Point to a function with an already - provided arguments

Asked May 21 '20 at 17:28

Active May 21 '20 at 17:28

Viewed 17 times

I want to point to a function with arguments, except the argument is provided before invoking. It is not possible for me to provide the arguments while invoking, as I am dealing with different function with different amounts of arguments. The specific function I want to invoke without arguments is exit(), so if there is version of the function exit with no arguments (not matter what the return code is) it will solve my problem as well. This is what I am trying to do:

/*Point to exit() with argument provided 1. Am  not trying to call the function just yet.*/
void *func = &exit(1);
/*Dereference and invoke the function. Provide no argument as they were already provided.*/
*(((void)()) func)();

asked May 21 '20 at 17:28

avivgood2

1

Can you use `void my_exit1(void) { exit(1); } ... func = my_exit1;`? – chux - Reinstate Monica May 21 '20 at 17:30
@chux-ReinstateMonica I thought about that solution, But it seem messy to create a function just for that – avivgood2 May 21 '20 at 17:32
2

No, C doesn't support closures. – KamilCuk May 21 '20 at 17:33
@KamilCuk and is there any exit function with no arguments? – avivgood2 May 21 '20 at 17:34
2

You'll need to write your own `exit_1()` function and use that. – tadman May 21 '20 at 17:35
At some point prior to calling `*(((void)()) func)();` the `exit()` arguments (just one arg in this case) needs definition. Could set globally and then use `void my_exit(void) { exit(global_my_exit_arg); }` (and risk the usual global variable gripes) – chux - Reinstate Monica May 21 '20 at 17:36

Point to a function with an already - provided arguments

0 Answers0