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# Modulo Operator (%) in C/C++ with Examples

• Difficulty Level : Easy
• Last Updated : 06 Jan, 2023

The modulo operator, denoted by %, is an arithmetic operator. The modulo division operator produces the remainder of an integer division.

Syntax: If x and y are integers, then the expression:

`x % y`

Produces the remainder when x is divided by y.

Return Value:

• If y completely divides x, the result of the expression is 0.
• If x is not completely divisible by y, then the result will be the remainder in the range [0, y-1]
• x mod y <x/2 if x>=y and x mod y=x if x<y
• If y is 0, then division by zero is a compile-time error.

Below is the C/C++ program to demonstrate the working of the modulo operator:

## C

 `// Program to illustrate the``// working of modulo operator`` ` `#include `` ` `int` `main(``void``)``{``    ``int` `x, y;`` ` `    ``int` `result;`` ` `    ``x = 3;``    ``y = 4;``      ``// using modulo operator``    ``result = x % y;``    ``printf``(``"%d"``, result);`` ` `    ``result = y % x;``    ``printf``(``"\n%d"``, result);``   ` `    ``// for different values``    ``x = 4;``    ``y = 2;``    ``result = x % y;``    ``printf``(``"\n%d"``, result);`` ` `    ``return` `0;``}`

## C++

 `// C++ Program to demonstrate the``// working of modulo operator``#include `` ` `using` `namespace` `std;`` ` `// Driver code``int` `main(``void``)``{``  ``int` `x, y;``   ` `  ``int` `result;``   ` `  ``x = 3;``  ``y = 4;``   ` `  ``// using modulo operator``  ``result = x % y;``  ``cout << result << endl;`` ` `  ``result = y % x;``  ``cout << result << endl;`` ` `  ``// for different values``  ``x = 4;``  ``y = 2;``   ` `  ``result = x % y;``  ``cout << result;`` ` `  ``return` `0;``}`` ` `//    This code is contributed by Mayank Tyagi`

Output

```3
1
0```

## Restrictions of the modulo operator

The modulo operator has quite some restrictions or limitations.

The % operator cannot be applied to floating-point numbers i.e float or double. If you try to use the modulo operator with floating-point constants or variables, the compiler will produce an error.

Below is the C/C++ program to demonstrate the restrictions of the modulo operator:

## C

 `// Program to illustrate the``// working of modulo operator`` ` `#include `` ` `int` `main(``void``)``{``    ``float` `x, y;`` ` `    ``float` `result;`` ` `    ``x = 2.3;``    ``y = 1.5;``   ` `    ``// modulo for floating point values``    ``result = x % y;``    ``printf``(``"%f"``, result);`` ` `    ``return` `0;``}`

## C++

 `// C++ Program to demonstrate the``// restrictions of modulo operator``#include ``using` `namespace` `std;`` ` `// Driver code``int` `main()``{``  ``float` `x, y;`` ` `  ``x = 2.3;``  ``y = 1.5;``   ` `  ``// modulo for floating point values``  ``result = x % y;``  ``cout << result;`` ` `  ``return` `0;``}`` ` `// This code is contributed by Harshit Srivastava`

Compilation Error:

```Compilation Error in C code :- prog.c: In function 'main':
prog.c:19:16: error:
invalid operands to binary % (have 'float' and 'float')
result = x % y;
^           ```

## Modulo Operator for negative operands

The sign of the result for the modulo operator is machine-dependent for negative operands, as the action takes as a result of underflow or overflow.

Below is the C/C++ program to demonstrate the modulo operator for negative operands:

## C

 `// C Program to illustrate the``// working of the modulo operator``// with negative operands`` ` `#include `` ` `int` `main(``void``)``{``    ``int` `x, y;`` ` `    ``int` `result;`` ` `    ``x = -3;``    ``y = 4;``   ` `    ``// modulo for negative operands``    ``result = x % y;``    ``printf``(``"%d"``, result);`` ` `    ``x = 4;``    ``y = -2;``    ``result = x % y;``    ``printf``(``"\n%d"``, result);`` ` `    ``x = -3;``    ``y = -4;``    ``result = x % y;``    ``printf``(``"\n%d"``, result);`` ` `    ``return` `0;``}`

## C++

 `// C++ Program to demonstrate the``// working of the modulo operator``// for negative operands``#include ``using` `namespace` `std;`` ` `// Driver code``int` `main(``void``)``{``  ``int` `x, y;`` ` `  ``int` `result;`` ` `  ``x = -3;``  ``y = 4;``   ` `  ``// modulo for negative operands``  ``result = x % y;``  ``cout << result << endl;`` ` `  ``x = 4;``  ``y = -2;``  ``result = x % y;``  ``cout << result << endl;`` ` `  ``x = -3;``  ``y = -4;``  ``result = x % y;``  ``cout << result;`` ` `  ``return` `0;``}`` ` `// This code is contributed by Harshit Srivastava`

Output

```-3
0
-3```

Note: Some compilers may show the result of the expression as 1 and other may show -1. It depends on the compiler.

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