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 iff they have opposite sign. In other words, XOR of x and y will be negative number number iff x and y have opposite signs. The following code use this logic.

C++

// C++ Program to Detect  
// if two integers have opposite signs. 
#include<stdbool.h> 
#include<stdio.h> 
  
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"
  
# This article is contributed by Prerna Saini. 

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. 

PHP

<?php 
// PHP Program to Detect if two  
// integers have opposite signs. 
  
function oppositeSigns($x, $y) 
{ 
    return (($x ^ $y) < 0); 
} 
  
    // Driver Code 
    $x = 100; 
    $y = -100; 
    if (oppositeSigns($x, $y) == true) 
    echo ("Signs are opposite"); 
    else
    echo ("Signs are not opposite"); 
      
// This code is contributed by vt_m.  
?> 

Output:

Signs are opposite

Source: Detect if two integers have opposite signs

We can also solve this by using two comparison operators. See the following code.

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

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.

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

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.

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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