Q: X plus plus operator overloading in c sharp?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

The answer is x + 6. The plus sign (+) is used to indicate sums. A sum is the result of addition and + indicates the addition operator.

0

The answer is 144. The compound operator for (a) and (b) in each pair is a * (a+b). For 9 and 7, this is 9 x (9+7) = 9 x 16 = 144.

x + x + 1 + x + 2 + x + 3 = -144x + 6 = 144x = 8x = 2

What is the variable of x plus 10 plus 4x-35?

Related questions

There is no unary plus in C, but if there were, it would have only one operand, unlike the binary plus which has two: x = a + b; /* binary plus */ x = + b; /* unary plus -- not in C*/ x = a - b; /* unary plus */ x = - b; /* unary minus */

An operand is the value that is being operated upon by an operator. For instance, the C++ increment operator (++) is a unary operator, which means it has only one operand, the variable that we wish to increment. This in the expression x++, x is the operand. The addition operator (+) is a binary operator and therefore has two operands. Thus in the expression x + y, x and y are the operands.

There is no need to overload the plus operator to achieve this. It is already overloaded with this functionality. #include<iostream> #include<string> int main() { int x=40; int y=2; int z=x+y; std::cout<<x<<" + "<<y<<" = "<<z<<std::endl; std::string s1="Hello "; std::string s2="world!"; std::string s3=s1+s2; std::cout<<"""<<s1.c_str()<<"" + ""<<s2.c_str()<<"" = ""<<s3.c_str()<<"""<<std::endl; return(0); }

You use the dot operator when the left side is the name of the object or a reference to an object, and you use the arrow operator when the left side is a pointer to an object. Example: struct foobar x, *p= &x; x.field = p->field; (&x)->field = (*p).field;

The '+=' operator behaves like a pre increment operator.

Yes, that would be F♯x (F triple sharp) - enharmonic with G sharp.

The left hand operand of a binary operator must represent a modifiable lvalue: void f (const int x, int y) { x = y; // error -- x is constant lvalue y = x; // ok }

The only disadvantage of operator overloading is when it is used non-intuitively. All operators must behave with predictable results, thus it makes no sense to implement the plus (+) operator so that it behaves like a subtract (-) operator, or a multiply (*) operator, or indeed anything other than the intuitive sum of two objects.

The answer is x + 6. The plus sign (+) is used to indicate sums. A sum is the result of addition and + indicates the addition operator.

I assume by 2 plus you really mean ++. This is the increment operator which is used to increment the operand. If placed before the operand, the operator evaluates the incremented operand (prefix increment). If placed after the operand, the operator evaluates the non-incremented operand (postfix increment). +++ and ++++ are meaningless but are assumed to mean incrementing an increment. If you wish to increment an increment, you must use the compound expression ++(++) or (++)++. Thus for the variable x, prefix incrementing twice would be achieved with ++(++x), while postfix incrementing twice would be achieved with (x++)++. You can also mix the two, such as ++(x++) or (++x)++, both of which would increment x twice but would evaluate the increment of x. If postfix increment is not a requirement, it would be much easier to use the compound expression x+=n, where n is the amount you wish to increment. This is the same as saying x=x+n.

Assuming that the first operator is 6x^2. The factors are (6x-5)*(x-1)

The address-of operator is a unary operator which returns the address of its operand: int x = 42; // instantiate a variable of type int std::cout << "Value of x: " << x << std::endl; // e.g., 42 std::cout << "Address of x: " << &x << std::endl; int* p = &x; // store the address of x in a pointer variable std::cout << "Value of p: " << p << std::endl; // e.g., the address of x std::cout << "Address of p: " << &p << std::endl; std::cout << "Value referred to by p: " << *p << std::endl; // e.g., 42