Function template linking error

Question

I've created a function template that allows me to get data for any data type but am receiving the error message on compilation:

Undefined symbols for architecture i386:
  "bool Json::getData<double>(double, Json&, std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&, DataType)", referenced from:
      Coupon::initCoupon(int const&, Json&)in libkuapay.a(Coupon.o)
ld: symbol(s) not found for architecture i386
collect2: ld returned 1 exit status
scons: *** [kuaposgw] Error 1
scons: building terminated because of errors.

The function is declared as:

 template < class T> static bool getData(T data, Json &jsonObject, const string &key, DataType dataType);

and called as:

 Json::getData (couponList[cpnCnt].discount, couponReader, "discount", realType);

where couponList[cpnCnt].discount is a double.

The code itself compiles fine in my "inner" directory but I get the error message above in the "outer" directory, where the latter is essentially a wrapper of the inner code.

Can you include the function definition? The error suggests something is wrong with it. — kichik, Dec 27 '11 at 18:43
`ld` is the linker, not the compiler. So when it complains that means you have a linker error, not a compiler error. — Omnifarious, Dec 27 '11 at 18:50

score 2 · Answer 1 · answered Dec 27 '11 at 18:44

2

The current state of templates generally requires that you have the function definition right there where you have the function declaration.

The way templates work, the compiler basically manufactures a custom version of your function for each variation on template arguments. Since the compiler can't know in advance what all those different template arguments will be (will it be int or double or some unknown type that's been declared in some other file?) it can't create those versions until the function is called.

This means that the whole function definition has to be available to the compiler when you call the function. In order to make that happen you should put the function definition in a header.

There are other ways of doing this. Explicit instantiation for class templates. Declaration of an overload in which there are no template arguments for a function. But in general, your whole template definition has to be in a header file.

answered Dec 27 '11 at 18:44

Omnifarious

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Hi thanks for the response, I'll try adding the function definition to the header. In the mean time, here's the definition: – jdeckman Dec 27 '11 at 19:01
template bool Json::getData(T data, Json &jsonObject, const string &key, DataType dataType) { if (!jsonObject.hasKey(key)) return false; if (jsonObject.value(key) == json_spirit::null_type) return false; if (dataType == realType) { if (jsonObject.value(key).type() == json_spirit::str_type) data = NumberParser::parseFloat(jsonObject.value(key).get_str()); else data = jsonObject.value(key).get_real(); } return true } – jdeckman Dec 27 '11 at 19:08
@user1118089: Well, that looks fine, though it'd be better if you updated your question with the definition and say which file it's in. – Omnifarious Dec 27 '11 at 22:35

score 1 · Answer 2 · answered Dec 27 '11 at 19:54

Templates are not instantiated automatically in C++, but rather when used implicitly or instantiated explicitly. You can trigger the former case by using the function when the template instantiation is available when the template is used (e.g. by putting it in the header file), like @Omnifarious described.

As an alternative, you can make the function non-static and explicitly instantiate it in a source file:

template bool getData<double>(double data, Json &jsonObject, const string &key, DataType dataType);

Function template linking error

2 Answers2