What happens when I assign value larger than byte?

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What happens when I assign value larger than byte?

According to official oracle documentation, Byte is

  • Byte: The byte data type is an 8-bit signed two's complement integer. It has a minimum value of -128 and a maximum value of 127 (inclusive). The byte data type can be useful for saving memory in large arrays, where the memory savings actually matters. They can also be used in place of int where their limits help to clarify your code; the fact that a variable's range is limited can serve as a form of documentation

My code is

public class B

    public static void main(String args[])
        byte b;




What happens when I assign value larger than byte. Java compiler will report an error. I will go if I cast this value to byte

byte b = (byte) 128;

I do not understand the output of this program?

For byte type, you only have 8 bits to store the value. You can have only 256 distinct values (2^8 = 256). Java represents negative values with '1' as the highest bit:

-128 (dec) => 10000000 (bit)
-127 (dec) => 10000001 (bit)
-1   (dec) => 11111111 (bit)
0    (dec) => 00000000 (bit)
+1   (dec) => 00000001 (bit)
+127 (dec) => 01111111 (bin)

When you try to set a value that needs more than one byte to store then setting the lowest byte to a byte value happens:

+129 (dec) => 00000000 00000000 00000000 10000001  (int representation)

but 10000001 (bit) is -127 (dec) in byte representation of java type (as described above)

To have a better understanding of the overflow problem in Java, see the article: https://medium.com/@jeanvillete/java-numeric-overflow-underflow-d6b206f96d88

What happens when a char is assigned a value too large to fit in a , In *cp = 1234; , *cp refers to just one byte, the first (lowest-addressed) byte of val2 . In an assignment, the value of the right-hand side is� Instead, if a value is out of range, it is divided by one greater than the largest number of the type, and only the remainder kept. The number 280 is too big to fit in our 1-byte range of 0 to 255. 1 greater than the largest number of the type is 256. Therefore, we divide 280 by 256, getting 1 remainder 24. The remainder of 24 is what is stored.

What happen when we exceed valid range of built-in data types in C , 1) Program to show what happens when we cross range of 'char' : as a result the first number from negative side of the range (i.e. -128) gets assigned to a. And for a Boolean variable anything else than 0 is 1 (or true). For example if we have 1 byte (We can use char and use %d as format specifier to� The answer to this question is that nothing happens because according to C99 standard if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.

Beacause byte datatype is of 1 byte as per name. so, its range is -127 to +126 (i.e capacity is 256). Hence +129 cannot be stored in byte datatype. so +129 is truncated to -127.

Chapter 2: Data types, They differ based on how large of an integer they can store and whether they In some systems, longs will be allocated more bytes than ints. The declaration statement can span multiple lines and you can assign values to the variables too. I can assign the character data by the byte. However, I'm not sure how I would assign a binary value to a certain location in a byte array where the value is greater than 255. I have an unsigned short and want to assign it to a specific location in the byte array. What is the best way to do this? thanks!

Variables -- declaring and assigning values, Variables -- declaring and assigning values. (if you need to write a line of code that is longer than the width of the window you can continue it on the next line by omitting the byte Integer variable allocated only 8 bits (i.e 1 byte) of memory. now if i use byte[] array={1,2,.} and i print array[2] i get the ascii value of ' . ' which is 46 then in that case if i print array[1] shouldnt i get the ascii value of 2,bur it prints the character 2?

4.5 — Unsigned integers, and why to avoid them, Compare this to the 1-byte signed integer range of -128 to 127. Trick question: What happens if we try to store the number 280 (which requires 9 bits Instead, if a value is out of range, it is divided by one greater than the largest isn't assigned a -2, but instead 2 is subtracted from his initial modifier of 1. I have to change value of byte[] dynamically. After first iteration of loop new value should assign to byte[]. Please guide me how I do this. I am doing the project of image processing. I have to match one image to all stored images (in sql database). I used the following piece of code to fetch images.

If the value represented by an integer literal exceeds UInt64.MaxValue, a compiler error CS1021 occurs. If the determined type of an integer literal is int and the value represented by the literal is within the range of the destination type, the value can be implicitly converted to sbyte, byte, short, ushort, uint, or ulong:

  • overflow happens
  • @CarlosHeuberger why is the program output is 127 with a minus sign. should not it be 127 only?
  • my previous comment was for the question before you changed it ("Error incompatible types: possible lossy conversion from int to byte. ") ! Documentation: 5.1.3. Narrowing Primitive Conversion
  • @CarlosHeuberger why is there a minus sign in output. Thanks for documentation link
  • Dr. Lanning: That, Detective, is the "right question." - I, Robot! because the highest bit of byte is set: 129 = 1000 0001 in binary - read the documentation: "this may cause the sign of the resulting value to differ from the sign of the input value"
  • not rounded, truncated - "A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits,"
  • @Sachin Maharjan Thanks for answer! Can you help with how rounding is done in java?
  • @CarlosHeuberger sorry, my bad.
  • Byte range is [-128..127].