带类的库如何工作？

Question

When learning C++ in school we never really talked about how to build libraries, so sorry for my rudimentary understanding. From what I've read online, it seems like a library is just a collection of code that is already compiled, and then there is a .h file that lists what functions are accessible in that library.

For example when I #include <cmath> I can now call sin(x) without having access to the cmath code to compile it. My question is if this works with classes that have data in them.

So can I create a library

//AccumulatorLibrary.h
class Accumulator
{
public:
    int num;
    int increment() {num++};
    void otherFunctions(); //otherFunctions defined in the .lib file
}

And then call it

//Main
#include "AccumulatorLibrary.h"
#include <stdio>
int main()
{
    Accumulator A(0); //initalize num to 0
    Accumulator B(7); //initalize num to 7
    cout<<A.increment;
    cout<<B.increment;
    cout<<A.increment;
}

and get an output of 1 8 2 ?

In summary, if I figure out how to put a bunch of classes into a library file can I access any data I want to, as long as that data has an access function in the .h file?

Or a more basic question, do a .h and .lib file work exactly the same as regular c++ code except that it doesn't have to be compiled when you use it, and you don't have access to the code in the .lib file?

Answer 1

From what I've read online, it seems like a library is just a collection of code that is already compiled, and then there is a .h file that lists what functions are accessible in that library.

Correct.

My question is if this works with classes that have data in them.

It does. A lot of C++ libraries expose classes and have their code precompiled in a library.

Or a more basic question, do a .h and .lib file work exactly the same as regular c++ code except that it doesn't have to be compiled when you use it...

Wait, wait. .h files still contain C++ code (declarations and sometimes even inline implementations). .lib files are dynamically linked libraries. They're the result of the compilation (and linkage) of the C++ source files.

...and you don't have access to the code in the .lib file?

You do have access to it: open it using a disassembler. It just won't be C++ anymore.

Answer 2

From what I've read online, it seems like a library is just a collection of code that is already compiled

Yes and no.

To me, a "library" is a body of code (one or more header ( .h) files and zero, one, or more source ( .cpp ) files), without the main() function, that can be independently compiled and linked (except for main() ).

A library may be available in various mechanisms:

• Available as source: Here you have to compile the library along with your application. Examples are C++ standard template library, Boost C++ libraries, a library that you or your colleague wrote, a library that you downloaded from sourceforge etc. (Caveat: Sometimes source-based libraries can be pre-compiled in a system as a compilation optimization.)

  Note that, being available as source means you can read it, but not necessarily modify it.

• Available as binary: Here, the library is already compiled and perhaps available in your system. Examples are C standard library, C++ standard library, C math library etc. From the question, it sounds like that this is the one you are referring to.

then there is a .h file that lists what functions are accessible in that library.

True, that is the case for a C library. For C++, the concept is naturally extended to include the classes and (public) member functions.

In summary, if I figure out how to put a bunch of classes into a library file can I access any data I want to, as long as that data has an access function in the .h file?

Yes!

Or a more basic question, do a .h and .lib file work exactly the same as regular c++ code except that it doesn't have to be compiled when you use it, and you don't have access to the code in the .lib file?

A library follows the same rules of C++, the only difference as mentioned above, is that it does not have a main() function.

Whether you need to compile or not depends on the way it is available to you (see above).

For many libraries, you have access to the source code (see above).

Following is the full code for your example:

// AccumulatorLibrary.h
class Accumulator {
  public:
    Accumulator( int x ) : num( x ) {}    // ctor with initializer
    int increment() {num++};
    int get() const;
    void set( int x );
  private:
    int num;
};
// AccumulatorLibrary.cpp
int Accumulator::get() const { return num; }
int Accumulator::set( int x ) { num = x; }

// Usercode.cpp
#include "AccumulatorLibrary.h"
#include <iostream>
using namespace std;
int main() {
    Accumulator A(0); //initalize num to 0
    Accumulator B(7); //initalize num to 7

    A.increment();
    cout << A.get() << endl;        // print 1

    B.increment();
    cout << B.get() << endl;        // print 8

    A.increment();
    cout << A.get() << endl;        // print 2
}