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INT_MAX and INT_MIN in C/C++ and Applications

  • Difficulty Level : Easy
  • Last Updated : 03 Sep, 2021

Most of the time, in competitive programming, there is a need to assign the variable, the maximum or minimum value that data type can hold, but remembering such a large and precise number comes out to be a difficult job. Therefore, C++ has certain macros to represent these numbers, so that these can be directly assigned to the variable without actually typing the whole number.

Depending upon the compiler and C++ standard, you may be required to include the header file <limits.h> or <climits> in your C or C++ source code respectively. So it is advisable to include this header file for using the INT_MAX, INT_MIN macros. For further reading on this header file, refer to this article

INT_MAX is a macro that specifies that an integer variable cannot store any value beyond this limit. 
INT_MIN specifies that an integer variable cannot store any value below this limit.

Values of INT_MAX and INT_MIN may vary
from compiler to compiler. Following are
typical values in a compiler where integers
are stored using 32 bits.

Value of INT_MAX is +2147483647.
Value of INT_MIN is -2147483648.


// C++ program to print values of INT_MAX
// and INT_MIN
#include <bits/stdc++.h>
using namespace std;
int main()
    cout << INT_MAX << endl;
    cout << INT_MIN;
    return 0;


// C program to print values of INT_MAX
// and INT_MIN
// we have to include limits.h for results in C
#include <limits.h>
#include <stdio.h>
int main()
    printf("%d\n", INT_MAX);
    printf("%d", INT_MIN);

Applications of INT_MAX and INT_MIN : 

1. Check for Integer overflow : 


// C++ code to check for Integer overflow while
// adding 2 numbers
#include <bits/stdc++.h>
// Function to check integer overflow
int check_overflow(int num1, int num2)
    // Checking if addition will cause overflow
    if (num1 > INT_MAX - num2)
        return -1;
    // No overflow occured
        return num1 + num2;
// Driver code
int main()
    // The sum of these numbers will equal INT_MAX
    // If any of them is incremented by 1, overflow
    // will occur
    int num1 = 2147483627;
    int num2 = 20;
    // Result is -1 if overflow occurred
    // Stores the sum, otherwise
    int result = check_overflow(num1, num2);
    // Overflow occurred
    if (result == -1)
        std::cout << "Integer overflow occurred";
    // No overflow
        std::cout << result;

Similarly, we can check for overflow while subtracting 2 numbers using INT_MIN.

2. Computing MIN in an array with large elements 
We usually assign a high value to MIN to compute minimum value in an array. But if an array has large elements, we must assign the highest possible value to the array.

Below is the C++ implementation : 


// C++ code to compute MIN element
#include <bits/stdc++.h>
// Function to compute minimum element in array
int compute_min(int arr[], int n)
    // Assigning highest value
    int MIN = INT_MAX;
    // Traversing and updating MIN
    for (int i = 0; i < n; i++)
        MIN = std::min(MIN, arr[i]);
    // Printing MIN element
    std::cout << MIN;
// Driver code
int main()
    // array with MIN to compute
    int arr[] = { 2019403813, 2147389580, 2145837140,
                  2108938594, 2112076334 };
    // size of array
    int n = sizeof(arr) / sizeof(arr[0]);
    // Function call to compute MIN
    compute_min(arr, n);

Similarly, MAX can be found in an array of large numbers using INT_MIN.

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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