I have a strange problem with hex2dec function in Matlab.
I realized in 16bytes data, it omits 2 LSB bytes.
hex2dec('123123123123123A');
dec2hex(ans)
Warning: At least one of the input numbers is larger than the largest integer-valued floating-point
number (2^52). Results may be unpredictable.
ans =
1231231231231200
I am using this in Simulink. Therefore I cannot process 16byte data. Simulink interpret this as a 14byte + '00'.
You need to use uint64 to store that value:
A='123123123123123A';
B=bitshift(uint64(hex2dec(A(1:end-8))),32)+uint64(hex2dec(A(end-7:end)))
which returns
B =
1310867527582290490
An alternative way in MATLAB using typecast:
>> A = '123123123123123A';
>> B = typecast(uint32(hex2dec([A(9:end);A(1:8)])), 'uint64')
B =
1310867527582290490
And the reverse in the opposite direction:
>> AA = dec2hex(typecast(B,'uint32'));
>> AA = [AA(2,:) AA(1,:)]
AA =
123123123123123A
The idea is to treat the 64-integer as two 32-bit integers.
That said, Simulink does not support int64 and uint64 types as others have already noted..
Related
I have a problem with the mod function output in Matlab. I am trying to perform some calculations for ECC double and add algorithm. I am reading data from a file and storing it in a variable and then performing some operations. All works smoothly except that I get 0 in temp1 when I use mod(X2,P). However if I put in values stored in X2(3.0323e+153) and P(1.1579e+77) on command window (mod( 3.0323e+153, 1.1579e+77)), I get the correct values. Can anyone please help me? Below is the part of script which is problematic.
P = hex2dec('FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F');
line = fread(fileID,[1,67],'*char');
while ~feof(fileID)
PX = line(4:67);
X = hex2dec(PX);
X2 = X^2;
temp1= mod(X2 , P)
end
line = fread(fileID,[1,69],'*char');
end
fclose(fileID);
I think the problem lies with how you're initializing P. From the documentation for hex2dec (emphasis mine):
d = hex2dec('hex_value') converts hex_value to its floating-point integer representation. The argument hex_value is a hexadecimal integer stored as text. If the value of hex_value is greater than the hexadecimal equivalent of the value returned by flintmax, then hex2dec might not return an exact conversion.
And the value of flintmax is:
>> flintmax
ans =
9.007199254740992e+15
Quite a bit smaller than your value for P. In fact, if we use num2hex to look at the two ways you initialize P, you can see a clear difference:
>> P = hex2dec('FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F');
>> num2hex(P)
ans =
4ff0000000000000
>> num2hex(1.1579e+77)
ans =
4fefffda293c30de
As it turns out, the inexact conversion done by hex2dec results in a number that evenly divides into 3.0323e+153, thus giving you a remainder of 0:
>> mod(3.0323e+153, P)
ans =
0
>> mod(3.0323e+153, 1.1579e+77)
ans =
8.795697942083107e+76
I want to know if it's possible in MATLAB to discard overflow digits in MATLAB when I add two binary numbers.
I've only been able to find how to have a least number of binary digits, but how to I set a maximum number of digits?
For example:
e = dec2bin(bin2dec('1001') + bin2dec('1000'))
That gave me:
e =
10001
How do I get only '0001'?
dec2bin will always give you the minimum amount of bits to represent a number. If you would like to retain the n least significant digits, you have to index into the string and grab those yourself.
Specifically, if you want to retain only the n least significant digits, given that you have a base-10 number stored in num, you would do this:
out = dec2bin(num);
out = out(end-n+1:end);
Bear in mind that this performs no error checking. Should n be larger than the total number of bits in the string, you will get an out of bounds error. I'm assuming you're smart enough to use this and know what you're doing.
