I got the following error:
features.cpp:(.text+0x4f6): undefined reference to `void Convolve<float>(CImageOf<float>, CImageOf<float>&, CImageOf<float>)'
So, in my features.cpp file, I added the following header file, with the content:
template <class T>
void Convolve(CImageOf<T> src, CImageOf<T>& dst,
CFloatImage kernel);
I'm compiling this with GCC on ubuntu 64bit. I realize that the error above is because what is in my header does not match the reference, but that's because of the template, and so I shouldnt be having this problem right? What else is going on?
I double checked my calls to convolve, and actually this is the types I'm using :
Convolve(CFloatImage, CFloatImage, CFloatImage);
I can't actually find any reference to the error above
If you put a function template into a .cpp file and use it in a different translation unit, you need to explicitly instantiate the template, e.g.:
template
void Convolve<float>(
CImageOf<float> src,
CImageOf<float>& dst,
CFloatImage kernel);
Basically, the compiler can instantiate a templte only when it has seen its definition. If you want the template to be instantiated automatically, you need to define it in a header.
As turned out in a chat with the OP the problem was probably in a Makefile that did not link correctly.
There was no definition but a declaration of the template in the .h file (as some of you suggested). I think the OP confused the terms.
However there was a definition of the template in the .cpp file along with a (non template) function definition that instantiated Convolve<float> (and others).
So to repeat it: The code was correct but the build routine was not.
Related
I'm currently writing a logger for my C++ project (I need it to have near to no dependences, so I don't want to take a prewritten one). It contains this function:
template <typename T>
Logger& Logger::operator<<(T f) {
m_file<<f;
return *this;
}
The logger.cpp code compiles, but when I call the log functions in my main.cpp, I have this compiler error:
/home/tuxer/prog/cpp/PRay/server/src/main.cpp:110: undefined reference to `Logger& Logger::operator<< <int>(int)'
for this line of code :
log<<lul; (lul being a int variable equals to 2)
The logger.o file is correctly included, as the Logger::init() function works properly and doesn't raise any linking error.
Thanks :)
Since you have non-inline templates, you need to force instantiation. See for example How do I force a particular instance of a C++ template to instantiate?.
The simple thing to do is to put the Logger::operator<< template in the header file. The compiler will automatically instantiate the versions it needs, and the linker will remove the duplicates (well, at least the ones that weren't inlined).
You shouldn't need to force instantiation unless your linker is old (e.g. gcc 2.7 or earlier).
This is the general rule with template code: put the definitions in the header file unless you have a good reason not to.
See Why can templates only be implemented in the header file? as well.
I try to make a generic function in C++, that creating an array of a type which depends on usage (when calling it), like array of int or char.
This is the code in my header file:
template<class T>
T** makeArray(...);
This is in the .cpp file:
template<class T>
T** ClassA::makeArray(...){
//...
}
And this is how I call it in the main file:
char** charArr = cg.makeArray<char>(...);
//...
int** intArr = cg.makeArray<int>(...);
But when compiling, I get these errors:
undefined reference to `char** ClassA::makeArray<char>( ...
undefined reference to `int** ClassA::makeArray<int>( ...
What do the errors mean and how do I fix it?
You have to include the entire body of a C++ template function in the header.
This is because of the way that C++ templates work: template parameter substitution happens only when you actually use the template, and the entire template definition has to be visible at that time. Another way of looking at it is to say that you cannot compile and link to a templateFunction<T>—you can only compile a template function once it's been instantiated as a templateFunction<char> or whatever. For this reason, you cannot put template function definitions in .cpp files, but rather have to put them in .h files.
The compiler needs to have access to the entire template definition (not just the signature) in order to generate code for each instantiation of the template, so you need to move the definitions of the functions to your header.
This is called the inclusion model. Also read this FAQ
In general, you cannot have template definitions in a source file; they should be in the header file so that the compiler can see them.
This comes up over and over again here, so I can't be bothered to give the full explanation again; I recommend you read this from the C++ FAQ: Why can't I separate the definition of my templates class from its declaration and put it inside a .cpp file?.
[Incidentally, why are you returning a pointer-to-pointer? A pointer should be enough...]
If your code is using the template on a diferent cpp file, the template should go on the header file.
I'm using Code::Blocks to build my project, which contains three files: main.cpp, TimeSeries.cpp, TimeSeries.h. TimeSeries.h provides declarations for the TimeSeries class as follows:
template<class XType, class YType> class TimeSeries {
public:
TimeSeries(void);
~TimeSeries(void);
};
Then TimeSeries.cpp contains:
#include "TimeSeries.h"
template<class XType, class YType>
TimeSeries<XType, YType>::TimeSeries(void) {
}
template<class XType, class YType>
TimeSeries<XType, YType>::~TimeSeries(void) {
}
And finally, main.cpp contains
#include "TimeSeries.h"
typedef TimeSeries<float, float> FTimeSeries;
int main(int argc, char** argv) {
FTimeSeries input_data;
return 0;
}
When building with C::B, I get the following error:
undefined reference to `TimeSeries<float, float>::TimeSeries()'
What can I do?
Thanks,
CFP.
Basically all templated code should be defined in a header, otherwise it will not be built since nothing uses it in the compiled unit.
Each cpp file is compiled as a separate unit, and thus constructor and destructor are not compiled. The compiler has no way of knowing what type of template argument you will use in main.cpp when it compiles TimeSeries.cpp.
The reason for splitting code into header- and source-files is so that the declaration and the implementation are separated. The compiler can translate the source-file (compilation unit) into an object file, and other compilation-units that want to use the classes and functions just include the header-file, and link the object file. This way, the code has to be compiled only once, and can be reused by linking it.
