programmerinprogress wrote:basically, chars can be any value from 0 to 255 (with it being an 8 bit number), each of these numbers represent a different action and/or character.
in ASCII, codes 48 to 57 represent the digits '0' to '9', what you have to do to convert these characters to digits is to turn the value 48 into 0, 49 into 1, 50 into 2 etc [using bitmasks]
...because subtracting 48 is simply too hard to understand :)
As for the floating-point numbers:
The number is always assumed to be "
+1 point something times 2 to the (something minus 127) power". So with the exponent-mantissa notation that netwatcher posted,
Click here to see the hidden message (It might contain spoilers)
, you take "1". You append the mantissa (binary decimal) to "1". You convert it the same way you do a decimal number. Recall, when you convert decimal numbers, that things like 0.4 are the same as 4 * 10^(-1).
So, a mantissa of
1100 0000 0000 0000 0000 000
is
1 * 2^(-1) + 1 * 2^(-2) + 0 * 2^(-3) + 0 * 2 ^ (-4) + .... + 0 * 2 ^ (-23)
Which is
1/2 + 1/4 + 0 + 0 + 0....
0.5 + 0.25
0.75
And since we make the assumption about "1", this means that the decimal that we multiply by "2 to the (something minus 127) power" is 1.75
Now we need the exponent part:
0000 0111
Which is
0 * 2^7 + 0 * 2^7 + 0 * 2^6 + ... + 0 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0
= 0 + 0 + 0 + ... + 0 + 1*4 + 1*2 + 1*1 = 7
The exponent is assumed to be in (theExponent - 127) form, and so, the exponent is 7 - 127 = -120. And again, the number as a whole is "1 point something times 2 to the (something minus 127) power". So our number is "1 point 75 times 2 to the negative 120th power" and becomes 1.75 * 2(-120) which hopefully comes out to 1.316554 * 10^(-36), since that's what the solution is in the example image above.
In short, floating-point notation has some rules -- such as "we will subtract 127 from the exponent specified" and "make the leading digit in the decimal be 1" which allows it to represent numbers in its limited number of bits, and books seem to neglect mentioning these. All this conversion is one reason that accuracy can be lost when dealing with floating-points.
Maybe the owners can make a few more dedicated threads wink wink...
