Baskaran Subramani

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Sunita Vontel

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Anonymous

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Sunita Vontel

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MahaAdd

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posted 16 years ago

Hi,

i couldn't get u..

can u tell me how could u arrive at oxFFFFFF81 in detail..

thanks

i couldn't get u..

can u tell me how could u arrive at oxFFFFFF81 in detail..

thanks

Originally posted by Tian:

Although you cast 0x81 to byte, it still gets promoted to int before shift. As the highese bit of 0x81 is 1, when it's promoted to int, it takes the 1 and propagates it all the way to the leftmost bit, which oxFFFFFF81.

Hope that helps.

Stephanie Grasson

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Posts: 347

posted 16 years ago

Hi all.

I think there may have been a mistake somewhere that propagated through this post.

The original question is:

"what is the result of the following .."

(byte)0x81 >> 2

The correct answer is 0xFFFFFFE0 (NOT 0xFFFFFF81).

You can test it for yourself with the following little program:

Here are the steps to get this result:

0x81 (hexadecimal) is equal to 129 (decimal) which is

10000001 (binary).

So there is the byte value.

Next, before the right shift occurs, the byte is promoted to an int. Because the most significant bit in the original byte is a 1, 1's are used to fill in the upper bits of this integer. So we get this:

11111111 11111111 11111111 10000001

Now we do our right shift by 2, and we get this:

11111111 11111111 11111111 11100000

Finally, we convert the binary to hexidecimal:

1111 1111 1111 1111 1111 1111 1110 0000

F F F F F F E 0

So the answer is:

0xFFFFFFE0

Hope this helps.

Stephanie

I think there may have been a mistake somewhere that propagated through this post.

The original question is:

"what is the result of the following .."

(byte)0x81 >> 2

The correct answer is 0xFFFFFFE0 (NOT 0xFFFFFF81).

You can test it for yourself with the following little program:

Here are the steps to get this result:

0x81 (hexadecimal) is equal to 129 (decimal) which is

10000001 (binary).

So there is the byte value.

Next, before the right shift occurs, the byte is promoted to an int. Because the most significant bit in the original byte is a 1, 1's are used to fill in the upper bits of this integer. So we get this:

11111111 11111111 11111111 10000001

Now we do our right shift by 2, and we get this:

11111111 11111111 11111111 11100000

Finally, we convert the binary to hexidecimal:

1111 1111 1111 1111 1111 1111 1110 0000

F F F F F F E 0

So the answer is:

0xFFFFFFE0

Hope this helps.

Stephanie