how to convert byte[] to int in Java?

[Jumpem]

Is there a straight forward way to convert two bytes into it's integer equivalent?

 

[CTho9305]

Does java support unions?
In C:

union foo {
byte java[2];
int sucks;
}

void myfunc()
{
union foo bar;
int stuf;
bar.java[0] = 0;
bar.java[1] = -1;
stuff = bar.sucks;
}

edit: by the way, on most systems, int = 4 bytes, short = 2 bytes.

 

[Kilrsat]

int temp = new Byte(tmpByte[0]).intValue();
int temp2 = new Byte(tmpByte[1]).intValue();

Find more byte references:
http://java.sun.com/j2se/1.4.2/docs/api/java/lang/Byte.html

Bah, just realized, you want to combine two bytes into 1 int, hold on a min....

Ok, two bytes to 1 32-bit integer:

int temp = ((int)tmpByte[0] << 8) + (tmpByte[1] & 0xff);

Three bytes:

int temp = ((int)tmpByte[0] << 16 ) + ((tmpByte[1] & 0xff) << 8 ) + (tmpByte[2] & 0xff);

Four bytes:

int temp = ((tmpByte[0] & 0xff) << 24 ) + ((tmpByte[1] & 0xff) << 16 ) + ((tmpByte[2] & 0xff) << 8 ) + ((tmpByte[3] & 0xff) );

 

[Jumpem]

What does the "<< 8" portion do in your example?

int temp = ((int)tmpByte[0] << 8) + (tmpByte[1] & 0xff);

 

[CTho9305]

Originally posted by: Jumpem
What does the "<< 8" portion do in your example?

int temp = ((int)tmpByte[0] << 8) + (tmpByte[1] & 0xff);

shift by 8, so if your original bit pattern for byte0 was 0000000010101010, the new pattern would be 1010101000000000. Then the addition adds the lower byte, producing the complete number.

 

[Jumpem]

Ok, I understand that. What does the "& 0xff" portion do?

 

[EagleKeeper]

Originally posted by: Jumpem
Ok, I understand that. What does the "& 0xff" portion do?

Ensure that any garbage in the upper byte is cleaned off before use.