Here is one program taken from a textbook featuring copy constructors:
#include <stdio.h>
#include <conio.h>
#include <iostream>
using namespace std;
class point
{
private:
int x,y;
public:
point(int ox =0, int oy =0)
{
cout << " Make object" << this << endl;
cout << " Using default constructor \n";
x = ox, y = oy;
}
point(const point &p)
{
cout << " Make object" << this << endl;
cout << " Using copy constructor \n";
x = p.x, y = p.y;
}
void move(int dx, int dy);
void display();
};
point fct(point a);
int main()
{
point a(5,2);
a.display();
point b = fct (a);
b.display();
getch();
return 0;
}
void point::move(int dx, int dy)
{
x += dx, y += dy;
}
void point::display()
{
cout << "Coordinates :" << x << " " << y << "\n";
}
point fct(point a)
{
point b=a;
//b.move(2,3);
return b;
}
It should be noted that the copy constructor is of the form : point (const point &p) instead of point (point &p) (the latter is what's in the textbook, but couldn't compile so I had to switch to the first one, but still I can't understand why :( )
The textbook said that there will be 3 lines of "Using copy constructor" , corresponding to 3 calls to the copy constructor. I think the reason for that is, when you call: b = fct(a)
The function "fct" makes a copy of a (pass-by-value), therefor one call
The line : point b = a : again, a copy constructor is invoked
The return value is then copied into b (the variable in main, not the one in fct) this is 3rd one.
However upon execution, there is only 2 calls. Can anyone give me a good explaination on this?
Two copies occur because Named Return Value Optimization (NRVO) elides one of the copies.
point fct(point a)
{
point b=a;
return b;
}
point b = fct (a);
The first copy is the one from the argument a in main to the parameter a in fct. This occurs because the point is taken by-value.
The second copy is from a to b inside func.
The copy which is elided is the one in returning b by value. In this case, b can be directly allocated into the b at the call site, so no copy takes place.
Related
I have to solve this " create class for describing triangle and trapeze with ability to return values and finding S of the figures.. declare function which allows comparing S of the both figures.. in main function declare object triangle and trapeze and compare their areas .. " - im trying to translate it from Bulgarian to English sorry if its not translated correctly ..
Anyways I came up with a solution, but when it asks me to enter value for trapeze x2 times and I can't understand why... it always takes the first 3 entered numbers but I want it to ask for input only once .. sorry if the answer is obvious
//
// main.cpp
// compare S of 2 figures
//
// Created by Георгиос Семерджиев on 17/05/22.
//
#include <iostream>
#include <cmath>
using namespace std;
class Trap // trap class with declared functions inside
{
protected:
double a, c, h;
void setValueTrap();
public:
Trap();
void Print();
virtual double S();
}; // end trap class
class Triangle : public Trap // triangle class with declared function for finding s() print and setting value
{
double b;
void setValueTriangle();
public:
Triangle();
void Print();
virtual double S();
double p(); // returning P/2
}; // end triangle class
// trap functions ...
Trap:: Trap()
{
setValueTrap();
}
void Trap::setValueTrap() // trap input function
{
cout << "Trap enter a = "; cin >> a;
cout << "Trap enter c = "; cin >> c;
cout << "Trap enter h = "; cin >> h;
cout << endl;
}
double Trap::S() // trap calculating and returning s()
{
return ( (a+c) * h ) / 2;
}
void Trap::Print() // printing S() for trap
{
cout << "Trap S = " << S();
cout << endl;
}
// Triangle functions ..
Triangle::Triangle():Trap()
{
setValueTriangle();
}
void Triangle::setValueTriangle() // setting value for triangle a,b,c
{
cout << "Triangle a = "; cin >> a;
cout << "Triangle b = "; cin >> b;
cout << "Triangle c = "; cin >> c;
cout << endl;
}
double Triangle::p() // returning P / 2 = p
{
return (a+b+c) / 2;
}
double Triangle::S() // calculating S() of triangle
{
return sqrt(p() * (p() - a) * (p()-b) * (p()-c));
}
void Triangle::Print()
{
cout << "Triangle S = " << S();
cout << endl;
}
// creating function to compare Trap S() and Triangle S()
bool Compare (Trap *F1, Trap *F2)
{
return F1 -> S() < F2 -> S();
} // compare function
int main()
{
Trap* trap = new Trap();
Triangle* triangle = new Triangle();
trap -> Print(); triangle -> Print();
cout << endl;
if (Compare(trap, triangle))
cout << "the Trap S < Triangle S.." << endl;
else
{
cout << "The Trap S > Triangle S.." << endl;
}
return 0;
}
For the Triangle constructor you have:
Triangle::Triangle():Trap()
{
setValueTriangle();
}
That will explicitly invoke the Trap constructor which read input. Then you will read the input for the triangle.
