C++: "expected ;" in declaration in template

Question

I've been running into the following problem inside a member function of a templated class:

#include <map>
using std::map;
template <typename A,typename B>
class C {
  public:
    B f(const A&,const B&) const;
  private:
    map<A,B> D;
};
template <typename A,typename B>
B C<A,B>::f(const A&a,const B&b) const {
   map<A,B>::const_iterator x = D.find(a);
   if(x == D.end())
     return b;
   else
     return x->second;
}

When I have g++ compile this I get the following error:

Bug.C: In member function 'B C<A,B>::f(const A&, const B&) const':
Bug.C:12: error:expected ';' before 'x'
Bug.C:13: error: 'x' was not declared in this scope

However, when I make a non-templated version of the class and function, with A and B both being int it compiles without a problem. The error is a little mystifying since I can't imagine why it wants a ';' before the 'x'.

wow, how quickly can you get 3 people to post 3 blocks of identical code? — Dave, Aug 09 '13 at 21:04
@HappyYellowFace hopefully whichever one breaks whatever pact they have and updates with an explanation. — Dave, Aug 09 '13 at 21:08
possible duplicate of [Where and why do I have to put the "template" and "typename" keywords?](http://stackoverflow.com/questions/610245/where-and-why-do-i-have-to-put-the-template-and-typename-keywords) — Praetorian, Aug 09 '13 at 21:14
BTW, naming a C++ source file `Bug.C` is just asking for trouble, and in future please tell us your compiler version. — Lightness Races in Orbit, Aug 09 '13 at 21:39
What version of gcc did you use? When I try your example, I get a very clear error: error.cpp: In member function ‘B C::f(const A&, const B&) const’: error.cpp:12:4: error: need ‘typename’ before ‘std::map ::const_iterator’ because ‘std::map ’ is a dependent scope error.cpp:12:29: error: expected ‘;’ before ‘x’ error.cpp:13:7: error: ‘x’ was not declared in this scope — Escualo, Aug 09 '13 at 21:52
@Arrieta: I just tried it on my macbook at home. It's running 4.2.1 (I know that that's pretty old, but that's what comes with xcode). I don't know what version I have at work (I can't access that computer now). But I'm glad to see that newer versions of gcc have a more comprehensible error message. — Victor Miller, Aug 10 '13 at 01:23

score 13 · Accepted Answer · edited May 23 '17 at 11:55

You're missing a typename:

typename map<A,B>::const_iterator x = D.find(a);

Please read Where and why do I have to put the “template” and “typename” keywords?. The reason you need typename here is because A and B are template parameters which means that the meaning of ::const_iterator depends on what A and B are. While to a human the name const_iterator makes it obvious that this is an iterator type, to a compiler it doesn't know if this is a type, a data member, etc.

The compiler will do a syntax check on the first pass before templates are instantiated and by adding typename you letting the compiler know to parse the map<A,B>::const_iterator as a type.

Also, there is a special rule in C++ (shamefully stolen from the linked question):

A name used in a template declaration or definition and that is dependent on a template-parameter is assumed not to name a type unless the applicable name lookup finds a type name or the name is qualified by the keyword typename.

If you do not add typename, the compiler has to assume that it is not a type.

The answer's comprehensive - I also find it interesting that the error message given in the question is so misleading. C++ compilers are known for sometimes bizarre error messages, but on my machine g++ 4.7.2 gives a far more helpful `error: need ‘typename’ before ‘std::map::const_iterator’ because ‘std::map ’ is a dependent scope` — DUman, Aug 09 '13 at 21:19

score 2 · Answer 2 · edited Aug 09 '13 at 21:38

2

You have missed the keyword typename, which is required before a qualified name that refers to a type and depends on a template parameter:

typename map<A,B>::const_iterator x = D.find(a);

edited Aug 09 '13 at 21:38

Lightness Races in Orbit

378,754
76
643
1,055

answered Aug 09 '13 at 21:04

Rahul Tripathi

168,305
31
280
331

score 1 · Answer 3 · edited Aug 09 '13 at 21:38

1

You have to add typename:

typename map<A,B>::const_iterator x = D.find(a);

Explanation:
typename states that the name that follows should be treated as a type. Otherwise, names are interpreted to refer to non-types.

edited Aug 09 '13 at 21:38

Lightness Races in Orbit

378,754
76
643
1,055

answered Aug 09 '13 at 21:03

Zagatho

523
1
6
22

C++: "expected ;" in declaration in template

3 Answers3