array of routine pointers
Author
Hi. I have a strange question.
Is there any way to create a lis or an array of pointers to functions.
I mean, i need a list in which each member of the list is a pointer and points to a routine with all the same arguments, but different code.
Thanks
Sure.
// C++
// C
[edit]
Usage in both cases (C as well as C++): sampleFunArray[index](arg0, arg1);
[/edit]
// C++
#include <boost/function.hpp>typedef boost::function<void (int, int)> sampleFun;void foo(int x, int y) { }void bar(int apples, int oranges) { }sampleFun sampleFunArray[2] = { foo, bar };// C
typedef void (*sampleFun)(int,int);void foo(int x, int y) { }void bar(int apples, int oranges) { }sampleFun sampleFunArray[2] = { foo, bar };[edit]
Usage in both cases (C as well as C++): sampleFunArray[index](arg0, arg1);
[/edit]
Quote:Original post by darookie
typedef boost::function<void (int, int)> sampleFun;
O_O How on Earth is the boost::function template declaration done such that you can put *that* in between the angle brackets?
Quote:Original post by ZahlmanQuote:Original post by darookie
typedef boost::function<void (int, int)> sampleFun;
O_O How on Earth is the boost::function template declaration done such that you can put *that* in between the angle brackets?
That's just a regular type parameter. void (int, int) is a function type.
Quote:Original post by Polymorphic OOPBut functions aren't first-class objects in C++, I think. It must be some kind of specialized syntax. Weird.Quote:Original post by ZahlmanQuote:Original post by darookie
typedef boost::function<void (int, int)> sampleFun;
O_O How on Earth is the boost::function template declaration done such that you can put *that* in between the angle brackets?
That's just a regular type parameter. void (int, int) is a function type.
Anyway, here's how you do it without boost:
If your function is like this:
Ok, it's 3:30 AM, so it might be messed up. [wink]
If your function is like this:
int myfunction1(int a, char b) { ...}int (*pointers[42])(int,char);Ok, it's 3:30 AM, so it might be messed up. [wink]
Quote:Original post by Tron3k
But functions aren't first-class objects in C++, I think. It must be some kind of specialized syntax. Weird.
No, this requires no special syntax. All you are doing is passing a type as an argument, as I stated.
typedef void poo(int, int);template< typename a >class foo{ typedef a test;};int main(){ foo< void (int, int) > test;}should work
Since when is void poo(int, int) a proper type, sure that you don't meant void (*poo)(int, int)?
At least GCC won't let me compile that example..
At least GCC won't let me compile that example..
Quote:Original post by doynax
Since when is void poo(int, int) a proper type, sure that you don't meant void (*poo)(int, int)?
At least GCC won't let me compile that example..
void poo(int, int) isn't a proper type, void (int, int) is the proper type. The statement I provided typedefs the void (int, int) type to poo. If GCC doesn't allow it, then it is a fault with the compiler. Try Comeau or VC++ 7.1.
Another fun example that is fully standard that uses function types is:
template< void function() > // Expects a function that returns nothing and takes no parametersclass foo{public: void do_stuff() { /* ... */ function(); /* ... */ }};void bar(){ /* ... */}int main(){ foo< bar > object; object.do_stuff();}This topic is closed to new replies.
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