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Computing INT_MAX and INT_MIN with Bitwise operations

Last Updated : 28 Jun, 2022
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Prerequisites : 
INT_MAX and INT_MIN in C/C++ and Applications. 
Arithmetic shift vs Logical shift
Suppose you have a 32-bit system : 
The INT_MAX would be 01111111111111111111111111111111 and INT_MIN would be 10000000000000000000000000000000. 0 & 1 in most-significant bit position representing the sign bit respectively.
Computing INT_MAX and INT_MIN In C/C++ : 
The number 0 is represented as 000…000(32 times). 
 

  • We compute the NOT of 0 to get a number with 32 1s. This number is not equal to INT_MAX because the sign bit is 1, i.e. negative number.
  • Now, a right shift of this number will produce 011…111 which is INT_MAX.
  • INT_MIN is NOT of INT_MAX.

Note : 
0 should be taken as unsigned int.
Reason : 
If 0 is signed, during Step 2, right shift of 111..111 will yield 111…111. This is because arithmetic right shift preserves the sign of the number. 
In Java, we have the feature of logical right shift available to us. 
 
 

C++




// CPP code to compute INT_MAX and INT_MIN using
// bitwise operations
#include <bits/stdc++.h>
using namespace std;
 
void printMinMaxValues()
{
    // 0 saved as unsigned int
    unsigned int max = 0;
 
    // Computing NOT of 0
    max = ~max;
     
    // 1 time arithmetic right shift
    max = max >> 1;
     
    // Computing INT_MIN
    int min = max;
     
    // INT_MIN = ~INT_MAX
    min = ~min;
     
    // Printing the result
    cout << "INT_MAX : " << max
        << " INT_MIN : " << min;
}
 
// Driver code
int main()
{
    printMinMaxValues();
    return 0;
}


Java




// Java code to compute INT_MAX and INT_MIN using
// bitwise operations
public class Solution
{
    static void printMinMaxValues()
    {
        int max = 0;
     
        // Computing NOT of 0
        max = ~max;
 
        // 1 time logical right shift for INT_MAX
        max = max >>> 1;
         
        // Computing INT_MIN
        int min = max;
         
        // INT_MIN = ~INT_MAX
        min = ~max;
     
        // Printing the result
        System.out.println("INT_MAX " + max +
                           " INT_MIN " + min);
    }
 
    public static void main(String[] args)
    {
        printMinMaxValues();
    }
}


Python3




# Python3 code to compute INT_MAX and INT_MIN using
# bitwise operations
def printMinMaxValues():
  
    # 0 saved as unsigned int
    max = 0
 
    # Computing NOT of 0
    #to signed integer to unsigned integer
    #in Python3, add 1 << 32 to the integer
    max = ~max + (1 << 32)
     
    # 1 time arithmetic right shift
    max = max >> 1
     
    # Computing INT_MIN
    min = max;
     
    # INT_MIN = ~INT_MAX
    min = ~min
     
    # Printing the result
    print("INT_MAX :", max, "INT_MIN :", min)
 
# Driver code
printMinMaxValues()
 
# This code is contributed by phasing17


C#




// C# code to compute INT_MAX and INT_MIN using
// bitwise operations
 
using System;
 
public class GFG
{
    static void printMinMaxValues()
    {
        // 0 saved as unsigned int
        uint max = 0;
     
        // Computing NOT of 0
        max = ~max;
         
        // 1 time arithmetic right shift
        max = max >> 1;
         
        // Computing INT_MIN
        int min = (int)max;
         
        // INT_MIN = ~INT_MAX
        min = ~min;
     
        // Printing the result
        Console.WriteLine("INT_MAX " + max +
                           " INT_MIN " + min);
    }
     
    //Driver code
    public static void Main(string[] args)
    {
        //Function call
        printMinMaxValues();
    }
}
 
//This code is contributed by phasing17


Javascript




<script>
 
// Javascript code to compute INT_MAX and INT_MIN using
// bitwise operations
 
    function printMinMaxValues()
    {
        let max = 0;
       
        // Computing NOT of 0
        max = ~max;
   
        // 1 time logical right shift for INT_MAX
        max = max >>> 1;
           
        // Computing INT_MIN
        let min = max;
           
        // INT_MIN = ~INT_MAX
        min = ~max;
       
        // Printing the result
        document.write("INT_MAX - " + max +
                           ", INT_MIN " + min);
    }
     
// driver program
     
        printMinMaxValues();
 
// This code is contributed by code_hunt.
</script>


Output: 
 

INT_MAX 2147483647 INT_MIN -2147483648

Time Complexity – O(1)

Space Complexity – O(1)

Asked in : Google 

 



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