Maximum Value of Int

What is the maximum value for an int32?

It's 2,147,483,647. Easiest way to memorize it is via a tattoo.

java maximum value of int

As per my knowledge ++i should increment the value of i by 1 and throw exception, because 2147483648 is not allowed in int.

Your understanding is incorrect. Integer arithmetic overflows silently in Java.

From section 15.18.2 of the JLS:

If an integer addition overflows, then the result is the low-order bits of the mathematical sum as represented in some sufficiently large two's-complement format. If overflow occurs, then the sign of the result is not the same as the sign of the mathematical sum of the two operand values.

Highest value of int type in C/C++. Why can't I calculate it correctly in this example?

Signed Integer Overflow.

It's undefined behavior. So getting 0 is a perfectly valid undefined result.

The largest integer in an int.

It's 2^31 - 1, not 2^31. Note that INT_MAX is an odd number.

Two's Complement

Most systems that you will ever encounter won't use Signed Magnitude to represent signed numbers. Instead they will use Two's Complement.

How can I write a program to find maximum value of integer variable

As ı know the maximum value of a integer variable is 2147483647

That's a wrong assumption. The maximum value of an int can vary across systems. For example, on a 16-bit machine maximum value of int is not 2147483647.

You don't need to find it yourself. There are pre-defined macros which you can use from <limits.h>. For example, INT_MAX represents the maximum value an int can hold.

Solution to go past the max value of integer?

The simple solution is to use long, however you can use int for larger numbers (and save work) by instead calculating fac(a)/fac(b) which is more efficient.

public static void main(String... args) {
    int[][] bino = new int[15][]; //Create 2d array for pascal pyramid
    for (int i = 0; i < bino.length; i++) {
        bino[i] = new int[i + 1];
        for (int j = 0; j < i + 1; j++) {
            bino[i][j] = nOverk(i, j);
        }
    }
}

static int nOverk(int n, int k) {
    int min = Math.min(k, n - k);
    int max = Math.max(k, n - k);
    return fac(n, max) / fac(min, 1);
}

static int fac(int hi, int lo) {
    if (hi == 0 || hi == 1)
        return 1;

    int res = 1;
    for (int i = lo + 1; i <= hi; i++)
        res *= i;
    return res;
}

maximum value of int

In C++:

#include <limits>

then use

int imin = std::numeric_limits<int>::min(); // minimum value
int imax = std::numeric_limits<int>::max();

std::numeric_limits is a template type which can be instantiated with other types:

float fmin = std::numeric_limits<float>::min(); // minimum positive value
float fmax = std::numeric_limits<float>::max();

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In C:

#include <limits.h>

then use

int imin = INT_MIN; // minimum value
int imax = INT_MAX;

or

#include <float.h>

float fmin = FLT_MIN;  // minimum positive value
double dmin = DBL_MIN; // minimum positive value

float fmax = FLT_MAX;
double dmax = DBL_MAX;

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