the IBM shaort floating point representation uses base6 ,one sign bit ,a 7-bit excess 64 exponent ,and a normalized 24 bit fraction (normalization :the point is on the left side of the leftmost non-zero digit )
what number is represent by the following bit pattern ?
1 0111111 011100000000000000000000
show your answer in decimal
我主要唔明 fraction 轉換個part
thx
show 一show 個step ，plx
computer art. 高手請入
2010-10-20 4:28 pm
回答 (2)
2010-10-21 6:30 am
✔ 最佳答案
I believe there is a mistake in the base which should be base 16.圖片參考：http://img843.imageshack.us/img843/286/15960110.png
2010-10-20 22:30:38 補充：
http://img843.imageshack.us/img843/286/15960110.png
2010-10-20 7:27 pm
Given two upper triangular matrices,
A = [ a1 * . . * ], B = [ b1 * . . * ]
0 a2 . . * 0 b2 . . *
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still upper triangular, with the diagonal entries simply multiplied together (the other entries are more complicated).
AB = [ a1b1 * . . * ]
0 a2b2 . . *
: : :
0 0 . . anbn
The similar statement is true for the multiplication of two lower triangular matrices.
For two diagonal matrices,
A = [ a1 0 . . 0 ], B = [ b1 0 . . 0 ]
0 a2 . . 0 0 b2 . . 0
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still diagonal, with the diagonal entries simply multiplied together.
AB = [ a1b1 0 . . 0 ]
0 a2b2 . . 0
: : :
0 0 . . anbn
A = [ a1 * . . * ], B = [ b1 * . . * ]
0 a2 . . * 0 b2 . . *
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still upper triangular, with the diagonal entries simply multiplied together (the other entries are more complicated).
AB = [ a1b1 * . . * ]
0 a2b2 . . *
: : :
0 0 . . anbn
The similar statement is true for the multiplication of two lower triangular matrices.
For two diagonal matrices,
A = [ a1 0 . . 0 ], B = [ b1 0 . . 0 ]
0 a2 . . 0 0 b2 . . 0
: : : : : :
0 0 . . an 0 0 . . bn
their multiplication is still diagonal, with the diagonal entries simply multiplied together.
AB = [ a1b1 0 . . 0 ]
0 a2b2 . . 0
: : :
0 0 . . anbn
收錄日期: 2021-04-23 23:24:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101020000051KK00208