Subexponential coding from c to matlab.

I am having a C coding for the subexponential coding.Please tell me the matlab code written by me is correct and please suggest some ideas to implement the final step.

binary b= {k, if n<2^k

            {floor(log2(n),if n>=2^k.

unary u=  {0  if n<2^k

          {b-k+1 if n>=2^k

Once these values are obtained, the code can be constructed by coding u in unary and continuing with the b LSB of n.

The matlab code written by me is as follows

function x=sec[n,k] %the value ok k is already estimated

if n<bitshift(2,k)

b=k;
u=0;

else

b=floor(log2(n));u=b-k+1;

x=[]%Once these values are obtained, the code can be constructed by coding u in unary and continuing with the b LSB of n.

I don't know to calculate x value in matlab.

The c code for calculating the code is given below

x =(((((1 < <( u )) -1) < < 1) < <( b )) | (n & ((1 < <( b )) -1)));

we build the unary code of u, which can be obtained by shifting a 1 bit u positions to the left and subtracting 1 to the result. A zero stop-bit can be added by shifting another position to the left the resulting value. The next operation is to append to this value the b LSB of n expressed in plain binary code. In order to do this the non-signicant bits of n must be discarded that is, theall-zero Most Signicant Bits (MSB) | by applying a mask that keeps the b LSB. A bit-wise OR operation with the truncated value of n will produce the required output.Once b and u are available, the subexponential code can be actually computed.

Please help me. Thankyou..

[Jan]

Please do not try to push me by email to answer your question. I really do not like this. I got 7 emails like yours in the last two days. But the time I spend in the forum is really enough. I'm going to replace my email contact address to an invalid one now.

[Pooja]

Oh sorry for disturbing you.You usually help us.I thought that if you didn't see this link i will miss your help.Sorry again.

[dpb]

Not to mention you then came after me..._not_ how to make friends and win help... :(

[Jan]

@Pooja: Contributors, who do not like to be pushed, might tend to ignore your question as a reaction. Pushing causes the impression, that somebody thinks, that his time is more important than the time of others. But it is not surprising, that the others have a different opinion.