Q:

Converting from byte to int in java

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

A:

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

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.

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.

A:

byte b = (byte)0xC8;
int v1 = b;       // v1 is -56 (0xFFFFFFC8)
int v2 = b & 0xFF // v2 is 200 (0x000000C8)

Most of the time v2 is the way you really need.

A:

if you want to combine the 4 bytes into a single int you need to do
int i= (rno[0]<<24)&0xff000000|
       (rno[1]<<16)&0x00ff0000|
       (rno[2]<< 8)&0x0000ff00|
       (rno[3]<< 0)&0x000000ff;

I use 3 special operators | is the bitwise logical OR & is the logical AND and << is the left shift
in essence I combine the 4 8-bit bytes into a single 32 bit int by shifting the bytes in place and ORing them together 
I also ensure any sign promotion won't affect the result with & 0xff

