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 operationspublic 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 operationsusing 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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