Converting binary to decimal and vice versa, please help!!
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I'm having trouble converting strings that represent binary numbers to decimal and vice versa. The main issue is that I'm unsure of how to deal with the decimal point, say converting -11.11 to decimal. This is a homework assignment, so I'm not allowed to use the built in functions.
Here's my code for one of the two:
function decimal= mybin2real(binarystring)
decimal = 0;
for i = 1 : length(binarystring)
decimal = decimal + str2num(binarystring(i)) * 2^(length(binarystring) - i);
end
decimal
end
I was thinking about finding where the decimal point was using strfind, but i'm not sure how to implement it in the matlab code.
6 个评论
Daniel Shub
2012-11-2
编辑：Daniel Shub
2012-11-2
You realize that your example code is littered with built in functions. As bin2dec is not a builtin function, why not just use it. Or maybe retype the function yourself. Have you looked at how bin2dec solves the problem?
SB
2012-11-2
Daniel Shub
2012-11-2
I will repeat myself: Have you looked at how bin2dec solves the problem. In general the code included in MATLAB is of pretty high quality.
SB
2012-11-2
Daniel Shub
2012-11-2
At the command line type
type bin2dec
or
edit bin2dec
this will let you see the source code.
SB
2012-11-2
采纳的回答
Here I have simply fixed your code. You were really close!
function decimal= mybin2real(binarystring)
% Converts an input string of 1s and 0s to a number.
Ind = strfind(binarystring, '.');
L = length(binarystring);
if isempty(Ind)
Ind = L+1;
end
Num1 = binarystring(1:Ind-1);
LN1 = length(Num1);
Num2 = binarystring(Ind+1:end);
LN2 = length(Num1);
dec1=0;
for ii = 1 : LN1
dec1 = dec1 + str2double(Num1(LN1-ii+1)) * 2^(ii-1);
end
dec2=0;
for ii = 1 : length(Num2)
dec2 = dec2 + str2double(Num2(ii)) * 2^-(ii);
end
decimal=dec1+dec2;
6 个评论
SB
2012-11-2
What input did you give?
>> mybin2real('10101.101')
ans =
21.625
>> mybin2real('1010.101')
ans =
10.625
>> mybin2real('1010.101111')
ans =
10.734375
>> mybin2real('11010111.101111001')
ans =
215.736328125
SB
2012-11-2
Matt Fig
2012-11-2
Well that's an easy fix, yes? As the first step, simply check if there is a negative sign. If so, remove it and negate the number at the end.... No Big Deal....
SB
2012-11-2
Matt Fig
2012-11-2
No. As the very first thing....
if binarystring(1)=='-'
binarystring = binarystring(2:end);
F = -1;
else
F = 1;
end
Then at the end:
decimal = decimal*F;
更多回答（2 个）
Andrei Bobrov
2012-11-2
编辑：Andrei Bobrov
2012-11-3
decimal = (binarystring-'0')*pow2(size(binarystring,2)-1:-1:0).';
ADD
binarystring = '-11.11';
b = binarystring - '0';
[~,s] = ismember([-3,-2],b);
p = b(b >= 0);
decimal = sign(b(1))*p*pow2(diff(s) + (-2:-1:-1-numel(p)))';
prince kumar
2018-7-15
编辑：KSSV
2019-10-23
n=input('input binary no.=');
n=int2str(n); s=0;
for i=1:numel(n)
s=s+(str2num(n(i))*(2.^(numel(n)-i)));
end
s
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