The first part: hex conversion
Binary: constituted by 0 and 1 every 2 into an
octal: consists of 0-7 into 1 every 8
hex: a 0 ~ 9, A ~ F configuration, feed 1 every 16
Two basic concepts
Base: n-ary base is n
123.4 = 110^2 + 210^1 + 310^0 + 410^-1
Bit right:
left of the decimal k-bit right as the base ^ k-1
k-bit right of the decimal point to the right as the base ^ -k
Other binary decimal turn (the right to expand by summation method)
Decimal = 123.4 1 10 ^ 2 + 10 ^ 3 1 + 10 ^ 0 + 4 10 ^ -1
binary 1011.1 1 = 2 ^ 3 + 0 2 + 1 2 ^ 2 ^ 1 + 1 2 + 1 ^ 0 2 ^ - 1 = 11.5
octal 123.4 = 1 . 8 ^ 2 + 2 . 8 ^ 1 + 3 . 8 ^ 0 + 4 . 8 -1 = 83.5
hex = 123.4 1 16 2 + 2 16 ^ 1 + 3 16 + 4 ^ 0 -1 * 16 = 291.25
Decimal turn the other band
The integer part of: the base division, modulo reverse
fractional part: multiplying the base sequence Rounding
Decimal -> Binary
integer division modulo 2 reverse (in units of 4 bits, 0 is less than the high bit)
the decimal rounding by 2 positive sequence (met take an integer, decimal and then multiplied by 2)
Among other binary conversion
- Binary octal (3-bit binary octal = 1 corresponds to "421")
01011010 == 132
000 = 0100 = 4
001 = 1101 = 5
010 = 2110 = 6
011 = 3111 = 7
- Binary hexadecimal (4-bit binary Hex = 1 corresponds to "8421")
0101 1010 ==. 5A
0000 1000 = 0. 8 =
0001 = 1. 9 1001 =
0010 A = 1010 = 2
0011 = 1011. 3 = B
0100 1100 is. 4 = C =
0101. 5 = 1101 = D
0110 = 1110. 6 E =
0111. 7 = 1111 = F.
Part II: data representing
Classification shaping data
1. Unsigned integer: all bits represent the size.
Range Data N bit unsigned integer is represented by: 0. 1 ~ 2N-
2. Signed integer data: indicates the highest bit symbols: n is 0, 1 negative. Common original code, anti-code complement representation.
Positive: three yards the same.
Example: Given x = + 76D, three write code representation x (8)
Solution: x = + 76D = + 1001100B
as x> 0
Therefore: [x] = the original [x] trans = [x] = complement 01001100B
negative:
Example 1: known x = -76D, three write code representation x (8)
x = -76D = -1001100B: Solutions
[x] original 1001100B 1 =
[x] trans = 1 0110011B (the original code negation)
[X] complement = 1 0110100B (plus a last one)
Example 2: Given: [X] original = [y] trans = [z] complement = 1100 1011B, calculated x, y, z of the magnitude relationship .
[x] of the original 1100 is 1011B X = -100 = 1011B
[Y] 1100 is trans 1011B =
[Y] 1011 0100B original Y = -011 = 0100B
[Z] 1011 0100B fill =
[Z] Original = 1100 1100B z = -100 1100B
y> x> z