CODEWITHSUNDEEP
javaintegerbyteunsignedsigned
Pim
Pim

Reputation: 433

Java: 4 bytes to 32 bit integer

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?

Upvotes: 1

Views: 1647

Answers (2)

aioobe
aioobe

Reputation: 421040

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

Upvotes: 5

Louis Wasserman
Louis Wasserman

Reputation: 198133

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.

Upvotes: 3

Related Questions