Arithmetic Operators in C++

C++ has 5 basic arithematic operators. They are −

• Addition(+)
• Subtraction(-)
• Division(/)
• Multiplication(*)
• Modulo(%)

Example

These operators can operate on any arithmetic operations in C++. Lets have a look at an example −

#include <iostream>
using namespace std;

main() {
   int a = 21;
   int b = 10;
   int c ;

   c = a + b;
   cout << "Line 1 - Value of c is :" << c << endl ;

   c = a - b;
   cout << "Line 2 - Value of c is  :" << c << endl;
   c = a * b;
   cout << "Line 3 - Value of c is :" << c << endl;

   c = a / b;
   cout << "Line 4 - Value of c is  :" << c << endl;

   c = a % b;
   cout << "Line 5 - Value of c is  :" << c << endl;
   return 0;
}

Output

This will give the output −

Line 1 - Value of c is :31
Line 2 - Value of c is  :11
Line 3 - Value of c is :210
Line 4 - Value of c is  :2
Line 5 - Value of c is  :1

There are some compound arithmetic operators also available in c++. The compound arithmetic assignment operators are specified in the form e1 op= e2, where e1 is a modifiable l-value not of const type and e2 is one of the following:

• An arithmetic type
• A pointer, if op is + or –

The e1 op= e2 form behaves as e1 = e1 op e2, but e1 is evaluated only once.

The following are the compound arithmatic assignment operators in C++ −

Let's have a look at an example using some of these operators −

#include<iostream>
using namespace std;

int main() {
   int a = 3, b = 2;

   a += b;
   cout << a << endl;

   a -= b;
   cout << a << endl;

   a *= b;
   cout << a << endl;

   a /= b;
   cout << a << endl;
   return 0;
}

This will give the output −

5
3
6
3

Note that Compound assignment to an enumerated type generates an error message. If the left operand is of a pointer type, the right operand must be of a pointer type or it must be a constant expression that evaluates to 0. If the left operand is of an integral type, the right operand must not be of a pointer type.