Dynamic array of function pointers without typedef [duplicate]

Question

I've got this:

void (**funcs)();
funcs = new void (*)()[n];

And am getting errors about void, parenthesis, etc. What is the right syntax?

I know about vectors but I just want to know how it would be done this way.

@jogojapan I've seen that question just before asking this actually--it's just missing the second part, eg. where [n] is (or at least as far as I can decipher). — idlackage
– idlackage, Commented Feb 7, 2013 at 3:51

Charles Salvia · Accepted Answer · 2013-02-07 03:55:10Z

4

If you insist on using raw pointers for some reason, or you're just curious, the syntax is:

void (**funcs)() = new (void(*[100])());

But seriously, just use std::vector:

std::vector<void(*)()> vec;

answered Feb 7, 2013 at 3:44

Charles Salvia

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2 Comments

I don't think this is answering the OP's question.

I do have a reason, but thank you for giving the alternate method too for when I'll feel less masochistic.