I can understand binary operation 11 & x, for example if x = 1011, the operation will take out 10 from x and left x to be 11. However, when it comes to hexadecimal, I am very confused. What is the math and reasoning behind the similar effect of 0xff & x? I can only understand this if I convert them all to binary.
0xFF & 0xABCD = 0xCD ... why?
Because:
A = 1010
B = 1011
C = 1100
D = 1101
F = 1111
So the 0xFF = 0x00FF = 0000 0000 1111 1111
The 0xABCD = 1010 1011 1100 1101
-------------------
0xFF & 0xABCD = 0000 0000 1100 1101
As with most things, once you work with hex for a while, you'll learn some tricks for remembering the values.
Related
I really really really can't understand how 15hex converted in Binary form gives me 10101bin.
That should be easy but I can't get it 😰
0x15 == 1*16 + 5*1 == 21
21 == 1*16 + 0*8 + 1*4 + + 0*2 + 1*1 == 10101 (binary)
What's not to love?
Well it's simple. In decimal base, number 15 means
10 + 5, because the number 1 means 1 * 10, and number 5 means 5 * 1.
And in hex, number 15 means:
1 * 16 + 5 * 1, meaning its 21. 21 in binary is 10101.
How to convert hex to binary
Convert each hex digit to 4 binary digits according to this table:
Hex Binary
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
A 1010
B 1011
C 1100
D 1101
E 1110
F 1111
Example #1
Convert (4E)16 to binary:
(4)16 = (0100)2
(E)16 = (1110)2
So
(4E)16 = (01001110)2
Example #2
Convert(4A01)16 to binary:
(4)16 = (0100)2
(A)16 = (1010)2
(0)16 = (0000)2
(1)16 = (0001)2
So
(4A01)16 = (0100101000000001)2
My task is to convert a hexadecimal number in double precision format (IEEE 754) on a paper.
I've converted a hexadecimal number: 0x40790A0000000000 to a binary 64bit format so far and now I have:
0 10000000111 1001000010100000000000000000000000000000000000000000
For the next step I am not totally sure what to do. I have to convert it into a decimal number and I've tried out several ways, but never got the right result.
Hope you can help me and thank you.
Going from https://en.wikipedia.org/wiki/Double-precision_floating-point_format,
4 0 7 9 0 A 0 0 0 0 0 0 0 0 0 0
0100 0000 0111 1001 0000 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
0 - Sign bit (this is a positive number)
100 0000 0111 - Exponent
1001 0000 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 - Fraction
The exponent's value is 1031. Because it's nonzero the fractional part is given by the expression 1 + sum from i = 1 to 52 bit_(52-i) * 2^(-i).
The fraction's value is therefore 1 + 1/2 + 0/4 + 0/8 + 1/16 + ... ~= 1.56
From there you should be able to figure the rest out.
(Not solving this completely because this looks like homework.)
I'm wondering if I get a question like:
"Convert a decimal number to two's complement, then give your answer in Hex".
Is the path below, how one would do it?
Decimal number: -23
23 = 00010111 = in hex 17 = -17
-23 = 11101001 = in hex E9
So to convert it to Hex would the answer be -17 or E9?
Thanks
The -17 has no relevance here, since according to your task, you have to return the two's complement as HEX and that is E9.
Your conversion path in general looks correct to me.
DEC to BIN without the sign:
23 → 0001 0111
Negate the BIN string:
0001 0111 → 1110 1000
Add 1 to the negated BIN result:
1110 1000 + 0000 0001 → 1110 1001
Verify the correct two's complement calculation:
-128 + 64 + 32 + 8 + 1 = -23 → correct
Convert final BIN string to HEX:
1110 1001 → 0xE9
Q=A/B , Q is a real number expressed as a pair of 8 bits:
most significant 8 bits for the integer part
least significant 8 bits for the fractional part
the number is unsigned
for example:
0 0 1 0 1 1 0 1 . 0 1 0 1 0 0 0 0
Can you find the remainder of division if you know B, on a paper? how?
I'll give an example:
2/172 :
0000 0010 . 0000 0000 /
1010 1100 . 0000 0000
=0000 0000 . 0001 0010
0000 0000 . 0001 0010 *
1010 1100 . 0000 0000
=0000 0000 . 1100 0001 (should be 2, or at least something greater than 1.5)
There are two algorithms: restoring and non-restoring. This is very well described in Division Algorithms and Hardware Implementations by Sherif Galal and Dung Pham. And here is about implementation in VHDL.
i found out, that you can do modulo using this :
x % m == (x + x / m) & m
but i cannot understand why its working...
like for 8 % 7 == (8 + 8 / 7) & 7, this is
x = 8 = 0001 0000
x / 7 = 1 = 1000 0000
x + x / 7 = 9 = 1001 0000
9 & 7 = 1001 0000 & 1110 0000 = 1000 0000 = 1
N = 7k + m, m<7
N/7 = k
N + N/7 = 8k + m
(N + N/7) & 7 = (8k + m) & 7
= m & 7
= m
It works for any 2n-1 number, not just 7.