ld symbol not found - c++

Question

Looked around and found a few similar questions but none of them were the same. Most had to do with the constructor or destructor. This issue is, more than likely, a result of my rusty C++ linker memory (picking it back up after a few years).
I'll keep it real simple since this is probably a basic misunderstanding of the linker:
data.h
#pragma once
namespace test {
class Data_V1 {
public:
// some getters/setters
int getData() { return _d; }
void setData( int d ) { _d = d; }
private:
// some data
int _d;
};
}
builder.h
#pragma once
namespace test {
template <class V>
class Builder {
public:
void build();
};
}
builder.cpp
#include <iostream>
#include "builder.h"
namespace test {
template<class V>
void Builder<V>::build() {
std::cout << "Insert building logic" << std::endl;
}
}
main.cpp
#include "builder.h"
#include "data.h"
using namespace test;
int main(int argc, char* argv[]) {
Builder<Data_V1> b;
b.build();
}
compiling:
g++ -Wall -ansi -pedantic -c builder.cpp
g++ -Wall -ansi -pedantic -c main.cpp
g++ -Wall -ansi -pedantic -o main main.o builder.o
Link error:
Undefined symbols for architecture x86_64:
"test::Builder<test::Data_V1>::build()", referenced from:
_main in main.o
ld: symbol(s) not found for architecture x86_64
collect2: ld returned 1 exit status
Any help would be appreciated!

Answer 1

Template definitions need to be visible to all translation units. Move the definition from the cpp to the header.
Builder.h
#pragma once
namespace test {
template <class V>
class Builder {
public:
void build();
};
template<class V>
void Builder<V>::build() {
std::cout << "Insert building logic" << std::endl;
}
}
Before you ask, no, there's no way to hide the implementation unless you know all possible specializations beforehand.
Templates represent a generalized form for the creation of a new class. If the implementation is not visible, when you attempt to specialize the template, the compiler can't know what code to generate.