So for your example:
e = dec2bin(bin2dec('1001') + bin2dec('1000'));
n = 4, and so:
>> n = 4;
>> e = e(end-n+1:end)
e =
0001
Here is a more robust (but less efficient, I fear) way: What you describe is exactly the modulo operation. A 4-bit binary number is the remainder after a division by 0b10000 = 16. This can be done using the mod function in MATLAB.
>> e = dec2bin(mod(bin2dec('1001') + bin2dec('1000'),16),4)
e =
0001
Note: I added 4 as additional argument to the dec2bin function, so the output will always be 4-bit wide.
This can of course be generalized to any bit width: If you want to add 8-bit numbers, you will need the remainder of the division by 0b1'0000'0000 = 256, for example
>> e = dec2bin(mod(bin2dec('10011001') + bin2dec('10001000'),256),8)
e =
00100001
Or for shorter numbers, e.g. 2-bit wide, it is 0b100 = 4:
>> e = dec2bin(mod(bin2dec('10') + bin2dec('11'),4),2)
e =
01
realmax on my machine is:
1.7977e+308
I know I have to write my code in a way to avoid long integer calculations, but is there any way to increase the limit?
I mean something like gmp library in C
You may find vpa (variable- precision arithmetic) helpful:
R = vpa(A) uses variable-precision arithmetic (VPA) to compute each element of A to at least d decimal digits of accuracy, where d is the current setting of digits.
R = vpa(A,d) uses at least d significant (nonzero) digits, instead of the current setting of digits.
Here's an example how to use it:
>> x = vpa('10^500/20')
ans =
5.0e498
Note that:
The output x is of symbolic (sym) type. Of course, you shouldn't convert it to double, because it would exceed realmax:
>> double(x)
ans =
Inf
Use string input in order to avoid evaluating large input values as double. For example, this doesn't work
>> vpa(10^500/20)
ans =
Inf
because 10^500 is evaluated as double, giving inf, and then is used as an input to vpa.
I'm using the diff Matlab function to get the difference between two successive values. And as shown here in this vector nz in this link as shown in nz the difference between col 261 and 260 is -1342 but when I use this script the result of difference between this coloumns don't appear in the result dnz. So if anyone could advise why this is not working?
This is my attempt:
load('nz.mat');
dnz = diff(nz);
If you type class(nz) you see that your data is unit16. And MATLAB saturates the results when dealing with integer values, i.e. since 0 - 1342 is lower than zero (the smallest value in uint16) it returns zero:
>> dnz=diff(nz);
>> dnz(260)
ans =
0
If you convert it to a class that can accomodate -1342 like int16 you get
>> dnz = diff(int16(nz));
>> dnz(260)
ans =
-1342
I'm a MATLAB beginner. Here's the problem:
>> a = floor(7/2.5)
a =
2.00
>> b = rem(7,2.5)
b =
2.00
>> c = floor(b/2)
c =
0
c should be 1, right? Why is it 0???
It is different when b = 2 is entered directly as follows:
>> b = 2
b =
2.00
>> c = floor(b/2)
c =
1.00
In two words: truncation errors.
You're right, c should be 1.0 in exact arithmetic. However, since you used a float in the arguments of rem, you get the answer as a float. Apparently, b is not exactly 2, but 2.0, which means that it is a double very close to 2. Therefore, b/2 becomes the double 1.0, apparently in this case its value is slightly less than one, giving you a 0 as the integer value. If you want to prevent this, use both floor and ceil, and compare the values.
If you want to convert the answer to integer, just use round instead of floor.
If you add the line
d = b-a
to your example you'll see the result
d =
-4.4409e-016
meaning Matlab calculated a number close to, but not exactly, 2 for b. This comes up quite a bit in working with floating point numbers. Try
help eps
for more information.
Numerical issues of this sort are also dealt with in the MATLAB FAQ
Yeah that is a numerical issue. You should use such things with care. If you want exact arithmetic , you should try 'sym' for your number e.g.
b=rem(sym(7),sym(2.5))
Then you will not any such errors, but your computations will be much slower.