Matlab中数字的二进制表示

Question

是否有Matlab函数返回浮点数的二进制表示形式？

Answer 1

在Matlab中，可以使用Java JDK函数。

在 Matlab 中将 float（单精度 32 位数字）转换为二进制字符串表示形式的简短答案可能是：

flt=3.14
import java.lang.Integer java.lang.Float;
Integer.toBinaryString(Float.floatToIntBits(flt))

长答案：Matlab 中的 float（单精度 32 位数字）转换为二进制字符串表示形式

function out=float2binstring(flt)
% converts a float number to binary in matlab according to IEEE754
%
% Usage:
% float2binstring(-3.14)
%
% http://www.h-schmidt.net/FloatApplet/IEEE754.html
%
% Limitations:
% Rounding errors: Not every decimal number can be expressed exactly as a     floating
% point number. This can be seen when entering "0.1" and examining its binary     representation which is either slightly smaller or larger, depending on the last bit. 
%
% Andrej Mosat, nospam@mosi.sk
% 03/2012
% v0.0
% License: GNU GPL
%
% See also: BINSTRING2FLOAT


% this is a trick to use java JDK should be installed, tested on Matlab R2010
import java.lang.Integer java.lang.Float;

if ( ~isnumeric(flt) )
    error('input must be a number');
end

out=Integer.toBinaryString(Float.floatToIntBits(flt));
end

将二进制字符串转换为浮点型，需要一点开销：

function out=binstring2float(binstr)
    % converts a binary string to float number according to IEEE754
    %
    % Usage:
    % binstring2float('11000000010010001111010111000011')
    %   -3.14
    %
    % 
    % http://www.h-schmidt.net/FloatApplet/IEEE754.html
    %
    % Limitations:
    % Rounding errors: Not every decimal number can be expressed exactly as a floating
    % point number. This can be seen when entering "0.1" and examining its binary representation which is either slightly smaller or larger, depending on the last bit. 
    %
    % Andrej Mosat, nospam@mosi.sk
    % 03/2012
    % v0.0
    % License: GNU GPL
    %
    % See also: FLOAT2BINSTRING

    import java.lang.Long java.lang.Float;

    if isequal(class(binstr), 'java.lang.String')
            binstr=char(binstr);
    end

    if ( ~isstr(binstr) )
        error('input must be a binary string');                                             
    end
    % Error handling for binary strings should be added here

    % the sign is negative

    if binstr(2)=='1'
        binstr(2)='';
        isnegative=1;
    else
        isnegative=0;
    end

    out=Float.intBitsToFloat( Long.parseLong(  binstr  , 2) );
    if isnegative
     out=-out;
    end

end