Arithmetic Operators Examples:
C offers five binary operators and two unary operators.

Unary Operators:

++ This operator increases the value of the operand by 1.
– – This operator decreases the value of the operand by 1.

The code snippet below shows how we can implement unary operators:

 
#include <stdio.h>

int  main(void) {

   int  operand1 = 10;

   int  operand2 = 5;

   printf("Before using unary operators: \n");

   printf("Value of operand1: %d \n", operand1);

   printf("Value of operand2: %d \n", operand2);

   operand1++; //incrementing operand1 by 1

   operand2--; //decrementing operand2 by 1

   printf("After using unary operators: \n");

   printf("Value of operand1: %d \n", operand1);

   printf("Value of operand2: %d \n", operand2);

   return  0;

}

The output of the above code snippet will be:

 
Before using unary operators:

Value of operand1: 10

Value of operand2: 5

After using unary operators:

Value of operand1: 11

Value of operand2: 4

  • We have created two integer variables operand1 and operand2.
  • We increment the value of operand1 by 1.
  • We decrement the value of operand2 by 1.

Binary operators:

+ Adds two operands
Subtracts the second operand from the first operand
* Multiplies two operands
/ Divides the first operand by the second operand
% Gives the remainder left after dividing the first operand by the second operand

The code snippet below shows how we can implement binary operators:

 
#include <stdio.h>


int  main(void) {

   int  operand1 = 10;

   int  operand2 = 5;

   int  sum = operand1 + operand2;

   int  difference = operand1 - operand2;

   int  product = operand1 * operand2;

   int  quotient = operand1 / operand2;

   int  remainder = operand1 % operand2;

   printf("Value of operand1: %d \n", operand1);

   printf("Value of operand2: %d \n", operand2);

   printf("Results: \n");

   printf("operand1 + operand2: %d \n", sum);

   printf("operand1 - operand2: %d \n", difference);

   printf("operand1 * operand2: %d \n", product);

   printf("operand1 / operand2: %d \n", quotient);

   printf("operand1 %% operand2: %d \n", remainder);
   
   return  0;

}

The output of the above code snippet is:

 
Value of operand1: 10

Value of operand2: 5

Results:

operand1 + operand2: 15

operand1 - operand2: 5

operand1 * operand2: 50

operand1 / operand2: 2

operand1 % operand2: 0

  • We have created two integer variables operand1 and operand2.
  • We have used all binary operators i.e. +, -, *, / and %, on the two operands and we have got the expected results.

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CopyRight © 2019