# Detect if two integers have opposite signs

Given two signed integers, write a function that returns true if the signs of given integers are different, otherwise false. For example, the function should return true -1 and +100, and should return false for -100 and -200. The function should not use any of the arithmetic operators.
Let the given integers be x and y. The sign bit is 1 in negative numbers, and 0 in positive numbers. The XOR of x and y will have the sign bit as 1 if they have opposite sign. In other words, XOR of x and y will be negative number if x and y have opposite signs. The following code use this logic.

## C++

 // C++ Program to Detect // if two integers have opposite signs. #include using namespace std;   bool oppositeSigns(int x, int y) {     return ((x ^ y) < 0); }   int main() {     int x = 100, y = -100;     if (oppositeSigns(x, y) == true)     cout << "Signs are opposite";     else     cout << "Signs are not opposite";     return 0; }   // this code is contributed by shivanisinghss2110

## C

 // C Program to Detect  // if two integers have opposite signs. #include #include     bool oppositeSigns(int x, int y) {     return ((x ^ y) < 0); }     int main() {     int x = 100, y = -100;     if (oppositeSigns(x, y) == true)        printf ("Signs are opposite");     else       printf ("Signs are not opposite");     return 0; }

## Java

 // Java Program to Detect  // if two integers have opposite signs.     class GFG {         static boolean oppositeSigns(int x, int y)     {         return ((x ^ y) < 0);     }             public static void main(String[] args)     {         int x = 100, y = -100;         if (oppositeSigns(x, y) == true)             System.out.println("Signs are opposite");         else             System.out.println("Signs are not opposite");     } }     // This code is contributed by prerna saini.

## Python3

 # Python3 Program to Detect  # if two integers have  # opposite signs. def oppositeSigns(x, y):     return ((x ^ y) < 0);     x = 100 y = 1     if (oppositeSigns(x, y) == True):     print ("Signs are opposite") else:     print ("Signs are not opposite")     #

## C#

 // C# Program to Detect  // if two integers have  // opposite signs. using System;     class GFG {         // Function to detect signs     static bool oppositeSigns(int x, int y)     {         return ((x ^ y) < 0);     }             // Driver Code     public static void Main()     {         int x = 100, y = -100;         if (oppositeSigns(x, y) == true)             Console.Write("Signs are opposite");         else             Console.Write("Signs are not opposite");     } }     // This code is contributed by Nitin Mittal.



## Javascript



Output

Signs are opposite

Time Complexity: O(1)
Auxiliary Space: O(1)

Source: Detect if two integers have opposite signs
We can also solve this by using two comparison operators. See the following code.

## CPP

 bool oppositeSigns(int x, int y) {     return (x < 0)? (y >= 0): (y < 0); }

## C

 //C program to detect whether two integers //have different sign or not   #include #include   bool oppositeSigns(int x, int y) {     return (x < 0)? (y >= 0): (y < 0); }   //this code is contributed by shruti456rawal

## Java

 class GFG{ static boolean oppositeSigns(int x, int y) {     return (x < 0)? (y >= 0): (y < 0); } }   // This code contributed by umadevi9616

## Python3

 def oppositeSigns(x, y):       return (y >= 0) if (x < 0) else (y < 0);   # This code is contributed by shivanisingjss2110

## C#

 using System; class GFG{ static boo oppositeSigns(int x, int y) {     return (x < 0)? (y >= 0): (y < 0); } }   // This code contributed by shivanisinghss2110

## Javascript



Time Complexity: O(1)
Auxiliary Space: O(1)

The first method is more efficient. The first method uses a bitwise XOR and a comparison operator. The second method uses two comparison operators and a bitwise XOR operation is more efficient compared to a comparison operation.
We can also use following method. It doesn’t use any comparison operator. The method is suggested by Hongliang and improved by gaurav.

