# i get an error.whats wrong? why the values of 'B' in intger instead of decimal.

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Sultan Mehmood on 8 Jul 2019
Edited: Stephen23 on 10 Jul 2019
II=[14 23;44 15];
R=II(:)';
x=0.3;
p=0.343;
for n=2:4;
if x(n-1)>=0 & x(n-1)<=p
x(n)=x(n-1)/p;
else
x(n)=(1-x(n-1))/(1-p);
end
end
A=sort(x);
[A,T]=sort(x);
Y=R(T);
C=reshape(Y,[2,2]);
r = 3.8;
L(1)= 0.234;
for i=2:4
L(i) = r*L(i-1)*(1-L(i-1));
end
mm=min(L);
nn=max(L);
oo=nn-mm;
Z=uint8(8*((L-mm)/oo))+1;
%for i=1:9
%K(i)=mod((abs(L(i))-floor(abs(L(i))))*1e14,50);
%end
CC(1)=bitxor(mod(157,50),mod(Y(1)+Z(1),50));
for i=2:4
CC(i)=bitxor(mod(CC(i-1),50),mod(Y(i)+Z(i),50));
end
ZZ=reshape(CC,[2,2]);
B=ZZ/10
why the value of B in intger.?

KALYAN ACHARJYA on 8 Jul 2019
Edited: KALYAN ACHARJYA on 8 Jul 2019
See your zz data is integer type (uint8).
>> whos ZZ
Name Size Bytes Class Attributes
ZZ 2x2 4 uint8
Hence
ZZ =
2×2 uint8 matrix
31 18
10 35
Now ZZ/10
B =
2×2 uint8 matrix
3 2
1 4
Souce stackoverflow{dor}com: Converting both integer values to a double precision value will ensure that MATLAB performs floating point division instead of integer division (which will round off the result). See this link
But when you convert zz to double
B=double(ZZ)/10
format shortg
See the results
B =
3.1 1.8
1 3.5
##### 2 CommentsShowHide 1 older comment
Stephen23 on 10 Jul 2019
Edited: Stephen23 on 10 Jul 2019
The MATLAB documentation explains the expected output classes for operations on integers: