What does int (*func)() do in C?

Question

What does this line do in C?

 int (*func)();

I extracted it from the following program that is supposed to execute byte code (assembly instructions converted to its corresponding bytes)

char code[] = "bytecode will go here!";
int main(int argc, char **argv)
{
  int (*func)();
  func = (int (*)()) code;
  (int)(*func)();
}

It's interesting that the example compiled and run produces a segmentation fault. Why `code` is referenced as a function... — Eugeniu Rosca, Jun 17 '15 at 20:04
@EugeniuRosca that's because the array doesn't (yet) contain executable code. So it crashes. — Weather Vane, Jun 17 '15 at 20:05
@EugeniuRosca I believe the reason for this is because the stack and the heap is not executable in most modern OSes (for security reasons). You would have to turn this functionality off for this code to work. — SivaDotRender, Jun 17 '15 at 20:06
@SivaDotRender so why did you ask the question if you know what it does? — Weather Vane, Jun 17 '15 at 20:09
How the name of an array can be assigned to a function pointer? Is it legal? — Eugeniu Rosca, Jun 17 '15 at 20:09
@WeatherVane that was not my question though. My question was about the int (*func)(); and exactly what that line does. — SivaDotRender, Jun 17 '15 at 20:11
@WeatherVane: Give me an example of data that can be put in the `code` array without producing a segmentation fault. — Eugeniu Rosca, Jun 17 '15 at 20:13
@EugeniuRosca it wasn't my question, but with no valid code it certainly won't run. — Weather Vane, Jun 17 '15 at 20:15
@EugeniuRosca have you tried to disable DEP on your machine? The code will be specific to your processor. — SivaDotRender, Jun 17 '15 at 20:20
@WeatherVane: That code is not human-readable for sure. I believe you must put there some stuff compiled *prior* to compiling this `main` function we see in the example. — Eugeniu Rosca, Jun 17 '15 at 20:23
@EugeniuRosca; You need to provide the byte code like `char code[] = "\x48\x31\xc0";` and then the function at this address will be executed. — haccks, Jun 17 '15 at 20:27
That is a pretty damn interesting example, i must confess. thanks for your clarifications. — Eugeniu Rosca, Jun 17 '15 at 20:31
@EugeniuRosca; [This](http://stackoverflow.com/q/28668138/2455888) would help you dig it more. — haccks, Jun 17 '15 at 20:38
@EugeniuRosca Change to `char const code[] = "\xb8\x05\x00\x00\x00\xc3";` and there should be no segmentation fault. `const` goes to `.rodata` section which is executable under Linux. The code just returns 5. — 4566976, Jun 17 '15 at 20:40
as a side note, that last line of main can be rewritten as `func();` everything else there is extra — Steve Cox, Jun 17 '15 at 20:42
haccks: you could link the 2 questions :D @4566976: it worked! — Eugeniu Rosca, Jun 17 '15 at 20:43

mastov · Accepted Answer · 2015-06-17T20:49:31.003

5

The line in question declares a function pointer for a function with unspecified arguments (an "obsolescent" feature since C99) and with return type int.

The first line of your main declares that pointer, the second line initializes the pointer so it points to the function code. The third line executes it.

You can get a description of pointers to functions in general here.

edited Jun 17 '15 at 20:49

answered Jun 17 '15 at 19:51

mastov

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@haccks: Edited. I had confused it with `void` as argument specification. – mastov Jun 17 '15 at 20:01
@haccks: No downvote, but withdrawal. I'm more of a C++ guy, I should watch out better for the differences when answering C questions. – mastov Jun 17 '15 at 20:06
obsolescent? do you mean obsolete? or do you actually mean that its been becoming more obsolete since C99? – Steve Cox Jun 17 '15 at 20:40
@SteveCox: http://dictionary.reference.com/browse/obsolescent and http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf (6.11.6, p. 179) – mastov Jun 17 '15 at 20:44
@SteveCox: Apparently it's their way of saying "deprecated": *Certain features are obsolescent, which means that they may be considered for withdrawal in future revisions of this International Standard. They are retained because of their widespread use, but their use in new implementations (for implementation features) or new programs (for language [6.11] or library features [7.31]) is discouraged.* (page XVII) – mastov Jun 17 '15 at 21:06

haccks · Answer 2 · 2015-06-17T20:30:42.243

3

int (*func)(); declares func as a pointer to a function that returns an int type and expects any number of arguments.

edited Jun 17 '15 at 20:30

answered Jun 17 '15 at 19:53

haccks

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The arguments can be any type, not just int. No conversions(other than the normal function argument promotions) or checking will be done on any arguments passed to the function when it is called, but if they don't match what the function is expecting, the result is undefined. – Chris Dodd Jun 17 '15 at 20:29
@ChrisDodd; Agreed. Removed that line, but many C compilers assumes default type to be `int`. – haccks Jun 17 '15 at 20:32
The defaulting to `int` only applies to declarators with no type specifier. So if you said `int (*func)(a)`, that would be one argument of type `int` named `a`. Or if you said `(*func)();` in the global scope, the return type would default to `int` -- defaulting to `int` only occurs in scopes where expressions are not possible (so not on local vars within a function). – Chris Dodd Jun 17 '15 at 20:36
1

@ChrisDodd - to be more precise, arguments will undergo default argument promotion. More details for the curious are here: http://stackoverflow.com/questions/1255775/default-argument-promotions-in-c-function-calls I don't mean to be pedantic, but such argument promotion can cause hard-to-diagnose problems, with this particular example being prone to such issues. – Andrew Henle Jun 17 '15 at 20:39

What does int (*func)() do in C?

2 Answers2

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