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Hi, :

I'm sorry, if ask the wrong question, actually I don't know which domain it is, just know it's about math, logic, programming, which Matlab plays a very important part in the world.

My question is , in computer domain, there are floating point, eg: single/double precision floating point, which store variables in [sign, exponent, significand] format sized in [1, 11, 52] bits (for 64 bit precision double type) , refer to https://en.wikipedia.org/wiki/Double-precision_floating-point_format

How do those floating number stored in computer (normally, binary format) to represent in decimal (with fractional number , between 0 ~ 1), especially, those numbers been right of the '.' (radix).

For example, a floating number in binary, 1.111111 is in decimal = 1.9844, how do the computer represent the .9844 to us ? From many docs, they all say, it's ' 2^-1 + 2^-2 +...+ 2^-6', but that's not what I want to ask, it's most likely I am curious about how do the computer translate 0.5 + 0.25 + .... + 0.0156, in computer, they are binary, only 010101..., so when they do arithmetic, they are based on 0,1,0,1..., so that definitely won't really recognize the decimal, '0.5', '0.25', ...,etc, but when we do those simple directives in Matlab, bin2dec(), f_b2d, it quickly gives the answers, I am interested in 'who' do those binary to 'decimal-string-output' task, in which level ? compiler ? What's the domain talking about those domain ?

Thank you very much.

Best regards.

Guillaume
on 12 Jul 2019

Edited: Guillaume
on 12 Jul 2019

Yes, as Stephen said, be careful about the difference between binary numbers and IEEE storage. You can easily see the IEEE representation of a double number in matlab with:

>> n = 1.9844

>> dec2bin(typecast(n, 'uint64'), 64); %typecast to uint64 doesn't change the bits, so we can extract the bit pattern

ans =

'0011111111111111110000000001101000110110111000101110101100011100'

As for your question, the computer is never aware of the decimal representation of the number, and doesn't care about it. All math is done with the binary representation. Which is why people sometimes are surprised by the result they get (see why 0.3-0.2-0.1 is not 0) because the binary math differs from the decima math. It's only for display to the user/programmer that the numbers are converted to their textual decimal representation. This is done by well established code routine, if you want to know more search for the source code of fprintf (in C). Similarly, the textual decimal representation of numbers you enter is immediately converted to the binary representation.

Guillaume
on 12 Jul 2019

James Tursa
on 12 Sep 2019

Guillaume
on 12 Sep 2019

Hum, did I make it up? (!)

Indeed dec2bin doesn't handle properly integers above flintmax (because internally it converts to double). The doc does warn you but not the function. It's something I've complained about to mathworks. It's not even consistent with dec2hex (which throws a warning) or dec2base (which throws an error).

Another way to get the correct binary representation of a number (on a little endian computer), which is probably what I did originally:

strjoin(flipud(cellstr(dec2bin(typecast(n, 'uint8'), 8))), '')

James Tursa
on 12 Sep 2019

You can't use dec2bin( ) reliably for this conversion in all versions of MATLAB because it is limited by flintmax (see note at bottom of doc). I.e., even though it lists int64 and uint64 as acceptable data type inputs, it really can't handle all of the values properly. E.g., using a simple example where we expect all of the mantissa bits to be 1's

R2016a & R2019a WIN64:

>> num2hex(realmax)

ans =

7fefffffffffffff <-- We should expect lots of trailing 1's

>> dec2bin(typecast(realmax,'uint64'),64)

ans =

0111111111110000000000000000000000000000000000000000000000000000 <-- TOTALLY goofed up!

>> reshape(dec2bin(sscanf(num2hex(realmax),'%1x'),4)',1,[]) % do it one hex digit at a time

ans =

0111111111101111111111111111111111111111111111111111111111111111 <-- This is what we were expecting

Both R2016a and R2019a dec2bin( ) can't handle the large uint64 value properly.

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