Converting from byte to int in java

Question

I have generated a secure random number, and put its value into a byte. Here is my code.

SecureRandom ranGen = new SecureRandom(); byte[] rno = new byte[4];  ranGen.nextBytes(rno); int i = rno[0].intValue(); 

But I am getting an error :

 byte cannot be dereferenced 

Answer 1

Your array is of byte primitives, but you're trying to call a method on them.

You don't need to do anything explicit to convert a byte to an int, just:

int i=rno[0]; 

...since it's not a downcast.

Note that the default behavior of byte-to-int conversion is to preserve the sign of the value (remember byte is a signed type in Java). So for instance:

byte b1 = -100; int i1 = b1; System.out.println(i1); // -100 

If you were thinking of the byte as unsigned (156) rather than signed (-100), as of Java 8 there's Byte.toUnsignedInt:

byte b2 = -100; // Or `= (byte)156;` int = Byte.toUnsignedInt(b2); System.out.println(i2); // 156 

Prior to Java 8, to get the equivalent value in the int you'd need to mask off the sign bits:

byte b2 = -100; // Or `= (byte)156;` int i2 = (b2 & 0xFF); System.out.println(i2); // 156 

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Just for completeness #1: If you did want to use the various methods of Byte for some reason (you don't need to here), you could use a boxing conversion:

Byte b = rno[0]; // Boxing conversion converts `byte` to `Byte` int i = b.intValue(); 

Or the Byte constructor:

Byte b = new Byte(rno[0]); int i = b.intValue(); 

But again, you don't need that here.

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Just for completeness #2: If it were a downcast (e.g., if you were trying to convert an int to a byte), all you need is a cast:

int i; byte b;  i = 5; b = (byte)i; 

This assures the compiler that you know it's a downcast, so you don't get the "Possible loss of precision" error.