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Ranch Hand
Posts: 42
• • • • Thanks
Q5)
byte b;
double d =417.35;
b=(byte)d;
System.out.println(b);
output is 95
but how to convert a double value into byte type?

Ranch Hand
Posts: 68
• • • • The answer should be -95, not 95. Here is how I think is gotten.
1) double 417.35 fist is truncated to int 417
2) 417 = 256 + 128 + 32 + 1 and have a binary representation of
0000 0001 1010 0001
3) it then get truncated again, because byte has only 8 bits, to
1010 0001
4) The sign bit (first bit) at step 3) is 1, therefore it is a negative number. To find the number I use the following relationship.
~x + 1 = -x
~x is 1's complement of the 8 bit in step 3) and it is
0101 1110
This number = 64 + 16 + 8 + 4 + 2 = 94
so x = - (~x + 1 ) = -95.
5) to double check it

C:\JavaRanch>java TestByte
d = 417.35 b = -95

kaffo lekan
Ranch Hand
Posts: 42
• • • • please how did you do the binary representation

huiying li
Ranch Hand
Posts: 68
• • • • here is more detail on step 2)
1) It is good to remember
2^10 = 1024, 2^9 = 512, 2^8 = 256, 2^7 = 128, 2^6 = 64 and so on.
2) 417 > 256, and 417 <512, so 9bit or 2^8 is it is highest non-zero bit. 2^0 is the first bit.
3) 417 - 256 = 161, 161 is < 256 , > 128, so the next non-zero bit is the 8th bit you do this recursively, until you get all the bits

4) 417 = 256 + 128 + 32 + 1
= 1 * 2^8 +
1 * 2^7 + 0 * 2^6 + 1 * 2^5 + 0 * 2^4 +
0 * 2^3 + 0 * 2^2 + 0 * 2^1 + 1* 2^0
5) therefore 417 is represented by
0000 0001 1010 0001

Greenhorn
Posts: 20
• • • • huiying li,
very good explanation, found it very useful
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