## CPP

 bool oppositeSigns(int x, int y) {     return ((x ^ y) >> 31); }

## C

 //C program to detect whether two integers //have different sign or not   #include #include   bool oppositeSigns(int x, int y) {     return ((x ^ y) >> 31); }   //this code is contributed by shruti456rawal

## Java

 import java.io.*;   class GFG { static boolean oppositeSigns(int x, int y) {     return ((x ^ y) >> 31); }        }   // This code is contributed by shivanisinghss2110

## Python3

 def oppositeSigns(x, y):       return ((x ^ y) >> 31)     # this code is contributed by shivanisinghss2110

## C#

 using System;   class GFG { static bool oppositeSigns(int x, int y) {     return ((x ^ y) >> 31); }        }   // This code is contributed by shivanisinghss2110

## Javascript



Time Complexity: O(1)
Auxiliary Space: O(1)

The function is written only for compilers where size of an integer is 32 bit. The expression basically checks sign of (x^y) using bitwise operator ‘>>’. As mentioned above, the sign bit for negative numbers is always 1. The sign bit is the leftmost bit in binary representation. So we need to checks whether the 32th bit (or leftmost bit) of x^y is 1 or not. We do it by right shifting the value of x^y by 31, so that the sign bit becomes the least significant bit. If sign bit is 1, then the value of (x^y)>>31 will be 1, otherwise 0.

## C++

 // C++ Program to detect if two integers have opposite // signs. #include using namespace std;   bool oppositeSigns(int x, int y) {     long long product = 1ll * x * y;     return (product < 0); }   int main() {     int x = 100, y = -100;     if (oppositeSigns(x, y) == true)         cout << "Signs are opposite";     else         cout << "Signs are not opposite";     return 0; }   // This code is contributed by Aditya Kumar (adityakumar129)

## C

 // C Program to detect // if two integers have opposite signs. #include #include   bool oppositeSigns(int x, int y) {     long long product = 1ll * x * y;     return (product < 0); }   int main() {     int x = 100, y = -100;     if (oppositeSigns(x, y) == true)         printf("Signs are opposite");     else         printf("Signs are not opposite");     return 0; }   // This code is contributed by Aditya Kumar (adityakumar129)

## Java

 // Java program for the above approach import java.util.*;   class GFG {     static boolean oppositeSigns(int x, int y)   {     long product = 1*x*y;     return (product<0);   }     // Driver Code   public static void main(String[] args)   {     int x = 100, y = -100;     if (oppositeSigns(x, y) == true)       System.out.print( "Signs are opposite");     else       System.out.print("Signs are not opposite");   } }   // This code is contributed by sanjoy_62.

## Python3

 # Python Program to detect # if two integers have opposite signs.   def oppositeSigns(x,y):     product = x*y     return (product<0)   # driver code x = 100 y = -100 if(oppositeSigns(x, y) == True):   print("Signs are opposite")  else :   print("Signs are not opposite")     # this code is contributed by shinjanpatra

## C#

 // C# program for the above approach using System; class GFG{     static bool oppositeSigns(int x, int y)   {     long product = 1*x*y;     return (product<0);   }   // Driver Code public static void Main(String[] args) {     int x = 100, y = -100;     if (oppositeSigns(x, y) == true)       Console.WriteLine( "Signs are opposite");     else       Console.WriteLine("Signs are not opposite"); } }   // This code is contributed by avijitmondal1998.

## Javascript

 // JavaScript Program to detect // if two integers have opposite signs.   function oppositeSigns(x,y) {     const product = Number(x)*Number(y);     return (product<0); }   // driver code let x = 100, y = -100; if(oppositeSigns(x, y) == true) {     console.log("Signs are opposite"); } else console.log("Signs are not opposite");     // this code is contributed by shinjanpatra

Output

Signs are opposite

Approach: The basic approach is to calculate the product of the two integers, and as we know, two integers having opposite signs will always produce a negative integer, we need to just find out whether the product is negative or not.

Time Complexity: O(1)
Auxiliary Space: O(1)

Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.

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