The problem with templates is that, as long as there are no parameters provided for them, the compiler cannot compile them. The same template instantiated with different parameters results in different types. std::vector<int> and std::vector<float> are, from the compilers perspective, not related in any way. Because of this, template-classes usually have to reside completely in a header-file, because, when the template is used, the compiler needs the complete definition in order to generate the class depending on the parameters.
As #Gabriel Schreiber pointed out in his answer, you can tell the compiler that he should compile the template with a specific set of parameters, making that available to other compilation units just by linking. However, this does not make the template available for other parameter sets.
You need to add this in your .cpp-file (below the definitions):
template class TimeSeries<float, float>;
When the compiler compiles TimeSeries.cpp it doesn't know which for which types the template is need because it is used in another source file. You need to tell the compiler explicitly.
Read about Explicit Template Instantiation in your copy of the Stroustrup or on the internet.
Template code is not compiled until the template function is used. But where does it save the compiled code, is it saved in the object file from which used the template function in the first place?
For example,
main.cpp is calling a template function from the file test.h, the compiler generates an object file main.o,
Is the template function inside the main.o file? because template code is not inlined, is it?
It's totally compiler implementation dependant. Most compilers will generate code around, inline or in cpp-like files and then compile with that. Sometimes, with optimization setup, some compilers will even reuse the same code instead of recreate it for each cpp.
So you have to see your compiler's doc for more details.
Yes, the template function code is emitted in the main.o file. Some of it may be inlined, as any other code may be inlined, but in general, code for templates is emitted in any file in which the template is instantiated. Think of it as first instantiating the template code to produce normal non-template functions, and then compiling those functions with whatever inlining and optimization the compiler applies to any other function.
When the same template instantiation (e.g. std::vector<int>) occurs in multiple compilation units (.cpp files), there is a bit of difficulty, because the code is instantiated in each of them. There are various ways of handling this, sometimes involving a cleanup step in the linking phase where duplicate template instantiations are resolved into a single one; your compiler manual can provide more information on exactly how it handles that situation.
Template code is compiled even if it is never instantiated. Otherwise, compilers couldn't be required to emit a diagnostic for this:
template< typename T >
void f()
{
blah
}
Template compilation happens in two phases. Besides the basic checks, everything that depends on template parameters can only be checked when a template is instantiated and the formal parameters are filled in with actual types. For example, here
template< typename T >
void f()
{
typename T::nested_type x;
}
T::nested_type can only be checked after the template is instantiated and an actual type is given for T. However, a basic check ("given T::nested_type is a type, is this a valid variable definition?") is performed the moment the compiler encounters the template's definition. (That's why that typename is required, BTW. Depending on T, T::nested_type might just as well be the name of member of T or a static data member - which would make T::nested_type x; a syntax error. So we have to tell the compiler to treat T::nested_type as the name of a type.)
You mean instantiated, not compiled. At compile time the compiler finds out each and every version that your code uses and instatiates (in object files) all the required versions.
It is always inlined (meaning, it is always internal linkage, having inline semantics). It may in fact be not inlined after all, just as an inline function, however, template is not code. It is a "template for making code". Therefore, it will normally reside in a header, except special cases, see below.
There was an idea to make something else, codenamed "export keyword". It was removed from standard.
Special cases: you can compile template instantiations into an object file, without having them used. This is the only way to avoid having all template code inlined.
This is how it is done:
template class std::vector<MyClass>;
This will force the compiler to instantiate a template in the current location. C++0x will have a syntax to force compiler not to do it, and have the linker search for template instantiation elsewhere:
extern template class std::vector<MyClass>; // C++0x only
I have a function declared like so:
template <typename T>
T read();
and defined like so:
template <typename T>
T packetreader::read() {
offset += sizeof(T);
return *(T*)(buf+offset-sizeof(T));
}
However, when I try to use it in my main() function:
packetreader reader;
reader.read<int>();
I get the following error from g++:
g++ -o main main.o packet.o
main.o: In function `main':
main.cpp:(.text+0xcc): undefined reference to `int packetreader::read<int>()'
collect2: ld returned 1 exit status
make: *** [main] Error 1
Can anyone point me into the right direction?
You need to use the export keyword. However, I don't think G++ has proper support, so you need to include the template function's definition in the header so the translation unit can use it. This is because the <int> 'version' of the template hasn't been created, only the <typename T> 'version.'
An easy way is to #include the .cpp file. However, this can cause problems, e.g. when other functions are in the .cpp file. It will also likely increase the compile time.
A clean way is to move your template functions into its own .cpp file, and include that in the header or use the export keyword and compile it separately.
More information on why you should try and put template function definitions in its header file (and ignore export altogether).
The problem is that a function template is not a function. It's a template for creating functions as needed.
So for a template to work, the compiler intuitively needs two pieces of information: The template itself, and the type that should be substituted into it.
This is unlike a function call, which the compiler can generate as soon as it knows that the function exists. It doesn't need to know what the function does, just that it looks like void Frobnicate(int, float), or whatever its signature is.
When you declare the function template without defining it, you're only telling the compiler that such a template exists, but not what it looks like. That's not enough for the compiler to be able to instantiate it, it has to be able to see the full definition as well. The usual solution is to put the entire template in a header that can be included where needed.
The best practice with template functions is to define them in header files. They are created at compile time so compiler has to have definition around to do so.
When export for templates would be more supported this wouldn't be the case though but right now it still hardly can be used.
Is their any compiler support template separate compilation?
As I know the common practice is declare and implement template functions in the header file