Assignment Operators in C++

Assignment operators store a value in the object designated by the left operand. There are two kinds of assignment operations: simple assignment, in which the value of the second operand is stored in the object specified by the first operand, and compound assignment, in which an arithmetic, shift, or bitwise operation is performed prior to storing the result.

Example

Simple assignment operator example −

#include<iostream>
using namespace std;
int main() {
   int i;
   i = 10;    // Simple Assignment
   cout << i;
   return 0;
}

Output

This will give the output −

10

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;
}

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.

Ramu Prasad

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Updated on: 10-Feb-2020

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