algorithm - 自定义函数无法将小数转换为单精度浮点

这里是一个试图将实数转换为其近似单精度浮点表示的代码，用于教育目的，以演示中间结果：

function [IEEE] = myDec2IEEE (d)
% Accepts a decimal number, d.
% Returns 1x32 bit array holding its closest IEEE representation.

  s = getSign(d);

  [e, f] = getCharacteristic(d);

  binaryE = binStr2binArr(dec2bin(e));  % convert to binary integer array.

  if numel(binaryE) < 8                 % extend to exponent bits number.
    binaryE = [zeros(1, 8 - numel(binaryE)), binaryE];
  end

  binaryF = expandFractionToBinary(f);

  IEEE = [s , binaryE, binaryF];
end

function [B] = binStr2binArr (s)
% Accepts a binary character string, s.
% Returns binary integer array, B.

  len = numel(s);
  B = zeros(1, len);

  for i = 1 : len
    B(i) = s(i) - '0';
  end

end

function [B] = expandFractionToBinary(f)
% Accepts what has remained from the decimal number
% after the calculation of the exponent, i.e its fractional part.
% Returns a 1x23 binary array which is the approximation of the
% fractional part represented as a sum of negative powers of 2,
% (up to the 22 - nd power).

  singlePrecision = 22;
  B = zeros(1, singlePrecision); % binary string to store fraction of IEEE754.
  i = 1;                         % exponent of 2; (i+1) -index in binary array.

  while f != 0 && i <= singlePrecision

    ith = 1 / (2^(i));

    if ith <= f

      f = f - ith;
      B(i) = 1;                  % include coefficient in the sum.

    end

    i = i + 1;
  end

end

function [e, f] = getCharacteristic (d)
% Accepts a number is base10, d.
% Returns the exponent and fraction in d's IEEE754 representation, in base10.

% write d in base-2 scientific  notation
% i.e. factor it into number in the range [1, 2] and a power of 2.
  bias = 127;
  i = 1;
  f = 0;

  while ~(f >= 1 && f <= 2)

    f = d / (2^(-i));
    % pause; % if number > 1 the denominator -> 0 and (faster than f -> Inf)
    i = i + 1;
  end

  i = i - 1;  % last check is done after incrementation.

  e = bias - i;
  f = f - 1;
end

function [s] = getSign (d)
% Accepts a number in base10, d.
% Returns the sign bit of its IEEE754 representation.

  if d >= 0
    s = 0;
  else
    s = 1;
  end
end

输入：

IEEE = myDec2IEEE(0.085)

输出：

Columns 1 through 21:

0   0   1   1   1   1   0   1   1   0   1   0   1   1   1   0   0   0   0   1   0

Columns 22 through 31:

1   0   0   0   1   1   1   1   0   1

但是，它只适用于0问题
我做错什么了？
如何修改代码以正确返回ieee表示的数字>=1和d注：
基于关系d=（-1）符号*2exponent-bias*（分数+1）的实现，其中分数=和（1/2^n），n=0，…，22；偏差=127。

最佳答案

经证实，MATLAB使用IEEE754进行单精度浮点运算。
因此，为什么不让MATLAB处理内部事务，然后做这样的事情呢？:

s = dec2bin(typecast(single(0.085),'uint32'), 32)

它给出：

00111101101011100001010001111011

这与所需的输出（从您提到的here中）相匹配，也适用于值>1的情况。
如果您需要一个数字，而不是字符串，结果您可以这样转换：

x = s-'0'