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];          


        

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.