Max Value of Integer

Explanation on Integer.MAX_VALUE and Integer.MIN_VALUE to find min and max value in an array

but as for this method, I don't understand the purpose of Integer.MAX_VALUE and Integer.MIN_VALUE.

By starting out with smallest set to Integer.MAX_VALUE and largest set to Integer.MIN_VALUE, they don't have to worry later about the special case where smallest and largest don't have a value yet. If the data I'm looking through has a 10 as the first value, then numbers[i]<smallest will be true (because 10 is < Integer.MAX_VALUE) and we'll update smallest to be 10. Similarly, numbers[i]>largest will be true because 10 is > Integer.MIN_VALUE and we'll update largest. And so on.

Of course, when doing this, you must ensure that you have at least one value in the data you're looking at. Otherwise, you end up with apocryphal numbers in smallest and largest.

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Note the point Onome Sotu makes in the comments:

...if the first item in the array is larger than the rest, then the largest item will always be Integer.MIN_VALUE because of the else-if statement.

Which is true; here's a simpler example demonstrating the problem (live copy):

public class Example
{
    public static void main(String[] args) throws Exception {
        int[] values = {5, 1, 2};
        int smallest = Integer.MAX_VALUE;
        int largest  = Integer.MIN_VALUE;
        for (int value : values) {
            if (value < smallest) {
                smallest = value;
            } else if (value > largest) {
                largest = value;
            }
        }
        System.out.println(smallest + ", " + largest); // 1, 2 -- WRONG
    }
}

To fix it, either:

1. Don't use else, or

2. Start with smallest and largest equal to the first element, and then loop the remaining elements, keeping the else if.

Here's an example of that second one (live copy):

public class Example
{
    public static void main(String[] args) throws Exception {
        int[] values = {5, 1, 2};
        int smallest = values[0];
        int largest  = values[0];
        for (int n = 1; n < values.length; ++n) {
            int value = values[n];
            if (value < smallest) {
                smallest = value;
            } else if (value > largest) {
                largest = value;
            }
        }
        System.out.println(smallest + ", " + largest); // 1, 5
    }
}

What is max value can be entered as string for Integer.parseInt(String)

2147483647 is the maximum value parseInt() can parse without giving an exception. 1000000000000 > 2147483647(=Integer.MAX_VALUE, Max value of int).1000000000000 is no longer be an int value(can't represent as an int without integer overflow). So partInt() will give you NumberFormatException

You can use Long.parseLong() for larger values like here.

Eg:

Long.parseLong("1000000000000")

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;
}

Is there a constant for max value for integer type?

VBA does not provide a MAXINT constant. But you can derive that value easily:

MAXINT = (2 ^ 15) -1
Debug.Print MAXINT
 32767

Or you could define it as a Public constant with this in the Declarations section of a standard module:

Public Const MAXINT As Integer = (2 ^ 15) - 1

Then MAXINT would be available for the rest of your VBA code in that application.

And for Long Integer, the maximum value is ...

MAXLONG = (2 ^ 31) -1
Debug.Print MAXLONG
 2147483647 

What is the maximum value for an int32?

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

Why Overflow in MAX_VALUE in Int makes it negative but keeps it positive in Byte?

When you add two numbers together, the operands undergo binary numeric promotion.

The operands are first unboxed, if necessary; then the first matching rule applies:

• If one operand is a double, the other is widened to double
• If one operand is a float, the other is widened to a float
• If one operand is a long, the other is widened to long
• Otherwise, both are widened to int.

Since 1 is an int (because it's an int literal), adding it to a byte means that the last rule applies, so the byte is widened to an int; and the result of the addition of two ints is an int.

Because 128 is within the range of int, no overflow occurs.

Note that the rules go no narrower than int, so even adding two bytes will result in an int:

System.out.println(Byte.MAX_VALUE + (byte) 1); // 128

Note also that if you used a pre/post-increment:

byte myMaxByteValue = Byte.MAX_VALUE;
++myMaxByteValue;

then the value of maxByte would be -128. This is because pre-increment is equivalent to:

myMaxByteValue = (byte) (myMaxByteValue + 1);

i.e. there is an implicit cast back to the variable type.

What's the maximum value for an int in PHP?

From the PHP manual:

The size of an integer is
platform-dependent, although a maximum
value of about two billion is the
usual value (that's 32 bits signed).
PHP does not support unsigned
integers. Integer size can be
determined using the constant
PHP_INT_SIZE, and maximum value using
the constant PHP_INT_MAX since PHP
4.4.0 and PHP 5.0.5.

64-bit platforms usually have a maximum value of about 9E18, except on Windows prior to PHP 7, where it was always 32 bit.

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.

The maximum value for an int type in Go

https://groups.google.com/group/golang-nuts/msg/71c307e4d73024ce?pli=1

The germane part:

Since integer types use two's complement arithmetic, you can infer the
min/max constant values for int and uint. For example,

const MaxUint = ^uint(0) 
const MinUint = 0 
const MaxInt = int(MaxUint >> 1) 
const MinInt = -MaxInt - 1

As per @CarelZA's comment:

uint8  : 0 to 255 
uint16 : 0 to 65535 
uint32 : 0 to 4294967295 
uint64 : 0 to 18446744073709551615 
int8   : -128 to 127 
int16  : -32768 to 32767 
int32  : -2147483648 to 2147483647 
int64  : -9223372036854775808 to 9223372036854775807

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