It's often not a good idea to read input in a constructor. It's usually better to only do basic initialization in the constructor and then get input using the fully constructed object.
Note that even if you don't have the explicit invokation of the Trap constructor, this will happen implicitly.
Triangle inherits from Trap, and both of their constructors are called when constructing Triangle object. To avoid this set up a virtual function setValue and call it only in Trap constructor.
I wanted to ask that when I don't write any copy constructor explicitly so the compiler automatically generates the copy constructor which performs shallow copy by default right?
So in the main() program when I changed the values of integers a, b and pointer p only the value of p changed and values of a and b remain unchanged in the copied object. Why the values of a & b didn't change too? My code is:
#include <iostream>
#include <string.h>
using namespace std;
class Dummy {
private:
int a, b;
int *p;
public:
Dummy() {
p = new int;
}
void setData(int x, int y, int z) {
a = x;
b = y;
*p = z;
}
void showData() {
cout << "a = " << a << " b = " << b;
cout << " p = " << *p << endl;
}
~Dummy() {
delete p;
}
};
int main() {
Dummy d1;
d1.setData(3, 4, 5);
Dummy d2 = d1;
d1.showData();
d2.showData();
d1.setData(6, 7, 8);
d1.showData();
d2.showData();
return 0;
}
The output of my program is:
a = 3 b = 4 p = 5
a = 3 b = 4 p = 5
a = 6 b = 7 p = 8
a = 3 b = 4 p = 8
What I'm saying is while the pointer of object d2 changed when I changed the values of object d1 then why didn't the values of a & b of object d2 changed too?
Also I'm using delete keyword in the destructor to delete the dynamically allocated pointer:
~Dummy() {
delete p;
}
But it's crashing my program instead. Why is that?
You've totally got it wrong - The idea of shallow copy. Actually, c++ does not have anything called deep copy built into itself. So, calling something shallow copy is a bit wrong. And just the use of these words shallow copy creates a lot of confusion too.
Now, let me explain, what happens when cpp performs initialization using assignment. cpp or c(while copying struct) has a concept called bitwise copy. In this concept, all the member variables of one object(struct object/class object - you can say either) is identically copied to another object. Now, it's totally wrong idea that, both objects point to same memory location. In actual, both object has their own memory location and of course, their variables occupy different memory spaces. For you, I have write some tests regarding memory. You would understand perfectly, if you just see the test and it's output:
#include <iostream>
#include <string.h>
using namespace std;
class Dummy {
int a, b;
int *p;
public:
Dummy() {
p = new int;
}
void setData(int x, int y, int z) {
a = x;
b = y;
*p = z;
}
void showData() {
cout << "a = " << a << " b = " << b;
cout << " p = " << *p << endl;
cout << endl; // an extra new line for readability of output
}
void showMemory() {
cout << "addr(a) = " << &a << " addr(b) = " << &b;
cout << " addr(p) = " << &p << endl;
}
~Dummy() {
*p = 100;
delete p;
}
};
// testing memory
void memoryTest() {
cout << "testing d1:" << endl;
Dummy d1;
d1.setData(3, 4, 5);
cout << "addr(d1) = " << &d1 << endl;
d1.showMemory();
cout << endl ;
cout << "testing d2:" << endl;
Dummy d2 = d1;
cout << "addr(d2) = " << &d2 << endl;
d2.showMemory();
}
int main() {
// memoryTest
memoryTest();
return 0;
}
And the output of the test was:
testing d1:
addr(d1) = 0x6dfed4
addr(a) = 0x6dfed4 addr(b) = 0x6dfed8 addr(p) = 0x6dfedc
testing d2:
addr(d2) = 0x6dfec8
addr(a) = 0x6dfec8 addr(b) = 0x6dfecc addr(p) = 0x6dfed0
This clearly shows that, the memory occupied by those two objects d1 and d2 are totally different.
