how i can convert binary 128 bit to decimal??

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nail
nail 2014-5-30
评论： Radwa 2015-1-14
i work in data encryption, but i don't know how to convert a 128 binary to decimal. i try bin2dec and its not work and the error massage tell me : "Binary string must be 52 bits or less". please help
a=dec2bin(7,128)
b=bin2dec(a)
Error using bin2dec (line 36)
Binary string must be 52 bits or less.
  1 个评论
nail
nail 2014-6-1
编辑：nail 2014-6-1
Thank you all
nothing works
but i change my code variables to make the output range (8-52) if i know how to make it bigger (64,128,.....) i will change just the conditions :)
you are help me so much
but i want to get 128 bits of it as binary of string data example
1= 00110001
2= 00110010
a= 01100001
b= 01100010
A= 01000001

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James Tursa
James Tursa 2014-5-31
编辑：James Tursa 2014-5-31
Double precision numbers only have a 52-bit mantissa (+1 hidden leading bit), hence the error. A uint64 has 64 bits, and that is the max you can have for numeric types. To deal with this many bits in a number you will need to use something else. E.g either the Symbolic Toolbox, or use the Variable Precision Integer FEX submission by John D'Errico:
http://www.mathworks.com/matlabcentral/fileexchange/22725-variable-precision-integer-arithmetic
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nail
nail 2014-5-31
编辑：nail 2014-5-31
Thank you James Tursa but i want to get something help me in my code. are you want to say its impossible ?? thank you again
James Tursa
James Tursa 2014-5-31
I don't understand your comment. If you want a single variable to hold the decimal value of a 128-bit binary string, you can't do it with numeric types in MATLAB because there aren't enough bits available. You could use one of the methods I proposed, or perhaps break up your problem into 64-bit chunks and use uint64 types.

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更多回答（1 个）

Hind Thanoon
Hind Thanoon 2014-5-31
Hello nail.. try the following code:
a=dec2bin(7,128); b=bin2dec(a(1:24))*2^104+bin2dec(a(25:76))*2^52+bin2dec(a(77:128))*2^0;
this will help you to solve your problem you will convert each 52 bit separately and then accumulate the result like the following example: a=1001=9, if I need to convert each 2bit then I will say bin2dec(10)*2^2+bin2dec(01)*2^0 {bin2dec(a(1:2))*2^2+bin2dec(a(3:4))*2^0} the result will be: 2*2^2+1*2^0=9
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James Tursa
James Tursa 2014-5-31
There are not enough bits in a double variable to hold the exact value of a 128-bit binary string, so your method of breaking things up and accumulating the result will not work in general.
Radwa
Radwa 2015-1-14
I work in encryption also, how u solve ur problem?

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