I'm wondering what a call to an object is supposed to display. I have a class called big_number that has a few different constructors. In another method, I'm declaring an object 'a' using class big_number as follows:
big_number a;
cout << "Default constructor gives " << a << endl;
And my constructor is:
big_number::big_number()
{
head_ptr = 0;
tail_ptr = 0;
positive = false;
digits = 0;
base = 10;
}
(Although I'm sure that this constructor is wrong).
The full code of the testing file:
int main()
{
int n1, n2;
unsigned int base;
string s;
char choice;
do
{
cout << "Type 'd' to test default constructor" << endl;
cout << "Type 'i' to test int constructor" << endl;
cout << "Type 's' to test string constructor" << endl;
cout << "Type 'a' to test assignment" << endl;
cout << "Type '>' to test input operator" << endl;
cout << "Type '=' to test comparison operators" << endl;
cout << "Type 'q' to quit" << endl;
cin >> choice;
if (toupper(choice) == 'D')
{
big_number a;
cout << "Default constructor gives " << a << endl;
}
//More Code
If by "call to an object" you mean your call of the operator<< on the object a with cout as the stream argument: It displays whatever you define it to display (in a member function of big_number or a free function). There is no "default" operator<< for user-defined classes. So, if you define it like
#include <iostream>
struct big_number
{
template<typename T>
friend T& operator<< (T& stream, const big_number& bn);
};
template<typename T>
T& operator<<(T& stream, const big_number& bn) {
return (stream << "Hello World");
}
int main()
{
big_number a;
std::cout << a << std::endl;
}
... it will just display "Hello world".
The purpose of making it a friend function is so it can access the private data members of big_number (since you usually want it to display something that depends on the data stored in the object, and not a constant "Hello world"). So, within the operator<< definition, you would probably iterate through the digits in your linked list and push them to the stream (if I understand correctly what you are trying to do).
Related
I have a class of light bulbs. There are methods and constructors in this class. There is even a destructor) The problem is that I have to determine and display information about class members with type "n" in the TEST() method (LED lamps).
To implement this task, he developed the gettype() method, which returns the type of an object, and, in fact, the TEST() method, which displays information about light bulbs.
The problem is that nothing works for me. I tried a lot of things, but it doesn’t work out for me to implement this task. I'm new to programming (
Code:
#include <iostream>
using namespace std;
class lamp
{
public:
// methods
void TEST(void);
char* gettype (void);
void INIT(void);
void SHOW(void);
// construcrors
lamp();
lamp(const char *t, int powe, const char *c, double cos);
lamp(const lamp & obj);
// destructor
~lamp();
private:
// data
char type[100]; // LED, energy-saving or incandescent lamp
int power; // LED lamp - "n"
char color[100];
double cost;
};
lamp::lamp() {
cout << "This object was created in the default constructor.\n";
strcpy(type, "");
power = 0;
strcpy(color, "");
cost = 0;
}
lamp::lamp(const char *t, int powe, const char *c, double cos) {
cout << "This object was created in the constructor with parameters.\n";
strcpy(type, t); //*t
power = powe;
strcpy(color, c); //*c
cost = cos;
}
lamp::lamp(const lamp & obj) {
cout << "This object was created in the copy constructor.\n";
strcpy(type, obj.type);
power = obj.power;
strcpy(color, obj.color);
cost = obj.cost;
}
lamp::~lamp() {
cout << "Deletion of object by destructor.\n";
}
void lamp::SHOW(void) {
cout << "Lamp Information:\n";
cout << "\nType > " << type;
cout << "\nPower > " << power;
cout << "\nColor > " << color;
cout << "\nCost > " << cost << endl;
}
void lamp::INIT(void) {
cout << "Enter lamp information:\n";
cout << "\nType (if LED, then n) > "; cin >> type;
cout << "\nPower > "; cin >> power;
cout << "\nColor > "; cin >> color;
cout << "\nCost > "; cin >> cost;
}
char* lamp::gettype (void) {
return type;
}
void lamp::TEST(void) {
cout << "\nType > " << type;
cout << "\nPower > " << power;
cout << "\nColor > " << color;
cout << "\nCost > " << cost << endl;
}
void main() {
setlocale(0, "");
// default constructor for 1 class instance
lamp l1;
cout << "Entering data for the first product." << endl;
l1.INIT();
// constructor with parameters for 2 class instances
cout << endl << "Information about the second object: \n";
