Typedef a container of function pointers

Question

Simple question; right now I have something like this:

typedef void(*MyFunctionPointer)(int);
typedef std::vector < MyFunctionPointer > MyFunctionPointerContainer;

However, I want to typedef this container in one row, skipping the first typedef, how can I do this?

Do you want to save one code line? Two typedefs look much better. — Alex F
– Alex F, Commented Aug 24, 2010 at 15:11

a1ex07 · Accepted Answer · 2010-08-24 15:09:54Z

10

typedef std::vector < void(*)(int) > MyFunctionPointerContainer;

answered Aug 24, 2010 at 15:09

a1ex07

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4 Comments

Billy ONeal Over a year ago

Damn you! Beat me by 4 seconds; +1 and nuking my duplicate.

John Dibling Over a year ago

+1, but I would still go with the typedef because I'd much rather write for( MyFunctionPointerContainer::iterator it = ... ) than for( std::vector < void(*)(int) >::iterator it = ...)

a1ex07 Over a year ago

typedef MyFunctionPointerContainer::iterator MyFunctionPointerIterator; would also work...

David Over a year ago

With the new C++0x auto keyword, this will be a moot point and was defined for these types of convoluted type typing.

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