Alternative method to dec2bin function
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I need help in the the code with alternative to bin2dec function which gives output in the double class
I have been trying but no success. Can anyone help
6 个评论
Jan
2022-3-7
This sounds like a homework question. Then please post, what you have tried so far and ask a specific question.
Hint: To get the first bit, use:
x = 17;
rem(x, 2)
To get the others, you can divide by 2 successively and round.
Haseeb Hashim
2022-3-7
Haseeb Hashim
2022-3-7
编辑：Haseeb Hashim
2022-3-7
Jan
2022-3-7
This is the floor() function. Another hint: You do not need loops:
x = 17;
y = x ./ [1, 2, 4, 8, 16]
With floor() you get the natural numbers and rem() extracts the first bit. You can express [1,2,4,8] as pow2(0:3).
Haseeb Hashim
2022-3-7
编辑：Rik
2022-3-7
Rik
2022-3-7
If you want to improve this code: try to determine the number of bits you need. That way you can use array operations. That removes the need for a loop and prevents dynamically growing your bin array.
采纳的回答
Some simplifications:
% Replace:
flag = true;
while flag == true
% by
while true
If you have checked for:
if bin(i) == 0
an
else
is enough and there is no reason to check for
elseif bin(i) ~= 0
again. But in your case there is no need to distinguish the 2 cases, because num = floor(num/2) works in all cases. Instead of catching the last bin manually, you can do this in the loop also:
num = 29;
i = 1;
while true
bin(i) = rem(num, 2);
num = floor(num / 2);
if num == 0
break
end
i = i + 1;
end
flip(bin)
Or even simpler: Check the condition instead of breaking an infinite loop:
num = 29;
bin = [];
while num > 0
bin(end+1) = rem(num, 2);
num = floor(num / 2);
end
flip(bin)
The Matlabish way is to omit the loop and to determine the number of bits in advance:
nBit = floor(log2(num)) + 1;
bin = rem(floor(num .* pow2(1-nBit:0)), 2);
Take the time to understand, what's going on here. Run it in pieces:
1-nBit:0
pow2(1-nBit:0)
num .* pow2(1-nBit:0)
floor(num .* pow2(1-nBit:0))
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