Which is the easiest method for shifting binary digits to the right?

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Darsana P M
Darsana P M 2018-2-15
评论： Walter Roberson 2018-2-23
x1 = {'1' '0' '1' '1' '0' '1' '0' '0' '0' '1' '1' '1' '0' '0' '1' '1'};
x=hex2dec(x1);
Thus in command window,
x =
1
0
1
1
0
1
0
0
0
1
1
1
0
0
1
1
Hence to do shift operation, which is the simplest way?
by one place, 0110100011100110
next, 1101000111001100
next , 1010001110011000
and so on?
Can somebody help me?

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采纳的回答

Walter Roberson
Walter Roberson 2018-2-15
[x(2:end), 0]
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Walter Roberson
Walter Roberson 2018-2-18
>> x = [ 1
0
1
1
0
1
0
0
0
1
1
1
0
0
1
1]
x =
1
0
1
1
0
1
0
0
0
1
1
1
0
0
1
1
>> [reshape(x(2:end), 1, []), zeros(1, min(1, length(x)))].'
[reshape(x(2:end), 1, []), zeros(1, min(1, length(x)))].'
ans =
0
1
1
0
1
0
0
0
1
1
1
0
0
1
1
0
>> length(ans)
ans =
16
Or you could be lazy in your error checking and just use
[x(2:end); 0]

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更多回答（1 个）

Darsana P M
Darsana P M 2018-2-23
for i=1:1:16
if x(i) == 0
% yj = y;
Z = Z
else x(i) == 1
Z = bitxor(Z,v);
end
if lsb == 0
v = [v(i:end); 0];
else lsb == 1
v = [v(i:end); 0];
v = bitxor(v,R);
end
end
I got an error stating this: Error using bitxor Inputs must have the same size.
Error in shifty (line 38) Z = bitxor(Z,v);
When I checked the size of v it was, 17
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Walter Roberson
Walter Roberson 2018-2-23
You can get around the first problem by changing
v = [v(i:end); 0]
to
v = [v(i+1:end); 0]
However, this will just postpone the problem one iteration: on the next iteration you will be generating a shorter v (because i has increased) and that shorter v will fail the Z = bitxor(Z,v) since Z did not also get shorter.
I think you need to revise your algorithm. One of the major revisions you need is that you need to document the code -- the purpose of the code, and a description of what you expect each step to accomplish.

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