Now, you may have another question remain: Then, why, when i write *p=8, it affects both d1 and d2?:
When you assign, Dummy d2 = d1;, we may say something happended like below(though, it's not actually happen when bitwise copy is applied, it's just for clarity):
d2.p = d1.p
So, we know that, d1.p and d2.p contains the same memory location(note: d1.p is a pointer. so, it does not contain any integer, rather it contains memory address of an int).
So, when you write *p = 8, you are telling the program to go to the memory location targeted by p and change the value of that memory location to 8.(note, here, you didn't change the content of d1.p, d1.p still contains the same memory location. rather, you just changed that memory location's content from 5 to 8). That's why when you call d2.p, you get the changed value. cause, d2.p contains the same memory location as d1.p.
Now, there may have one more question: Why your code crashes when you freed p in destructor?:
Now, let me first ask you, can you free a memory what is already freed. You can write the code, but the behavior is undefined. It may crashes your program or it may do nothing.
Well, in Dummy destructor you've written delete p;. Now, either d2 or d1 would be destroyed first. Let's assume, d2 is destroyed first. So, when d2's destroyer is called, p is freed. Then, d1's destroyer will be called and it'll also try to free p. But p is already freed. And in your case, the program meets a crash for this reason.
Hope, everything is clear to you now.
If anything is not clear about what I've described above, then ask questions, I will try my best to answer them too.
I am very new to programming, and am near the end of this program, but cannot quite finish the last detail, which I have been stuck on. I am attempting to switch what shape pointer *sp is pointing to, and it seems to me that what I am doing should work, since rectangle and circle both are shapes; however, when I compile, only the value of the color changes. The area of the circle prints instead of the area of the rectangle and the perimeter prints 0. Any help would be greatly appreciated!
#include <iostream>
#include <string>
using namespace std;
double const pi = 3.1519;
class shape {
public:
shape() {};
shape(string);
virtual double getCircumference() {
return 0;
};
virtual double getPerimeter() {
return 0;
};
virtual double getArea() {
return 0;
};
string getColor();
protected:
string color;
};
string shape::getColor() {
return color;
}
class circle : public shape {
public:
circle(double r, string c) {
radius = r;
color = c;
};
double getArea();
double getCircumference();
private:
double radius;
};
double circle::getCircumference() {
return pi * radius * 2;
}
double circle::getArea() {
return pi * radius * radius;
}
class rectangle:public shape {
public:
rectangle(double w, double l, string c) {
width = w;
length = l;
color = c;
};
double getArea();
double getPerimeter();
private:
double length;
double width;
};
double rectangle::getPerimeter() {
return width * 2 + length * 2;
}
double rectangle::getArea() {
return length * width;
}
void change(shape *sp, shape *sp1) {
*sp = *sp1;
}
int main() {
circle mary(3.2, "Green");
shape *sp = new circle(4.5, "Yellow");
cout << "Circle #1 is " << mary.getColor() << endl;
cout << "Circle #1 has an area of " << mary.getArea() << endl;
cout << "Circle #1 has a circumference of " << mary.getCircumference() << endl << endl;
cout << "Circle #2 is " << sp->getColor() << endl;
cout << "Circle #2 has an area of " << sp->getArea() << endl;
cout << "Circle #2 has a circumference of " << sp->getCircumference() << endl << endl;
shape *sp1 = new rectangle(1.0, 2.1, "Red");
change(sp, sp1);
cout << "Rectangle #1 is " << sp->getColor() << endl;
cout << "Rectangle #1 has an area of " << sp->getArea() << endl;
cout << "Rectangle #1 has a perimeter of " << sp->getPerimeter() <<endl<< endl;
}
It’s important to keep in mind what is meant by various different ways of using pointers. In your program, sp refers to the pointer itself—that is, a memory location telling the computer where to find an object. The asterisk in *sp is a ‘dereference’ operator; it takes a pointer and gives you the thing that it is pointing to.
With this in mind, your line *sp = *sp1; is saying, ‘take the thing that sp is pointing to, and set it to be equal to the thing that sp1 is pointing to.’ In other words, you are changing the value of the object pointed to by sp, not the value of sp itself. To point sp at the object pointed to by sp1, you need sp = sp1; with no asterisks.
