The constructor code#

The constructor code#

It can see the template Mењњlass, but it can't emit code for that (it's a template, not a class). So when foo.cpp is compiled, the compiler can't see bar.cpp to know that Mењњlass is needed. A template cannot be compiled into code, only the result of instantiating the template can be compiled. So a template is literally a template a class template is not a class, it's a recipe for creating a new class for each T we encounter. The intention of C++ templates is to avoid having to write nearly identical class Mењњlass_int, class Mењњlass_float, etc, but to still be able to end up with compiled code that is mostly as if we had written each version separately. That would add overhead such as boxing, needing to pass function pointers to allocators and constructors, etc.

'Instantiation-style polymorphism' means that the template Mењњlass isn't really a generic class that can be compiled to code that can work for any value of T. foo.cpp could even be compiled into a dynamic library, distributed somewhere else without foo.cpp, and linked with code they write years after I wrote foo.cpp. It's only the linker that needs to handle the entire program at once, and the linker's job is substantially easier.Д«ar.cpp doesn't even need to exist when I compile foo.cpp, but I should still be able to link the foo.o I already had together with the bar.o I've only just produced, without needing to recompile foo.cpp. The compiler does all the hard work of analysis, optimization, and code generation on each compilation unit completely independently we don't need to do whole-program analysis. Separate compilation means I should be able to compile foo.cpp independently from bar.cpp. defines the implementation of class Mењњlass.declares the interface of class Mењњlass.

Lets get a little closer to concrete for an explanation. It's because of the requirement for separate compilation and because templates are instantiation-style polymorphism. If my explanation isn't clear enough, you can have a look at the C++ Super-FAQ on this subject. You will only be able to use Foo with int or float Alternative solutionДЄnother solution is to keep the implementation separated, and explicitly instantiate all the template instances you'll need:ДЇoo.cpp // implementation of Foo's methods

This way, implementation is still separated from declaration, but is accessible to the compiler. tpp), and include this implementation file at the end of the header. If these implementations were not in the header, they wouldn't be accessible, and therefore the compiler wouldn't be able to instantiate the template.ДЄ common solution to this is to write the template declaration in a header file, then implement the class in an implementation file (for example. Void doSomething(T param) Д¬onsequently, the compiler needs to have access to the implementation of the methods, to instantiate them with the template argument (in this case int). If for a class C, you have multiple fields X, Y, Z, etc.Caveat: It is not necessary to put the implementation in the header file, see the alternative solution at the end of this answer.ДЄnyway, the reason your code is failing is that, when instantiating a template, the compiler creates a new class with the given template argument. In case of parameterized constructor, you can use following syntax to initialize the fields в€’ДЄbove syntax is equal to the following syntax в€’

Using Initialization Lists to Initialize Fields When the above code is compiled and executed, it produces the following result в€’ Member functions definitions including constructorД¬out << 'Object is being created, length = ' << len << endl Д¬out << 'Length of line : ' << line.getLength() <len) // This is the constructor