lamp l2("n", 950, "yellow", 1580);
// copy constructor for the third object
cout << endl << "Information about the third object: \n";
lamp l3(l2);
// Derived information about all the lamps using the method SHOW
l1.SHOW();
l2.SHOW();
l3.SHOW();
// I create an array of two objects using the default constructor
lamp la[2];
I enter data into an array of objects using the method INIT
cout << "Fill an array of objects with 2 elements." << endl;
for(int i = 0; i < 2; i++) {
la[i].INIT();
}
// I output data from an array of objects using the method SHOW
cout << "Showing items." << endl;
for (int i = 0; i < 2; i++) {
la[i].SHOW();
}
// looking for and displaying information about LED lamps
cout << "Search and display information about LED lamps." << endl;
for (int i = 0; i < 3; i++) {
if (la[i].gettype() == "n") {
cout << endl << " lamp number : " << (i + 1) << endl;
la[i].TEST();
cout << endl;
}
}
system("pause");
}
There are several errors in your code：
strcpy is included in <cstring> which is missed. You need to add it in the beginning:
#include <cstring>
main() function should be declared as int main() and you need to add a return statement
int main() {
//YOUR CODE HERE
return 0;
}
You missed a comment sign at line 104
lamp la[2];
//I enter data into an array of objects using the method INIT
cout << "Fill an array of objects with 2 elements." << endl;
After fixed, your code should be able to run.
I hope this time my question is better formulated and formatted.
Here's the code that produces two separate outputs when I think it should not since I use everytime (I think) the overloaded operator<< for an enum type.
#include <iostream>
using namespace std;
enum Etat { Intact = 5 };
class Ship {
public:
Etat etat_;
Ship ( Etat t = Intact) : etat_(t) {}
~ Ship() {}
ostream& description ( ) const { return cout << "Etat: " << etat_ << " --- ";}
//---------------------------------------ˆˆˆˆ----
};
ostream& operator<< ( ostream& s, const Etat& etat_ )
{
switch ( etat_ )
{
case Intact: s << "intact"; break;
default: s << "unknown state";
}
return s;
}
ostream& operator<< ( ostream& s, Ship n ) { return s << "Etat: " << n.etat_ ; }
int main()
{
Etat etat_ = Intact;
cout << endl << endl << "Etat: "
<< etat_ << " \"cout << etat_\"" << endl << endl;
cout << Ship(etat_)
<< " \"cout << Ship(etat_)\"" << endl << endl;
cout << Ship(etat_).description()
<< " \"cout << Ship(etat_).description()\"" << endl << endl;
return 0;
}
This is what I get in the terminal:
Etat: intact "cout << etat_"
Etat: intact "cout << Ship(etat_)"
Etat: 5 --- 1 "cout << Ship(etat_).description()"
Can anyone explain to me why, in the last case, not only it takes the integer value of the enum attribut, but also adds a "1" after the test string " --- "???
The only thing I can think of is because I used an unorthodox return method in description(), ie 'return cout << ..", but it seems to work since the test string appears.
Is there a way to force the use of the operator<< overload in description()?
Thanks
In the description() function you are returning a reference to std::cout and use it in the std::cout call in main function. There is a reason why operator<< takes an ostream reference as it's first argument. You should modify your function like this and all should work:
ostream& description(ostream& os) const {
return os << "Etat: " << etat_ << " --- ";
}
The random "1" printed out there is caused likely due to the ostream in your example trying to print out reference to itself.
We know that the fascinating class iostream is something too powerful.
it has overloded the insertion operator "<<" to take many datatypes:
ostream& operator(ostream&, int),
ostream& operator(ostream&, char)...
we cannot instantiaate ostream: ostream print;
because ostream because its most CTORSs are "protected-socoped" (cannot be accessed from outside).
the only Constructor we can call is ostream(streambuf*) which takes a pointer to another class object ( class streambuf);
I just wanted to mess up with this class:
#include <ostream>
using namespace std;
int operator << (ostream& out, int* x)
{
out << "invoked!" << endl;
cout << *x; // works well!
cout << x; // normally works well and it prints the address that x points to but instead the program get in infinite loop or crushes!
return *x;
}
int main()
{
system("color 1f");
int* pVal = new int(57);
cout << *pVal << endl;
int* pX = new int(7);
cout << *pX << endl;
cout << *pVal << ", " << *pX << endl;
//cout << pVal << endl; // this doesn't work because my operator returns
//int and not a reference to ostream.
