Retrieving function pointers from container

Question

I think this will make more sense if I put the code first:

I have a member variable defined as such:

std::queue<void (*)()> fptrs;

...and am trying to pop a pointer back out of it like this:

void (*f)() = fptrs.pop();

The 'alert' I get is this:

Assigning to 'void (*)()' from incompatible type 'void'

It seems that I'm not putting function pointers into my queue, or they're popping back out magically transformed. Either way, I'm not sure how to fix this, and I've fiddled with it for a bit.

Thanks for the help.

juanchopanza · Accepted Answer · 2013-06-08 09:10:48Z

1

std::queue::pop() returns void. You need a call to std::queue::front(), followed by one to std::queue::pop().

I would suggest using a typedef for the function pointer, for the sake of readability.

answered Jun 8, 2013 at 9:10

juanchopanza

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2 Comments

Tyler Over a year ago

I knew I'd feel stupid when this got answered. Thanks a lot, I'll read the docs better next time!

juanchopanza Over a year ago

@Tyler The interface is a bit non-intuitive. If you're interested, it is like this for reasons of exception safety.

BlackMamba · Accepted Answer · 2013-06-08 10:25:08Z

0

#include <iostream>
#include <queue>
using namespace std;

typedef void (*pFun)();

void fun1(){ printf("fun1\n"); }
void fun2(){ printf("fun2\n"); }
int main()
{
    queue<pFun> d;
    d.push(fun1);
    d.push(fun2);

    pFun fun = d.front();
    fun();
}

answered Jun 8, 2013 at 10:25

BlackMamba

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