Array of function pointer pointers

Question

If I want an array of pointers to something, I declare it like this:

Type** var = new Type*[8];

and use it like this:

if(var[0] != NULL)
    // Do something

But how can I have an array of function pointers in a similar fashion? Something like this maybe:

typedef bool (*Handler)(int, int);
Handler** list = new Handler*[8];

...

Handler* func = list[0];
if(func != NULL)
    *func(6, 5);

SeedmanJ · Accepted Answer · 2013-04-29 05:53:19Z

2

You would have:

typedef    bool (*Handler)(int, int);

Handler* list = new Handler[8];

Handler func = list[0];
if (func != NULL)
    func(6, 5);

Just don't put an extra *

answered Apr 29, 2013 at 5:53

SeedmanJ

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3 Comments

user600003 Over a year ago

I love you. Now I feel silly... (will accept answer when SO allows me)

SeedmanJ Over a year ago

Function pointers syntax can be sometimes confusing, methods pointers is another funny thing :D

john Over a year ago

Nitpick but you can put in the * you just have to bracket it correctly, (*func)(6, 5);. This was the original C syntax (which is why I still find myself using it) but at some point the * became optional.

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