Compound Assignment Operators in C++

The compound 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 assignment operators in C++ −

Example

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;
}

Output

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.

Govinda Sai

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Updated on: 11-Feb-2020

3K+ Views

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