Visual C++ 2012: compiler complains about "missing ';'"

Question

Could you please point out to me what is wrong with the following template class?

#include <vector>

template <typename T, typename C>
struct pQueue{
    pQueue():currEnd(c.end()){};
    ~pQueue(){c.~vector();}
    void insert(T& t);
    void remove(T& t);
    bool find(T& t);
    T head(void);
private:
    std::vector<T> c;
    std::vector<T>::iterator currEnd;
};

The compiler isn't very happy about the std::vector<T>::iterator currEnd; line, and produces the following error messages:

error C2146: syntax error : missing ';' before identifier 'currEnd'

error C4430: missing type specifier - int assumed. Note: C++ does not support default-int

Thank you very much!

[Relevant](http://stackoverflow.com/questions/610245/where-and-why-do-i-have-to-put-the-template-and-typename-keywords). By the way, GCC and Clang give extremely straightforward errors for this. — chris, Aug 27 '13 at 13:02
Luckily, gcc suggests the solution: http://ideone.com/tUQz2B — mfontanini, Aug 27 '13 at 13:03
Not related to the errors, but you are explicitly invoking a destructor from `~pQueue()`. Why? Since `c` is an actual vector and not a pointer, its destructor would run anyway. — Frédéric Hamidi, Aug 27 '13 at 13:04

Walter · Accepted Answer · 2013-08-27T13:19:51.240

3

the compiler doesn't recognise std::vector<T>::iterator as a type. You have to tell it:

typename std::vector<T>::iterator currEnd;

typename is used as disambiguitator in templated context. It just tells the compiler that the identifier std::vector<T>::iterator is a type rather than a static member. The standard requires this in the templated context, even though in most cases the compiler could work that out even before the template parameter T has been specified.

While the above answers your question, I can only emphasize what juanchopanza has pointed out about your code. Besides, there is std::queue for you already.

edited Aug 27 '13 at 13:19

answered Aug 27 '13 at 13:02

Walter

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It compiles indeed. But what exactly does "typename" keyword do in this context? – Einheri Aug 27 '13 at 13:04
@user2531913 It tells the compiler that vector::iterator is a type, and not a static member variable. – Kaz Dragon Aug 27 '13 at 13:17
It's a priority queue though. I can use the priority_queue, but I can't seem to figure out how to take things off it; I can also use std::set, but I guess I'd rather implement a proper one myself which does exactly what a priority queue is supposed to do. – Einheri Aug 27 '13 at 13:25
@user2531913 `std::priority_queue` has members `top()` to access the top element and `pop()` to remove it. `push()` and `emplace()` insert a new element (and sort the queue in log(N) time). You cannot remove others than the top element. – Walter Aug 28 '13 at 17:09
@Walter It's A* search, wouldn't I need to sometimes re-evaluate the same node and, if it turns out to have a shorter distance to the starting node than before, take it off the queue and insert it again? – Einheri Aug 29 '13 at 07:13

juanchopanza · Answer 2 · 2013-08-27T13:29:26.327

1

You need to tell the compiler the iterator is a type name, because it is a dependent name:

typename std::vector<T>::iterator currEnd;

Besides that, this class does not need a user-provided destructor. Your class definition could be simplified to this:

template <typename T>
struct pQueue
{
    pQueue() : currEnd(c.end()) {};
    void insert(const T& t);
    void remove(const T& t);
    bool find(const T& t);   // this should probably be a const method
    T head();                // this should probably be a const method
private:
    std::vector<T> c;
    typename std::vector<T>::iterator currEnd;
};

edited Aug 27 '13 at 13:29

answered Aug 27 '13 at 13:03

juanchopanza

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I want to use std::remove, so that I can take things out of the priority queue without actually freeing them while keeping track of where the end of queue actually is. Thanks for pointing out the consts, I have added them in. – Einheri Aug 27 '13 at 13:21
@user2531913 Oh I see, so you keep an `end` iterator that isn't necessarily the end of `c`? I think I might have done something similar with a queue once... – juanchopanza Aug 27 '13 at 13:23

score 0 · Answer 3 · edited Aug 27 '13 at 13:45

0

I think it should be typename std::vector<T>::iterator currEnd;.

Then the ~~compilcater~~ ~~compilator~~ compiler can figure out that iterator is actually a type and not a member (of std::vector).

edited Aug 27 '13 at 13:45

user1233963

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answered Aug 27 '13 at 13:03

0x26res

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The visual studio compiler is indeed a 'complicater' :) – user1233963 Aug 27 '13 at 13:04

score 0 · Answer 4 · answered Aug 27 '13 at 13:29

Just by way of addendum to the other answers:

Sadly, VisualStudio doesn't have very good error messages. You pretty much have to learn what they mean, rather than read them literally. For example:

missing ';' before identifier 'currEnd'

Is VisualStudio-ese typically means "I don't know what the identifier before currEnd is." Usually this is due to it being undefined at that point. 9 times out of 10, it will be a type/class name for a type/class that you neglected to #include the header for. In this case it looks like it was more a matter of the compiler getting confused with a template use in a template.

However, the principle stands. Had you started with, "The compiler doesn't know what the type/class before currEnd is.", you would have been on the right track.

Visual C++ 2012: compiler complains about "missing ';'"

4 Answers4