Can someone understand that what 1, 2, 3 is? And what {line l(5); } is in this program? Please, help me :)

Question

0.00/5 (No votes)

See more:

If someone know,let me know ,please :) Thanks in advance :)
Can someone understand that what 1, 2, 3 is? And what {line l(5); } is in this program? One more question that : *p=*obj.p what does this mean? and * means dereference?

C++

#include<iostream>
using namespace std;
class Line{
public:
	int getLength(void);
	Line(int len);//constructor
	Line(const Line&obj);//copy constructor
	~Line();//destructor
private:
	int *p;
};

Line::Line(int len)
{
	cout<<"Simple constructor"<<endl;
	p=new int;
	*p=len;
}
Line::Line(const Line&obj)
{
	cout<<"Copy constructor"<<endl;
	p=new int;
	*p=*obj.p;//object is type of line
}
Line::~Line()
{
	cout<<"Destructor"<<endl;
	delete p;
}
Line::getLength(void)
{
	return *p;//dereference p(int p,not adress p)
}
void show(Line obj)
{
	cout<<"2"<<endl;
	cout<<obj.getLength()<<endl;
}


int main()
{
	{Line l(5);}
	Line line(10);

	cout<<"1"<<endl;
	//cout<<line.getLength()<<endl;
	show(line);
	cout<<"3"<<endl;
	return 0;
}

What I have tried:

I read about copy constructor much,i understood it,but some things that i could not understand in this program,i asked in the question.

Posted 29-Aug-16 11:23am

Member 12702056

Updated 2-Sep-16 6:55am

v7

Add a Solution

Comments

Maciej Los 29-Aug-16 17:29pm

Is it a joke?

Member 12702056 29-Aug-16 17:32pm

what did you mean? no,of course. i am beginner in c++,i read many things about work of copy constructor,but some little things that i wrote,i can't understand and i want to know with helping.

Member 12702056 29-Aug-16 17:34pm

if you mean about uni.lesson,it was by mistake,i wrote it here wrongly .it was my message to my friend.i saw it,i edit it.it was by mistake.

Philippe Mori 2-Sep-16 16:59pm

Obviously, you should read books and learn to use a debugger.

Philippe Mori 2-Sep-16 17:04pm

Obviously, 1, 2, 3 are there so that you can see on the output in which order instructions are executed.

Philippe Mori 2-Sep-16 17:07pm

By the way, this sample code is broken because it does not override assignment operator. If you take that from a book, then maybe that book is good for garbage...

Member 12702056 2-Sep-16 17:17pm

But the code works on my computer.
I'm gonna learn ,but i am only beginner in c++ ,so firstly i try only to learn concepts and some simple programs what i must be able to write and then i'll learn debugging :)

Member 12702056 2-Sep-16 18:37pm

Philippe Mori,i searched about debugging,but which one is better for debugging?beginning with visual studio or dev c++?

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Patrice T · Accepted Answer · 2016-08-29T11:43:00

Seeing the program executing may help to understand what it is doing.

You should learn to use the debugger as soon as possible. Rather than guessing what your code is doing, It is time to see your code executing and ensuring that it does what you expect.

The debugger allow you to follow the execution line by line, inspect variables and you will see that there is a point where it stop doing what you expect.
Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
http://docs.oracle.com/javase/7/docs/technotes/tools/windows/jdb.html[^]
https://www.jetbrains.com/idea/help/debugging-your-first-java-application.html[^]

The debugger is here to show you what your code is doing and your task is to compare with what it should do.
When the code don't do what is expected, you are close to a bug.

Advice: Choose which language is this program, no need to list every single possible and impossible language. The program don't compile in C.

BillW33 · Accepted Answer · 2016-09-02T06:55:00

This section of code:

C++

{Line l(5);}

This line defines the variable l of type Line and calls its constructor. Braces, sometimes called wavy brackets, are used to create a separate block of code. A block of code defines the scope of the variables that are declared within it. If you do not manually call the destructor of a class then its destructor is called automatically when it goes out of scope. Including the wavy brackets around this line causes its destructor to be called when the "}" is reached.

The code calls the following method in the line class:

C++

Line::Line(int len)
{
	cout<<"Simple constructor"<<endl;
	p=new int;
	*p=len;
}

This method will point p to a new int storage then assign 5 to the contents of p.

The 1,2,3 output in the console window are from the cout statements and are just there to show where the program is in its execution. It would be more informative to output descriptive text instead of just a number.

Some of the more confusing things for people new to C++ are the & and * operators. The copy constructor is designed to make a copy of an object of the Line class to another instance of a Line object. The parameters of the Line copy constructor, "(const Line &obj)", says that the method takes the address of an instance of the Line class. An instance of a class is also known as an object so the parameter is called "obj" which is short for object.
p is initialized to hold an integer by the following line:

C++

p=new int;

Let's analyze the next line

C++

*p=*obj.p;

"*obj" means the contents of obj. obj holds an address pointing to a Line object; the contents of that address is a Line object.
".p" means the member "p" of the object before the "." therefore "*obj.p" means the member p of the object whose address was passed to the method.
Then that value of p is assigned to the contents of the local p via "*p=".

Can someone understand that what 1, 2, 3 is? And what {line l(5); } is in this program? Please, help me :)

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0