why i get an error ?

II=[114 223;176 155];
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);
r = 3.342448;
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(254*((L-mm)/oo))+1;
SS=mod((Y+Z),256)

problem in SS.

Answer 1

[KSSV]

Z=uint8(254*((L-mm)/oo))+1;
Y = uint8(Y) ; 
SS=mod((Y+Z),256)

Note that same class varibales only can be added. A integer shoud ld be added to integer. So ocnvert your variable Y, which is double to unit8.

[Sultan Mehmood]

suppose Y=114;

Z=164;

SS=mod((Y+Z),256);

ans =255

butt if you solve mannully the

ans =22;

why ?

[KSSV]

In this case your Y and Z are double.......convert them to uint8.

[Yash Totla]

Z has been typecasted to uint8. When you perform Y+Z it is also typecasted to uint8 datatype. Now if the value of the expression is capped at 255 because that is the max value that you can store in it. So when you take the modulo with 256, it returns 255.

You can try to typecast the expression Y+Z to a datatype whose max value is more than 255. Taking the modulo of a uint8 with 256 is basically returning the same number in uint8.

Check this link for more information on numeric datatypes in matlab.