/ circarea.cpp // demonstrates floating point variables #include <iostream> //for cout, etc. using namespace std; int main() { float rad; //variable of type float const float PI = 3.14159F; //type const float cout << "Enter radius of circle: "; //prompt cin >> rad; //get radius float area = PI * rad * rad; //find area cout << "Area is " << area << endl; //display answer return 0; } */ void main(void) { cout<<"### Programmed By Amahdy(MrJava) ,right restricted.\n"; cout<<"-------------------------------------------------------------------------------\n"<<endl; float rad; do{ cout<<"Enter radius of circle: "; cin>> rad; cout <<"Area is "<<circarea(rad); cout<<"\n\n !Press c to continue or any key to exit."<<endl<<endl; }while(getch()=='c'); } float circarea(float radius) { return 3.14159F*radius*radius;} Code:
/**2. Raising a number n to a power p is the same as multiplying n by itself p times. Write a function called power() that takes a double value for n and an int value for p, and returns the result as a double value. Use a default argument of 2 for p, so that if this argument is omitted, the number n will be squared. Write a main() function that gets values from the user to test this function.*/ #include<iostream> #include<conio.h> using namespace std; double power(double n, int p=2); void main(void) { cout<<"### Programmed By Amahdy(MrJava) ,right restricted.\n"; cout<<"-------------------------------------------------------------------------------\n"<<endl; double n; int p=2; do{ cout<<"Enter n: "; cin>> n; cout<<"Enter p: "; cin>> p; cout <<"The power is "<<power(n, p); cout<<"\n\n !Press c to continue or any key to exit."<<endl<<endl; }while(getch()=='c'); } double power(double n, int p){ for(int ret=1; p>0; p--) ret*=n; return ret;}
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/**3. Write a function called zeroSmaller() that is passed two int arguments by reference and then sets the smaller of the two numbers to 0. Write a main() program to exercise this function.*/ #include<iostream> #include<conio.h> using namespace std; bool zeroSmaller(int& n1, int& n2); void main(void) { cout<<"### Programmed By Amahdy(MrJava) ,right restricted.\n"; cout<<"-------------------------------------------------------------------------------\n"<<endl; int n1, n2;
If you are describing a circuit breaker then its electromagnetic trip component will operate in the event of a line*-to-ground (*not 'phase'!) short circuit. The thermal overload component will only operate in the event of a sustained overload.
An object is an instance of a class. A class is a data type that combines data and the specific methods that operate upon that data into a self-contained entity. To prove that objects are user-defined types, consider the following: class foo { }; int main() { foo a; foo b; foo c = a + b; // ERROR! } The reason there is an error is because the plus operator (+) only works with primitive data types (integral, arithmetic and built-in data types). foo is neither an integral, arithmetic nor built-in data type, thus the plus operator is not supported. However, if you provide a user-defined plus operator overload (foo::operator+) that accepts a constant reference to a foo object, the code will compile. It's up to you, the class designer, to provide the appropriate implementation.
with caution, commonsense, and if opportunity provides, the supervision of an experienced operator.
I think you mean operation overlord??? It is the American, Canadian and British offensive on Europe in World War 2. They landed in Normandy on 6th June 1944 (Commonly called D-Day, Day of Days or Deliverance Day) and progressed throughout France liberating Paris on the 25th August. This allowed the allies a foothold in Europe.
Generally operate machinery on a production line. It's a generic term covering a wide range of machinery and products.
All operators are built-in but not all operators can operate upon data types they know absolutely nothing about. There are some exceptions such as the new operator and the sizeof operator -- both will work on any datatype. However, for those that cannot, operator overloads allow you to cater specifically for those types. An operator overload is implemented just as you would overload a function, but it is not a function per se because operators have different calling conventions to functions. It is also important to keep in mind that just because you can overload an operator, it does not mean that you should. Operators should only be overloaded when the overload would allow a user to interact with your data type in an intuitive manner; a manner that is consistent with the operator's intended purpose. So while it can be amusing to overload the plus (+) operator to perform a subtraction (-), it could hardly be called intuitive. The assignment operator is the most overloaded operator of them all. This is because it is extremely useful to be able to copy the members of one object and assign those values to another object of the same type. We can also overload the assignment operator to cater for objects of different types, but such assignments are rarely intuitive. Does it make sense to assign the properties of a banana object to a person object? Probably not. Even if you could find a practical reason for doing so, would it be an intuitive operation? Definitely not. Therefore there's no point in providing an operator to cater for this. To create an operator overload, the declaration will often be placed inside the class it pertains to. However there are exceptions. The output stream insertion operator is a good example of this. The following example demonstrates how we can overload an internal operator (the assignment operator) as well as an external operator (output stream insertion operator). #include<iostream> // required to make use of I/O streams class A { private: unsigned m_data; public: // constructors... A (const unsigned data = 0): m_data (data) {} A (const A& copy): m_data (copy.m_data) {} // accessor function (interface) unsigned get_data() const { return m_data; } // operator overloads... A& operator= (const A& rhs) { m_data = rhs.m_data; } A& operator= (const unsigned rhs) { m_data = rhs; } }; std::ostream& operator<< (std::ostream& os, const A& a { os << a.get_data(); return os; } int main() { A a, b; // invoke default constructors a = 42; // call assignment operator overload b = a; // call default assignment operator overload // call insertion operator overload std::cout << a << std::endl; std::cout << b << std::endl; } Output: 42 42
he who operate the computer.
The licenced operator.
If you are describing a circuit breaker then its electromagnetic trip component will operate in the event of a line*-to-ground (*not 'phase'!) short circuit. The thermal overload component will only operate in the event of a sustained overload.
Only a licensed operator
Computer operate
A licensed operator.
Fuses don't operate anything,you do. fuses protect electrical circuits from overload
An object is an instance of a class. A class is a data type that combines data and the specific methods that operate upon that data into a self-contained entity. To prove that objects are user-defined types, consider the following: class foo { }; int main() { foo a; foo b; foo c = a + b; // ERROR! } The reason there is an error is because the plus operator (+) only works with primitive data types (integral, arithmetic and built-in data types). foo is neither an integral, arithmetic nor built-in data type, thus the plus operator is not supported. However, if you provide a user-defined plus operator overload (foo::operator+) that accepts a constant reference to a foo object, the code will compile. It's up to you, the class designer, to provide the appropriate implementation.
The noun forms of the verb to operate are operator, operation, operand, and the gerund, operating.
with caution, commonsense, and if opportunity provides, the supervision of an experienced operator.
with caution, commonsense, and if opportunity provides, the supervision of an experienced operator.