Answers1

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Neither matter, functions decay to function pointers. As with all things that don't matter, pick one and stick to it.

Additionally, there is no stylistic requirement that you call functions with (*myFunc)(input) if you get them with &function. There is no fundamental issue with getting function pointers using myFunc = &function and then writing myFunc(input), or getting function pointers with myFunc = function and then writing (*myFunc)(input).

Fun fact: the language never treats a function call as a call to a function value. Functions always undergo decay to a function pointer type, and the function pointer is what is called. In other words:

puts("hello world!");

is implicitly converted to:

((int (*)(const char*))&puts)("hello world!");

and so (*myFunc)(input) would, similarly, resolve to ((void (*)(int))&*myFunc)(input), which seems a bit silly, but unless you care deeply about the abstract syntax tree view of your program, as we said before, it doesn't matter. It does the exact same thing.

zneak
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