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How to interpret a typedef <>

Posted on 2013-06-25
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Last Modified: 2013-06-28
I have structure

struct ace {
     uint32_t type;
     uint32_t flag;
}

typedef ace  ace_attr<>

What does <>  mean? Can someone please illustrate with an example
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Question by:perlperl
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2 Comments
 
LVL 86

Accepted Solution

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jkr earned 500 total points
ID: 39275074
This defines 'ace' as type for the template specialization 'ace_attr<>' of the template class 'ace_attr<typename T>' - you should find a declaration like

template<typename T> 
class ace_attr {

  // ...
};

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somehere in your code as well. See also http://en.wikipedia.org/wiki/Template_(C%2B%2B)#Explicit_template_specialization and http://www.cprogramming.com/tutorial/template_specialization.html
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LVL 40

Expert Comment

by:evilrix
ID: 39275638
>> This defines 'ace' as type for the template specialization...
Indeed, that is the most logical conclusion; however, for that to be the case the parameters need to be swapped.

eg.

typedef ace_attr<> ace;  

perlperl, did you copy/paste that example or type it in? If the latter, is it possible you've transposed the parameters? If so then jkr is spot on and you should accept his answer. If not it would help if you could actually copy and past the code and the context in which it's used.

Additional (please don't award this any points it just buils on what jkr has already said):

The ace_attr<> is just referencing a template type that has a default value defined. In your code you'll have something like...

template<typename T = <sometype>>
class ace_attr { };

Note, this is what jkr suggested and all I've done is add the default parameter. This allows you to instantiate the template without a type because the default will be used. This is not too dis-similar to default arguments for function parameters.

void foo(int n = 1) {}

This can be called as such...

foo();

It's basically the same thing but with templates rather than function calls.

Cheers.
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