How to use structs, typedef, and template?

Question

I'm trying to implement the following code, but I keep getting an error because of typedef, can someone please help?

 template<class T>
 struct box{
 T data;
 box *link;
 };
 typedef box* boxPtr;

the error is:

Use of class template 'box' requires template arguments

It's a short snippet, but there is still a fair bit to unpack here. You may be better served by choosing and working through one of the books [here](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). — StoryTeller - Unslander Monica, Nov 09 '20 at 17:52
an answer to your title can fill several chapters of a book. Please try to be more specific. What exactly do you want to achieve? And what is the error? — 463035818_is_not_an_ai, Nov 09 '20 at 17:55
ok, now, what part of the error message do you not understand? `box` is a template, not a type — 463035818_is_not_an_ai, Nov 09 '20 at 17:55
I don't understand how I would fix it, I've tried typedef... but still the same — harry, Nov 09 '20 at 17:56
what is the aim? You want to declare a pointer to a box that has what type as data member? — 463035818_is_not_an_ai, Nov 09 '20 at 17:57
or do you want a template alias? Ie a template that you can instantitate to get the type of pointers to different `box` instantiations? — 463035818_is_not_an_ai, Nov 09 '20 at 17:57
what type should that box have as member? `T` is just a place holder that only existst in the template. To get a concrete pointer you need to choose what `T` should be — 463035818_is_not_an_ai, Nov 09 '20 at 17:58
if I remove typdfef, how can I in a class say define boxPtr as a type? — harry, Nov 09 '20 at 17:59
@harry Just don't define boxPtr as a type. You don't need it. Problem solved. — eerorika, Nov 09 '20 at 18:00
Side note: Usually a type alias is used to reduce the amount of typing necessary (`boxPtr` longer than `box*`) or to simplify and clarify (`boxPtr` is no clearer than `box*`) I don't see any gain from `boxPtr`. — user4581301, Nov 09 '20 at 18:03
you dont want `box` to hold data of arbitrary type? do you? From some of your comments I get the impression that you confuse templates with javas generics, which is a completely different story. I might be completely wrong, tough — 463035818_is_not_an_ai, Nov 09 '20 at 18:04

score 2 · Answer 1 · edited Nov 09 '20 at 18:04

2

Writing

template<class T>
struct box{
    T data;
    box *link;
 
    typedef box<T>* boxPtr; 
};

with

int main()
{
    box<int>::boxPtr use_of_a_bad_idea;
}

is one way, but having pointer types masquerading as object types is a recipe for memory leaks: don't do it as a rule of thumb. That's the best way of addressing this problem.

edited Nov 09 '20 at 18:04

cigien

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answered Nov 09 '20 at 17:59

Bathsheba

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2

*"having pointer types masquerading as pointer types"* - Ah... um... – StoryTeller - Unslander Monica Nov 09 '20 at 18:02
Not so long ago, I misread a question and wrote it off as a trivial matter of pass by value where pass by reference was required because the asker used a typedef to conceal that the type *was* a pointer. I could have just as easily given bad advice if I hadn't read through the code several times. the `Ptr` in `boxPtr` makes the pointer orders of magnitude more obvious than the other lost question, but still not as obvious as `box *` for anyone who's brain is conditioned to look for the `*`. – user4581301 Nov 09 '20 at 18:15
@StoryTeller-UnslanderMonica: Had a delivery today which required an excavator to make lots of turns over my once-pristine lawn. It now looks like a MOD training ground. Very stressful indeed. Hence the odd mistake or two creeping in. – Bathsheba Nov 09 '20 at 21:35

score 2 · Accepted Answer · 2020-11-10T21:50:58.600

2

box is a class template. This means that it needs an argument whenever you instantiate it:

box<int>   integerBox;
box<float> floatBox;
// etc

This is required anywhere you try to use a template class such as box. So a pointer to just a box doesn't make sense. It needs a type with it. For example:

template <class type>
using boxptr = box<type>*;

This does effectively the same thing as your typedef, but it allows you to specify he underlying template type.

Of course, if you do it this way, you will always need to specify a type when you use the pointer version:

boxptr<int>   integerBoxPtr;
boxptr<float> floatBoxPtr;
// etc

edited Nov 10 '20 at 21:50

answered Nov 09 '20 at 18:03

@harry You're welcome. Please note that on this site, the best way to thank someone is to up-vote their answer and/or accept it. We try to avoid comments in general on this site except when absolutely necessary. Thank you comments are usually not considered absolutely necessary--again we usually prefer you to up-vote and accept the best answer instead. – Nov 09 '20 at 22:58

score 1 · Answer 3 · answered Nov 09 '20 at 18:01

1

Use using:

template<class T>
struct box
{
    T data;
    box *link;
};

template <typename T>
using boxPtr = box<T>*;

Now boxPtr<int> is the same as box<int>*.

answered Nov 09 '20 at 18:01

score 0 · Answer 4 · answered Nov 09 '20 at 17:59

0

This will compile:

template<class T>
struct Box {
    T data;
    Box<T> *link;
};

typedef Box<int> * BoxPTR;

answered Nov 09 '20 at 17:59

Joseph Larson

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I recommend you highlight or clarify the difference between your code and the OP's. – Thomas Matthews Nov 09 '20 at 18:05
1

Remember that compiling is typically only the beginning. – user4581301 Nov 09 '20 at 18:15

How to use structs, typedef, and template?

4 Answers4