ImplicitExplicitCast doubt
Jas Oberai
Ranch Hand
Posts: 231
posted 11 years ago
Hi Friends,
I have some doubt with this ImplicitExplicitCast thing example on javaranch journal:
Here,I don't understand how come we lose our precious signbit after chopping off first 24bit.The 8th bitwon't that be our signbit??If this,is wrong what should be the correct answer then 7??
Thanks in Advance.
I have some doubt with this ImplicitExplicitCast thing example on javaranch journal:
Here,I don't understand how come we lose our precious signbit after chopping off first 24bit.The 8th bitwon't that be our signbit??If this,is wrong what should be the correct answer then 7??
Thanks in Advance.
SCJP 1.4 (88%)<br />SCWCD 1.4 (88%)
Animesh Shrivastava
Ranch Hand
Posts: 298
posted 11 years ago
Jas,
I am not able to understand what u r trying to say. But this is how u calculate the final result.
After chopping off the 24 bits u get the result as:
11111001
To calculate the actual value, which is ve ofcourse what u do is
00000110 (1's complement of the original value)
+1 (add 1 to it)

00000111 (this is ur result)= 7
since, thr result has to be negative, so it comes to 7.
Hope this is clear.
If ur doubt still persists do calrify it.
I am not able to understand what u r trying to say. But this is how u calculate the final result.
After chopping off the 24 bits u get the result as:
11111001
To calculate the actual value, which is ve ofcourse what u do is
00000110 (1's complement of the original value)
+1 (add 1 to it)

00000111 (this is ur result)= 7
since, thr result has to be negative, so it comes to 7.
Hope this is clear.
If ur doubt still persists do calrify it.
Joe Sondow
Ranch Hand
Posts: 195
posted 11 years ago
The statement
b1 >>>= 1;
is a compound statement that is equivalent to
b1 = (byte)(b1 >>> 1);
The part in the parentheses shows a byte variable getting bitshifted to the
right with the unsigned right shift operator. This means that b1 will be
promoted to an int so its bit pattern will be:
11111111111111111111111111110011
which is 13.
Then it will be shifted right one place, with a leading zero added on to
the left side, because that's how >>> works:
01111111111111111111111111111001
which is positive 2147483641 because the 0 on the left end is the sign bit
for this 32bit int value. The number is positive because the leftmost bit
is a 0.
So (b1 >>> 1) evalutes to 2147483641.
The statement can now be rewritten:
b1 = (byte)(2147483641);
The next step is to cast this value explicitly to a byte. Since its current
bit pattern is
01111111111111111111111111111001
casting it to a byte will discard the highorder 24 bits (the ones on the
left). Here's that value with x's for the bits that will be discarded:
xxxxxxxxxxxxxxxxxxxxxxxx11111001
After discarding those bits, the resulting byte has the bit pattern
11111001
which is now negative because the leftmost bit is 1. This value is 7.
The statement can now be rewritten:
b1 = 7;
If b1 had been an int or long instead of a byte, then you could assume that
the result of
b1 >>>= 1;
would be positive, since >>> puts a zero on the left. However,
because of the cast to byte, there's no certainty that the result will be
positive or negative until you work through the problem, since the sign bit
of the int gets discarded, and whatever bit happens to be in the 8th place
from the right will be the sign bit of the byte that results from the cast
to byte. In this case that bit is a 1, so the byte is negative.
[ May 05, 2005: Message edited by: Joe Sanowitz ]
[ May 05, 2005: Message edited by: Joe Sanowitz ]
b1 >>>= 1;
is a compound statement that is equivalent to
b1 = (byte)(b1 >>> 1);
The part in the parentheses shows a byte variable getting bitshifted to the
right with the unsigned right shift operator. This means that b1 will be
promoted to an int so its bit pattern will be:
11111111111111111111111111110011
which is 13.
Then it will be shifted right one place, with a leading zero added on to
the left side, because that's how >>> works:
01111111111111111111111111111001
which is positive 2147483641 because the 0 on the left end is the sign bit
for this 32bit int value. The number is positive because the leftmost bit
is a 0.
So (b1 >>> 1) evalutes to 2147483641.
The statement can now be rewritten:
b1 = (byte)(2147483641);
The next step is to cast this value explicitly to a byte. Since its current
bit pattern is
01111111111111111111111111111001
casting it to a byte will discard the highorder 24 bits (the ones on the
left). Here's that value with x's for the bits that will be discarded:
xxxxxxxxxxxxxxxxxxxxxxxx11111001
After discarding those bits, the resulting byte has the bit pattern
11111001
which is now negative because the leftmost bit is 1. This value is 7.
The statement can now be rewritten:
b1 = 7;
If b1 had been an int or long instead of a byte, then you could assume that
the result of
b1 >>>= 1;
would be positive, since >>> puts a zero on the left. However,
because of the cast to byte, there's no certainty that the result will be
positive or negative until you work through the problem, since the sign bit
of the int gets discarded, and whatever bit happens to be in the 8th place
from the right will be the sign bit of the byte that results from the cast
to byte. In this case that bit is a 1, so the byte is negative.
[ May 05, 2005: Message edited by: Joe Sanowitz ]
[ May 05, 2005: Message edited by: Joe Sanowitz ]
SCJA 1.0 (98%), SCJP 1.4 (98%)
Jas Oberai
Ranch Hand
Posts: 231
posted 11 years ago
Thanks Animesh and Joe,
Animesh i am trying to understand what Joe is trying to explain to me.Thanks Joe,after reading your post all my doubts got cleared....the only mistake i was making was i was taking the "unsigned right shift operator"...as a "signed right shift operator"..and then wondering what the author is tryin to prove.
Anywas,thanks for your help!!!
Animesh i am trying to understand what Joe is trying to explain to me.Thanks Joe,after reading your post all my doubts got cleared....the only mistake i was making was i was taking the "unsigned right shift operator"...as a "signed right shift operator"..and then wondering what the author is tryin to prove.
Anywas,thanks for your help!!!
SCJP 1.4 (88%)<br />SCWCD 1.4 (88%)
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