The other thing to bear in mind is that C++ by default passes function arguments by value: when the function is called, the arguments are copied, and the function operates on the copies. This means that the original arguments themselves cannot be changed by a function that works like this. Adding an ampersand to the argument declaration, like void change(shape *&sp, shape *sp1) causes the first argument to be passed by reference: the object operated on by the function is the same object that was passed in by the calling code. This allows the function to change objects passed as arguments, and for those changes to remain after the function has returned.
Sorry for the long answer: I could have given you a few lines that did what you wanted, but I thought you might appreciate an explanation of the reason why things work the way they do.
If you are trying to change the address of pointers, you must pass pointers by reference. Try this:
void change(shape *&sp, shape *&sp1)
A pointer is passed by value even if it's a pointer.
This means that you are actually passing the address by value, so the argument is a copy of the original argument.
Think about
void sum(int a, int b, int result);
void foo() {
int result;
sum(5,10,result);
}
While you expect to be able to store the result into the variable passed to the sum argument you won't be able to do it since result is passed by value and hence copied. Every modification you do to result inside the method will be local to the method.
That's exactly the same thing, a pointer is nothing more than an address, if you pass it by value then a copy of the address is passed but every modification to the local variable is just local.
That's why you must use references if you want to be able to modify their values, exactly as every other variable, so you would have
void sum(int a, int b, int& result);
void change(shape*& shape1, shape*& shape2);
This, under the hood, will pass the address to the variable which stores the address (a sort of shape**) so the function is able to know where the original argument is located and modify it directly.
Hello I have two questions. First, here is some code. I am new to c++. I have to calculate the square of rectangle by the x,y coordinates of upper left and lower right corner - downRightx, upperLeftx, downRighty, upperLefty, the diagonal, and the sides of rectangle. I must make a function print() that calls other private functions only to show the result. Everything is defined inside the class.
class rectangle {
private:
double uLx, uLy, dRx, dRy;
public:
rectangle() {
cout << "enter x coordinate of upper left corner" << uLx;
cout << "enter y coordinate of upper left corner" << uLy;
cout << "enter x coordinate of down right corner" << dRx;
cout << "enter y coordinate of down right corner" << dRy;
}
~rectangle() {
cout << "Deleting object" << endl;
}
private:
void sides() {
double a, b;
a = sqrt(pow((dRx - uLx), 2));
b = sqrt(pow((dRy - uLy), 2));
}
void facediag() {
double s, d;
d = sqrt(pow((dRx - uLx), 2) + pow((dRy - uLy), 2));
---- 1. //here must be the calculation of square s = a*b
}
public:
void print() {
--- 2. //here I must print the results
}
};
so the question is: How to call a and b parameters from side in facediag() function to calculate s = a*b And how to print the results. Can I write cout << a; cout << d; cout << s, etc. in sides() and facediag() and just call them in print? Or can I print them in print() without writing cout << ... in other functions, but otherwise, another access method.
void facediag(){
//code
cout << s;
cout << d;
}
void sides(){
// code
cout << a;
cout << b;
}
void print()
{
sides();
facediag();
}// not like this, is there another way?
Second question I let Cygwin to be installed at its complete form and at some point I realized that I will run out of hdd and the installation hangs, so I interrupted the installation. How can I uninstall it - just delete the folder or to step through the FAQ in the Cygwin site?
You cannot access local variables from other functions. They only exist while that function executes.
What you can do is define more member functions that compute the values you need, like
double height() const
{ return /* something */; }
double width() const
{ return /* something else */; }
and use those functions where you need a or b.
You CAN call the private function in your OWN class.
But, the variable a and b are local variables, so they will not exist out of the function sides(), you can do the same thing in function facediag() to calculate a and b
As I specified in the title, this is a problem that puzzled me enormously, and still I can not find a fix for it.
I created a Point class which has three private attributes (_id, _x and _y) with related public get and set methods.
For example, I created three points (G, H, R), where the point G is a "parent". Point H is the prototype of the point G (a clone) and point R will be the outcome of the meeting points (G+H) (addition of the points).
