Question

I am new to c++ programming and know the basics of c++. I am trying to link three files, but i am unable to get the output.I am studying the book c++ cookbook right now

Suppose we have the files a.hpp a.cpp, b.hpp b.cpp,c.hpp c.cpp with the following code:

a.hpp

#ifndef A_HPP_INCLUDED
#define A_HPP_INCLUDED
void a();
#endif

a.cpp

#include "a.hpp"
#include <iostream>
void a()
{
    std::cout<<"a \n ";
}

b.hpp

#ifndef B_HPP_INCLUDED
#define B_HPP_INCLUDED
void b();
#endif

b.cpp

#include "b.hpp"
#include <iostream>
void b()
{
    std::cout<<"b \n ";
}

c.hpp

#ifndef C_HPP_INCLUDED
#define C_HPP_INCLUDED
void c();
#endif

c.cpp

#include "a.hpp"
#include "b.hpp"
#include "c.hpp"
void c()
{
    a();
    b();

}

int main()
{
    c();
    return 0;
}

I have created all the files in one folder and commands i used to compile and link them are

$:g++ -c -Wall a.cpp b.cpp c.cpp
$:g++ -o -Wall a.o b.o c.o
$:./a.out

I was expecting the outout

a
b

but there was no output at all.Request you all to help me out with this.

Answer 1

Remove the -o from your g++ command line.

You are currently telling g++ to link your objects together to a file called -Wall. So the alternative solution would be to call ./-Wall instead of ./a.out.

Due to the spaces in your strings (" \n " vs. "\n"), you’ll still not be getting the exact expected output. Also you will probably want to replace:

std::cout << "something \n";

with

std::cout << "something" << std::endl;

if you want to have the data on screen immediately (flushing). See also C++: “std::endl” vs “\n”.