Modulo Operator (%) in C/C++ with Examples
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 <stdio.h> 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 <iostream> 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 |
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 <stdio.h> 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 <iostream> 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 <stdio.h> 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 <iostream> 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 |
-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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