Java: 4 bytes to 32 bit integer

Question

Im trying to understand the following piece of code:

int omgezetteTijd =
((0xFF & rekenOmNaarTijdArray[0]) << 24) | ((0xFF & rekenOmNaarTijdArray[1]) << 16) |((0xFF & rekenOmNaarTijdArray[2]) << 8) | (0xFF & rekenOmNaarTijdArray[3]);

What I do not understand is why do you AND it with OxFF, you're ANDING an 8 bit value with 8 bits like so (11111111), so this should give you the same result.

But, when I do not AND it with OxFF I'm getting negative values? Cant figure out why this is happening?

Why do you repliate the job of what a ByteBuffer can do for you? — fge
– fge, Commented Jan 22, 2015 at 20:29
Because I want to understand different data type's and know how to use them :) — Pim
– Pim, Commented Jan 22, 2015 at 20:35

aioobe · Accepted Answer · 2015-01-23 10:11:09Z

5

When you or a byte with an int the byte will be promoted to an int. By default this is done with sign extension. In other words:

                           //                      sign bit
                           //                         v
byte b = -1;               //                         11111111 = -1
int i = (int) b;           // 11111111111111111111111111111111 = -1
                           // \______________________/
                           //      sign extension

By doing & 0xFF you prevent this. I.e.

                           //                      sign bit
                           //                         v
byte b = -1;               //                         11111111 = -1
int i = (int) (0xFF & b);  // 00000000000000000000000011111111 = 255
                           // \______________________/
                           //     no sign extension

edited Jan 23, 2015 at 10:11

answered Jan 22, 2015 at 20:19

aioobe

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Louis Wasserman · Accepted Answer · 2015-01-22 20:24:50Z

3

0xFF as a byte represents the number -1. When it's converted to an int, it is still -1, but that has the bit representation 0xFFFFFFFF, because of sign extension. & 0xFF avoids this, treating the byte as unsigned when converting to an int.

answered Jan 22, 2015 at 20:24

Louis Wasserman

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