How do you declare a const array of function pointers? - c++

Question

Firstly, I've got functions like this.
void func1();
void func2();
void func3();
Then I create my typedef for the array:
void (*FP)();
If I write a normal array of function pointers, it should be something like this:
FP array[3] = {&func1, &func2, &func3};
I want to make it a constant array, using const before "FP", but I've got this error messages:
error: cannot convert 'void ( * )()' to 'void ( * const)()' inialization
PD: Sorry my bad English.
EDIT:
x.h
typedef void (*FP)();
class x
{
private:
int number;
void func1();
void func2();
void func3();
static const FP array[3];
}
x.cpp
const FP x::array[3] = {&x::func1, &x::func2, &x::func3};
My code is more large and complex, this is a summary

Answer 1

Then I create my typedef for the array:
void (*FP)();
Did you miss typedef before void?
Following works on my compiler.
void func1(){}
void func2(){}
void func3(){}
typedef void (*FP)();
int main()
{
const FP ar[3]= {&func1, &func2, &func3};
}
EDIT (after seeing your edits)
x.h
class x;
typedef void (x::*FP)(); // you made a mistake here
class x
{
public:
void func1();
void func2();
void func3();
static const FP array[3];
};

Answer 2

without typedef:
void (*const fp[])() = {
f1,
f2,
f3,
};

Answer 3

typedef void (*FPTR)();
FPTR const fa[] = { f1, f2};
// fa[1] = f2; You get compilation error when uncomment this line.

Answer 4

Which compiler are you using? This works on VS2005.
#include <iostream>
void func1() {std::cout << "func1" << std::endl;}
void func2() {std::cout << "func2" << std::endl;}
void func3() {std::cout << "func3" << std::endl;}
int main()
{
int ret = 0;
typedef void (*FP)();
const FP array[3] = {&func1, &func2, &func3};
return ret;
}

Answer 5

If you want the array itself to be const:
FP const a[] =
{
func1,
func2,
func3
};