Answer / rupinder

You overload a binary unary operator with either a nonstatic
member function that has one parameter, or a nonmember
function that has two parameters. Suppose a binary operator
@ is called with the statement t @ u, where t is an object
of type T, and u is an object of type U. A nonstatic member
function that overloads this operator would have the
following form:

return_type operator@(T)

A nonmember function that overloads the same operator would
have the following form:

return_type operator@(T, U)

An overloaded binary operator may return any type.

The following example overloads the * operator:

struct X {

// member binary operator
void operator*(int) { }
};

// non-member binary operator
void operator*(X, float) { }

int main() {
X x;
int y = 10;
float z = 10;

x * y;
x * z;
}

The call x * y is interpreted as x.operator*(y). The call x
* z is interpreted as operator*(x, z).

Answer / saranya

In overloading binary operators the object to the left of the operator is used to invoke the operator function while the operand to the right of the operator is always passed as an argument to the function.
eg) sum.x = x + real.x
here x is used to invoke the function +() and real.x is passed as argument to that function.

#include <iostream> using namespace std; int main() { int a = 2; int c[5][5]; for (int x=0;x<5;x++) { for (int y=0;y<5;y++) { c[x][y] = x*y; } } cout << c[a][c[1][4]]; }

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