How do I use "new" this way? Is it even possible?

[heebleworp]

Say I want to deal with an array of pointers to functions that take no arguments and return no arguments. I can declare an array of such pointers like so:
void (*pFuncs[5])(void);
But if I want to use new to allocate memory I get stuck:

void (**pFuncs)(void); //is this right?
pFuncs = new *what to put here?*

Is this even possible?

[Smac89]

Whenever you get the chance, use typedefs, they make things clearer

int main()
{
	typedef void(*p)();
	p * pfuncs;
	pfuncs = new p[5];
        delete [] pfuncs;
	return 0;
}

[cire]

In addition to using typedefs, use containers so you that you don't need to use new. And you don't end up with an extra level of indirection like Smac89 did. =P

#include <iostream>
#include <vector>

typedef void (*func_type)();

void func1() { std::cout << "func1\n"; }
void func2() { std::cout << "func2\n"; }
void func3() { std::cout << "func3\n"; }
void func4() { std::cout << "func4\n"; }
void func5() { std::cout << "func5\n"; }


template <typename iter_type>
void execute(iter_type beg, const iter_type& end)
{
    while (beg != end)
        (*beg++)();
}

void with_new_and_array()
{
    func_type* funcs = new func_type[5];
    funcs[0] = func1;
    funcs[1] = func2;
    funcs[2] = func3; 
    funcs[3] = func4;
    funcs[4] = func5;

    execute(funcs, funcs + 5);
    delete [] funcs;
}

void with_vector()
{
    std::vector<func_type> funcs = { func1, func2, func3, func4, func5 };
    execute(funcs.begin(), funcs.end());
}


int main()
{
    with_new_and_array();
    with_vector();
}

http://ideone.com/PIfq6i

[heebleworp]

typedef got it. Thanks.

use containers

Never heard the term "container" before, but guesing with_new_and_array() is the container in your example?

[Cubbi]

containers are classes in the C++ container library: http://en.cppreference.com/w/cpp/container

as for doing that new without a typedef, the little-known catch is that if the type-id you're new'ing includes any parentheses, it must be parenthesized:

void (**pFuncs)() = new (void(*[5])());

[Catfish666]

Cubbi wrote:

as for doing that new without a typedef, the little-known catch is that if the type-id you're new'ing includes any parentheses, it must be parenthesized:

I suspect that this is because of default-initialization versus value-initialization syntax:

int *a = new int; // default initialization, *a is garbage
int *b = new int(); // value initialization, *b is 0 

If there's another reason please clarify.

[Cubbi]

@Catfish it breaks at grammar level even without taking initializers into account: the standard's example in 5.3.4[expr.new]/4 is

new int(*[10])(); // error
is ill-formed because the binding is
(new int) (*[10])(); // error
Instead, the explicitly parenthesized version of the new operator can be used to create objects of compound
types:
new (int (*[10])());

remember, new is an operator, like sizeof or +. It can be anywhere in an expression.