Points sample:
Point G ("pG", 3.5f, 6.5f);
Point H (G.prototype ()); H.setId ("pH");
Point R;
R = G + H;
The program works correctly, but unfortunately after the operation {R=G+H}, _id attribute of point G becomes the clone attribute of point R, I do not understand why this is happening because the evaluation is done from right to left (H+, +G, =R).
The question is why G point attribute changes itself?
Code where the problem occurs:
#include <iostream>
#include <string>
using namespace std;
class Point
{
private:
string _id;
double _x;
double _y;
public:
Point(string id="undefined", double x = 0.00f, double y = 0.00f)
:_id(id), _x(x), _y(y){}
~Point(){}
public:
// Getters methods
string getId() const { return _id;}
double getX() const { return _x; }
double getY() const { return _y; }
void setId(string id) { _id = id; }
// Setters methods
void setX(double n = 0.00f) { _x = n; }
void setY(double n = 0.00f) { _y = n; }
Point prototype() { return Point(_id, _x, _y); }
Point operator+ (const Point& p)
{
cout << "Operator + called, returning a new object" << endl;
return Point
(
_id.append("+").append(p._id),
_x+p._x,
_y+p._y
);
}
Point operator= (const Point& p)
{
cout << "Operator = called, returning a new object Point(p._id, p._x, p._y)" << endl;
_id=p._id;
_x=p._x;
_y=p._y;
return *this;
}
};
int main()
{
Point G("G",10.0f, 10.0f);
Point H(G.prototype()); H.setId("H");
Point R;
R = G + H;
cout << "object Point {id: " << G.getId() << ", x: " << G.getX() << ", y: " << G.getY() << "}" << endl;
cout << "object Point {id: " << H.getId() << ", x: " << H.getX() << ", y: " << H.getY() << "}" << endl;
cout << "object Point {id: " << R.getId() << ", x: " << R.getX() << ", y: " << R.getY() << "}" << endl;
return 0;
}
Thanks in advance!
Point operator+ (const Point& p)
{
cout << "Operator + called, returning a new object" << endl;
return Point
(
_id.append("+").append(p._id), // HERE
_x+p._x,
_y+p._y
);
}
In this line, you modify _id property of this object. As a rule of thumb, binary + operator should be a const member or - IMO cleaner - a static method taking two const reference arguments.
BTW, you can add strings in C++ by just applying + operator to them (_id + "+" + p._id).
If you implement your binary arithmetic operators as members, you should consider making the operator a const member. This would catch the obvious modification in your implementation (the implementation is made out of line to avoid including unrelated code):
Point Point::operator+ (const Point& p) const
{
cout << "Operator + called, returning a new object" << endl;
return Point
(
_id.append("+").append(p._id),
_x+p._x,
_y+p._y
);
}
The operation _id.append("+") actually operates on this->_id, i.e., the _id member of the left hand operand. Since you member operator isn't const, the compiler lets you do the modification. This is probably not what you intended to do. You probably rather wanted to write:
Point Point::operator+ (const Point& p) const
{
cout << "Operator + called, returning a new object\n";
return Point
(
_id + "+" + p._id,
_x+p._x,
_y+p._y
);
}
... which creates a suitable temporary string with the desired value (I also replaced the excessive use of std::endl).
This line in the operator+ overload:
_id.append("+").append(p._id),
You have to be very careful here, you are currently in an object (in your example this would be the G object). append actually alters the local string, so your code is appending + then p._id then deep copying that to the returned value.
A quick fix is to change this to a temporary that holds what you need:
_id + "+" + p._id
To avoid problems like this in the future, you should declare operator overloads as a friend method so it's extremely clear what you're operating on.
friend Point operator+ (const Point& lhs, const Point& rhs)
{
cout << "Operator + called, returning a new object" << endl;
return Point
(
lhs._id.append("+").append(rhs._id), //COMPILER ERROR! We're modifying the const "lhs"!
lhs._x+rhs._x,
lhs._y+rhs._y
);
}
So let's change this to:
friend Point operator+ (const Point& lhs, const Point& rhs)
{
cout << "Operator + called, returning a new object" << endl;
return Point
(
lhs._id + "+" + rhs._id, // no longer modifying lhs -- all is well!
lhs._x+rhs._x,
lhs._y+rhs._y
);
}