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[Bak]

How would one declare a const array of function pointers? I played with this thing for 10 minutes and couldn't get it compiling. Had to settle for making a wrapper struct for the function pointer, then making an array of structs.

I guess a typedef would work too... but seems like it shouldn't be necessary.
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[Mr Ekted]

const int (*fp[10])(int);

Each function has the form:

int f0 (int) { }
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Code: Show/Hide #include <stdio.h> int main() {         void *fList[2];         fList[0] = printf;         fList[1] = main;         printf("Location of printf: %p\n", printf);         printf("Location of main: %p\n", main);         printf("Location of fList[0]: %p\n", fList[0]);         printf("Location of fList[1]: %p\n", fList[1]);         return 0; }

Code: Show/Hide Location of printf: 0x80482a8 Location of main: 0x8048384 Location of fList[0]: 0x80482a8 Location of fList[1]: 0x8048384

Worked fine for me. Requires you to typecast it correctly when you use it though. If all the functions are the same style, use Ekted's method.

[Bak]

That works.

[Bak]

Looks like I spoke too soon.

when you do const int (*fp[10])(int); it thinks the function is returning a const int, rather than the array is a const

For example this won't compile:

Code: Show/Hide int (*f)(int) = 0; int (*f2)(int) = 0; const int (*fp[2])(int) = {    f,    f2 };

got it to work though, you need

Code: Show/Hide int (*const fp[2])(int) = {    f,    f2 };

[Mr Ekted]

You could also typedef a function pointer, then declare a const array of that type.

[Cyan~Fire]