The copy-constructor doesn't work when defines an object with "=" operator(Copy-initialization)?

Question

0.00/5 (No votes)

See more:

The code:

C++

#include <iostream>
using namespace std;

class String
{

	char name[256];
public:

	String(char* str)
	{
		cout<<"Constructor" << endl;
		strcpy(name,str);
	}

	String(String &s)
	{
		cout<< "Copy Constructor"<< endl;
		strcpy(name,s.name);
	}

	String()
	{
		cout<<"Default Constructor"<<endl;
	}

	~String(){}

	String& operator=(const String&s)
	{
		strcpy(name,s.name);
		cout<<"Assign Operator"<<endl;
		return *this;
	}


	void display()
	{
		cout << "The String is:" << name << endl;
	}
};


int main()
{
	String mystr1 = "a";//Why does it only print "Constructor"?
	String mystr2;
	mystr2 = "aa";
	String mystr3("aaa");//Here just invoke String::String(char *str).
	return 0;
}

Why don't invoke String::String(String &s) while defining "mystr1" ? How compiler deals with the mystr1 definition?

Posted 14-Jan-12 0:09am

Bo Ye

Add a Solution

5 solutions

Solution 1

Your code only requires the default constructors as you are passing char*s as intialisation parameters. You need to pass objects to get the copy constructor and =op to be invoked. Try this:

C++

String mystr2;
mystr2 = mystr1;
String mystr3(mystr2);

Posted 14-Jan-12 0:20am

Richard MacCutchan

Comments

Bo Ye 14-Jan-12 7:08am

You misunderstand what I mean. This is my test code, mystr1 has no relationship with mystr2. I just would like to know how the compiler deals with the sigle sentence String mystr1 = "a";

Richard MacCutchan 14-Jan-12 7:15am

No, I think you misunderstand. The statement String mystr1 = "a"; calls the first constructor of your class. You are asking why it does not call the copy constructor or the assignment operator and I gave you some sample code to demonstrate it. Give it a try and see what happens, or better still, step through the code with your debugger and you will see each statement in turn.

Bo Ye 14-Jan-12 7:26am

ah... I know it doen't call the assignment constructor. I just wanna figure it out why it seemed that the copy constructor is not called.

Richard MacCutchan 14-Jan-12 7:33am

Because there is nothing to copy, you are passing an Rvalue to the constructor. Try the code that I suggested and you will see how (and when) it does get called.

Bo Ye 14-Jan-12 7:32am

I rewrite the main function:
int main()
{
String mystr1 = "a";//Why does it only print "Constructor"?
String mystr3("aaa");//Here just invoke String::String(char *str).
return 0;
}

Richard MacCutchan 14-Jan-12 7:34am

See my comment above.

Bo Ye 14-Jan-12 8:16am

I've tried it and konw how your code run...... I'am sorry for my bad English expression. I just what you to explain String mystr1 = "a"; in detail. As CPallini said, it generates String mystr1("a"); . Is it Compiler optimization?

Richard MacCutchan 14-Jan-12 9:23am

See my comments in my alternate solution entry.

Solution 3

mystr1 is constructed without parameter. So the default constructor String() is called. Additionally, there is no copy operation defined for char* You may add this:

C++

String& operator=(const char* s)
{
    ASSERT(s);
    strncpy(name, s, 256);
    name[255] = '\0';
    cout<<"char* assign Operator"<<endl;
    return *this;
    }

Edit:
CPallini is right and my answer is wrong.

Posted 14-Jan-12 0:44am

Jochen Arndt

Updated 14-Jan-12 0:49am

v2

Comments

Ashish Tyagi 40 14-Jan-12 9:49am

Exactly right, my 5.

Solution 4

Constructors are implicit converters,

C++

String mystr1 = "a";//Why does it only print "Constructor"?
     
is as good as (if you are using non explicit constructor taking char*)

String mystr1("a");//Why does it only print "Constructor"?

C++

 use explicit keyword and then try -


....
explicit String(char* str)
	{
		cout<<"Constructor" << endl;
		strcpy(name,str);
	}
....
String mystr1 = "a";// This won't compile now
                   // and no implicit conversions will take place.

This link will also help.

Posted 14-Jan-12 3:18am

Ashish Tyagi 40

Updated 16-Jan-12 5:30am

v3

Comments

Richard MacCutchan 14-Jan-12 9:37am

And a 5 for that, I forgot about the new keyword(s).

Ashish Tyagi 40 14-Jan-12 9:39am

:-)

Bo Ye 16-Jan-12 8:30am

No, "String mystr1 = "a";" is initialization, not assignment. operator=(const char*) is not called.

Ashish Tyagi 40 16-Jan-12 11:32am

Yes you are correct, I update my answer accordingly.

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)

Richard MacCutchan · Accepted Answer · 2012-01-14T03:22:00

OK, I think I see where you may be confused.

C++

String mystr1 = "a";

In this case the compiler can optimise the statement to an existing constructor (String(char* str)), because the creation and assignment happen together.

C++

String mystr2;
mystr2 = "aa";

In this case the creation of the String object does not have any initialiser so the default constructor is called. On the next line a value ("aa") is to be assigned to mystr2, and the way to do that is to construct a temporary String object (using an appropriate constructor), and then use the assignment operator to set the value of the previously created one (mystr2). It could be argued that the compiler should optimise this to the same as case 1, as the two operations are in sequence.

C++

String mystr3("aaa");

In this case the code is a constructor call so the constructor that takes a char* will be called.

C++

String mystr3(mystr2);

In this case the parameter is a String object so the compiler will generate a call to the copy constructor.

CPallini · Accepted Answer · 2012-01-14T00:39:00

Solution 2

Bo Ye wrote:
String mystr1 = "a";//Why does it only print "Constructor"?

The compiler short-cuts the passage of temporary object creation, i.e. instead of producing

C++

String mystr1 = String("a");

it generates

C++

String mystr1("a");

[update]
I used at least improper terminology in my answer, please see Emilio's remark.
[/update]

Posted 14-Jan-12 0:39am

CPallini

Updated 14-Jan-12 10:47am

v2

Comments

Bo Ye 14-Jan-12 7:13am

Is all compilers so do? like gcc or others?

CPallini 14-Jan-12 10:20am

All the ones I used (yes, gcc is one of them).

Emilio Garavaglia 14-Jan-12 13:50pm

Carlo, it is not an optimization: the standard explicitly says that the = after a declaration is "initialization", not "assignment", and hence the constructor is called. It is not "creating an empty string and then change it by give it an 'a', but "create a string that 'is like an' 'a' ".

CPallini 14-Jan-12 16:46pm

Yes, you are right. Thank you for pointing out.