After knowing the Java rules of Data Type Casting (Type Conversion), let us cast byte to int. The byte takes 1 byte of memory and int takes 4 bytes of memory. Assignment 1 byte value to 4 bytes is done implicitly by the JVM.
byte –> short –> int –> long –> float –> double
The left-side value can be assigned to any right-side value and is done implicitly. The reverse requires explicit casting.
Example:
short x = 10;
int y = x;
byte x = 10;
short y = x;
Following program on byte to int explains the syntax of implicit casting in Java.
public class Conversions
{
public static void main(String args[])
{
byte b1 = 10; // 1 byte
int i1 = b1; // one byte to int 4 bytes
System.out.println("byte value: " + b1); // prints 10
System.out.println("Converted int value: " + i1); // prints 10
}
}

Output screenshot of byte to int Java
1 byte value b1 is assigned to 4 bytes i1 and is type casted implicitly. This is known as implicit casting or widening conversion.
View all for 65 types of Conversions
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public static int byteToint(byte byt) {
int b = 0xff;
b = b & (byt);
return b;
}
Hi sir,
Thanks for providing java concepts with detailed explanation. Here my question is:
int a=130;
byte b=(byte) a;
Ans. -126
I can’t understand this program……. please explain this in detail……
Hi sir,
Thanks for providing java concepts with detailed explanation. Here my question is: The range of java byte is -128 to 127. If we given the value like byte b=129; we are getting an error called “possible loss of precision”. But at the same time it allows the below code:
int a=143;
byte b=(byte) a;
How it allows byte range more than 127?
You are casting it. Observe the output. It is not 143.