Binary
to octal numbers: two solutions in Matlab
To convert a value from binary
to octal,
we first need
to know what an octal number is.

A major numbering system
used in digital electronics is the
octal system, also named base 8. In this system, the numbers are
counted
from 0 to 7.
The following table shows
the meaning of all the symbols in
the bin and oct (for short) numbering systems. 
Equivalent decimal,
binary and octal numbers
To convert a value from binary
to octal, you merely
translate each 3bit binary group to its octal equivalent. For
example,
the binary number 001111101100010 (or, to be clearly seen, 001
111 101 100 010 translates into
the octal 17542
equivalent.
Solution 1. Using builtin functions
In Matlab, we can go from
binary to
decimal, and then, from
decimal to
octal (using dec2base). We can embed one instruction into
the other,
like this:
oct_str
= dec2base(bin2dec(bin_str), 8)
It’s
important to remember that both binary
numbers and octal
ones are treated
as strings
in Matlab.
So, we
can use this concept, like this:
bin_str
= '111101100010'
oct_str
= dec2base(bin2dec(bin_str), 8)
Matlab’s answer is:
oct_str = 7542
Solution 2. Separating in groups
Now, let’s say that we want to
explore how the binary groups
are separated to form the oct symbols and we want to manipulate our own
strings
(binary and octal).
We can develop a function
to translate the table shown
before. Our suggestion uses a switchcase
structure.
% Binary to octal
conversion
function o =
b2o(b)
switch b
case {'0', '00', '000'}
h = '0';
case {'1', '01', '001'}
h = '1';
case {'10', '010'}
h = '2';
case {'11', '011'}
h = '3';
case {'100'}
h = '4';
case {'101'}
h = '5';
case {'110'}
h = '6';
case {'111'}
h = '7';
end
Now, we have to call
the function for every 3bit group of binary numbers. One possible
solution to
separate the binary number into 3bit groups is shown here:
bin_str
= input('Enter
binary number: ', 's');
i =
length(bin_str);
n =
ceil(i/3);
for g = n :
1 : 1
if i > 3
oct_str(g) = b2o(bin_str(i3 : i));
i = i  3;
else
oct_str(g) = b2o(bin_str(1 : i));
end
end
oct_str
Let’s
try it.
Enter
binary number: 101010
oct_str
= 2A
Enter
binary number: 110001011110101
oct_str
= 62F5
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