// and it is equal to: 0 << endl; which generates the same error
cout << pVal; // this works
// cout << endl << endl << endl;
return 0;
}
I overloaded the insertion operator to take an lvalue as a reference to an ostream object and a pointer to int as rvalue, I popup a message inside my function to get sure that it is invoked.
Note that I intentionally overloaded it to return int value so that no one can write:
out << pInt << *pInt << endl;
... but just:
out << pInt;
My problem, as you can see in the inline-comments above, is that whilst cout << x normally works well, instead the program get in infinite loop or crushes!
return *x;
Can anyone explain why I am getting the error?
The problem hapens because if you just cout << x, it will call your overloaded function over and over. It never returns.
Here's the solution (cast x to void*)
int operator << (ostream& out, int* x)
{
out << "invoked!" << endl;
cout << *x; // works well!
cout << (void*)x;
return *x;
}
I'm trying to overload the << operator for the display function call.
Heres my code:
#include <iostream>
#include <cstring>
using namespace std;
// global variable
const int MAX = 3;
// class definition
class CString{
char str[MAX+1];
public:
CString(char* param){
if(param == nullptr){
str[0] = '\0';
return;
}
strncpy(str,param,MAX);
str[MAX] = '\0';
}
void display(ostream& os){
os << str;
}
};
// << operator overloading
ostream& operator << (ostream& os, CString& cs){
static int call = 0;
os << call << ": ";
cs.display(os);
call++;
return os;
}
void process(char* parm){
CString cs(parm);
// here is where my issue is
cs.display(cout);
cout << endl;
}
//----------------------------------------------------------------
int main(int argc,char *argv[]){
cout << "Command Liine : ";
for(int arg = 0; arg < argc ; arg++){
cout << " " << argv[arg];
}
cout << endl;
if( argc == 1){
cout << "Insufffiecentnumber of arguemnts (min1)" << endl;
return 1;
}
cout << " Maxium numver of characters stored: " << MAX << endl;
for(int arg = 1; arg < argc; arg++){
process(argv[arg]);
}
return 0;
}
EDIT:
Here is the correct output and the output I have:
Correct:
Command Line : w1 oop345 btp305
Maximum number of characters stored : 3
0: oop
1: btp
Mine:
Command Line : w1 OOP345 DBS305
Maxium number of characters stored: 3
OOP
DBS
I'm having an issue with my << operator not working, I can't seem to figure it out. The ostream& operator<<(ostream& os, CString& cs) does not seem to be loading its syntax.
Question:
Does anyone know where my mistake has been made?
You wrote a correct overloading of << operator, but in method process() you used a public method display() of class CString instead of using << operator directly.
Just change one line in method process():
cs.display(cout); to: cout << cs;
void process(char* parm){
CString cs(parm);
// here is where my issue is
cout << cs;
cout << endl;
}
P.S. you do not need method CString::display at all as you already overload << operator for this class.
class A
{
public:
ostream& operator<<(int string)
{
cout << "In Overloaded function1\n";
cout << string << endl;
}
};
main()
{
int temp1 = 5;
char str = 'c';
float p= 2.22;
A a;
(a<<temp1);
(a<<str);
(a<<p);
(a<<"value of p=" << 5);
}
I want the output to be: value of p=5
What changes should is do...and the function should accept all data type that is passed
There are 2 solutions.
First solution is to make it a template.
template <typename T>
ostream& operator<<(const T& input) const
{
cout << "In Overloaded function1\n";
return (cout << input << endl);
}
However, this will make the a << str and a << p print c and 2.22, which is different from your original code. that output 99 and 2.
The second solution is simply add an overloaded function for const char*:
ostream& operator<<(int string)
{
cout << "In Overloaded function1\n";
return (cout << string << endl);
}
ostream& operator<<(const char* string)
{
cout << "In Overloaded function1\n";
return (cout << string << endl);
}
This allows C strings and everything convertible to int to be A <<'ed, but that's all — it won't "accept all data type that is passed".
BTW, you have forgotten to return the ostream.