In C, the language itself does not determine the representation of certain datatypes. It can vary from machine to machine, on embedded systems the int can be 16 bit wide, though usually it is 32 bit.

The only requirement is that short int <= int <= long int by size. Also, there is a recommendation that int should represent the native capacity of the processor.

All types are signed. The unsigned modifier allows you to use the highest bit as part of the value (otherwise it is reserved for the sign bit).

Here's a short table of the possible values for the possible data types:

          width                     minimum                         maximum
signed    8 bit                        -128                            +127
signed   16 bit                     -32 768                         +32 767
signed   32 bit              -2 147 483 648                  +2 147 483 647
signed   64 bit  -9 223 372 036 854 775 808      +9 223 372 036 854 775 807
unsigned  8 bit                           0                            +255
unsigned 16 bit                           0                         +65 535
unsigned 32 bit                           0                  +4 294 967 295
unsigned 64 bit                           0     +18 446 744 073 709 551 615

In Java, the Java Language Specification determines the representation of the data types.

The order is: byte 8 bits, short 16 bits, int 32 bits, long 64 bits. All of these types are signed, there are no unsigned versions. However, bit manipulations treat the numbers as they were unsigned (that is, handling all bits correctly).

The character data type char is 16 bits wide, unsigned, and holds characters using UTF-16 encoding (however, it is possible to assign a char an arbitrary unsigned 16 bit integer that represents an invalid character codepoint)

          width                     minimum                         maximum

SIGNED
byte:     8 bit                        -128                            +127
short:   16 bit                     -32 768                         +32 767
int:     32 bit              -2 147 483 648                  +2 147 483 647
long:    64 bit  -9 223 372 036 854 775 808      +9 223 372 036 854 775 807

UNSIGNED
char     16 bit                           0                         +65 535
Answer from gaborsch on Stack Overflow
Top answer
1 of 11
450

In C, the language itself does not determine the representation of certain datatypes. It can vary from machine to machine, on embedded systems the int can be 16 bit wide, though usually it is 32 bit.

The only requirement is that short int <= int <= long int by size. Also, there is a recommendation that int should represent the native capacity of the processor.

All types are signed. The unsigned modifier allows you to use the highest bit as part of the value (otherwise it is reserved for the sign bit).

Here's a short table of the possible values for the possible data types:

          width                     minimum                         maximum
signed    8 bit                        -128                            +127
signed   16 bit                     -32 768                         +32 767
signed   32 bit              -2 147 483 648                  +2 147 483 647
signed   64 bit  -9 223 372 036 854 775 808      +9 223 372 036 854 775 807
unsigned  8 bit                           0                            +255
unsigned 16 bit                           0                         +65 535
unsigned 32 bit                           0                  +4 294 967 295
unsigned 64 bit                           0     +18 446 744 073 709 551 615

In Java, the Java Language Specification determines the representation of the data types.

The order is: byte 8 bits, short 16 bits, int 32 bits, long 64 bits. All of these types are signed, there are no unsigned versions. However, bit manipulations treat the numbers as they were unsigned (that is, handling all bits correctly).

The character data type char is 16 bits wide, unsigned, and holds characters using UTF-16 encoding (however, it is possible to assign a char an arbitrary unsigned 16 bit integer that represents an invalid character codepoint)

          width                     minimum                         maximum

SIGNED
byte:     8 bit                        -128                            +127
short:   16 bit                     -32 768                         +32 767
int:     32 bit              -2 147 483 648                  +2 147 483 647
long:    64 bit  -9 223 372 036 854 775 808      +9 223 372 036 854 775 807

UNSIGNED
char     16 bit                           0                         +65 535
2 of 11
84

In C, the integer(for 32 bit machine) is 32 bit and it ranges from -32768 to +32767.

Wrong. 32-bit signed integer in 2's complement representation has the range -231 to 231-1 which is equal to -2,147,483,648 to 2,147,483,647.

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reddit.com › r/learnjava › beginner here; please help with explaining why integer.max_value and integer.min_value is assigned as values in this code.
r/learnjava on Reddit: Beginner here; please help with explaining why Integer.MAX_VALUE and Integer.MIN_VALUE is assigned as values in this code.
July 4, 2022 -

I was searching for a way to find the min and max value within a number of integers and I came across this code:

Scanner in = new Scanner(System.in);

int maxNum = Integer.MIN_VALUE;
int minNum = Integer.MAX_VALUE;
while (scanner.hasNextInt()) {
  int num = scanner.nextInt();
  maxNum = Math.max(maxNum, num);
  minNum = Math.min(minNum, num);
}

System.out.println("The maximum number: " + maxNum);
System.out.println("The minimum number: " + minNum);

I am struggling to wrap my mind around Integer.MAX_VALUE and Integer.MIN_VALUE. Why are they assigned as values and how does it work with the Math. min & max methods. Why is the variable maxNum assigned the value Integer.MIN_VALUE and minNum assigned the value Integer.MAX_VALUE?

Thanks.

Top answer
1 of 3
3
Think of only one, you are looking for the lowest number in an array, you take the first number and compare it to your minValue, if it is smaller you choose that as new minValue and try the next. If you start minValue as the absolute MIN_VALUE, it will never find a smaller number in your array. SO you should start minValue as higher as possible and maxValue as lover. (Or you can just initiate them as the first number in your array)
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1
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W3Schools
w3schools.com › java › ref_math_max.asp
Java Math max() Method
System.out.println(Math.max(2.0, ... 22)); System.out.println(Math.max(96L, 2048L)); ... The max() method returns the number with the highest value from a pair of numbers....
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44

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.


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
    }
}
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19

Instead of initializing the variables with arbitrary values (for example int smallest = 9999, largest = 0) it is safer to initialize the variables with the largest and smallest values representable by that number type (that is int smallest = Integer.MAX_VALUE, largest = Integer.MIN_VALUE).

Since your integer array cannot contain a value larger than Integer.MAX_VALUE and smaller than Integer.MIN_VALUE your code works across all edge cases.

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What is Integer.MAX_VALUE?
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w3docs.com › java
max value of integer
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Java Integer max() Method
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Java Integer max() Method - Studytonight
Here, one positive and one negative ... { public static void main(String[] args) { int x = 5485; int y = -3242; int z = -5645; // print the larger number between x and y System.out.println("Greater number is " + Integer.max(x, y)); // print the ...