double转换为二进制

arctan 在verilog 里是1qn或2qn格式，所以要把浮点数转换成1qn格式

1、dec2bin（十进制整数变为二进制）

Convert decimal to binary number in string

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Syntax

str = dec2bin(d)

str = dec2bin(d,n)

Description

returns the

str = dec2bin(d) binary representation of d as a string. d must be a nonnegative integer smaller than 2^52.

str = dec2bin(d,n) produces a binary representation with at least n bits.

The output of dec2bin is independent of the endian settings of the computer you are using.

2参考：https://zhidao.baidu.com/question/431648565.html?qbl=relate_question_0

这两个都是我收集的，验证过没问题
方案一：
%函数（1）含小数的十进制，会调用函数二
function y=f_d2b(n)
 strn=strtrim(num2str(n));  
    p=length(strn);
    if isempty(find(strn=='.'))
        y=d2b(n);   %这里
        return;    
    else
        for i=1:p
        if find(strn(i)=='.')
            k=i;
            break;
        end
    end
    %Retrieving INTEGER and FRACTIONAL PARTS as strings
    i_part=strn(1:k-1);
    f_part=strn(k:end);
   %Converting the strings back to numbers
    ni_part=str2num(i_part);
    nf_part=str2num(f_part);
   ni_part=d2b(ni_part);
    strtemp='';
    temp=nf_part;
      t='1';s='0';
    while nf_part>= 0
        nf_part=nf_part*2;
        if (nf_part==1) || (nf_part==temp)
            strtemp=strcat(strtemp,t);
            break;
        elseif nf_part>1
            strtemp=strcat(strtemp,t);
            nf_part=nf_part-1;
        else
            strtemp=strcat(strtemp,s);
        end
    end
     if ni_part==0
        y=strcat('0.',strtemp);        
     else
        y=strcat(ni_part,'.',strtemp);
     end
   end
%%%%%%%%%%%
%函数（2）转换整数部分
function y=d2b(n)
strtemp='';
strn=strtrim(num2str(n));
if n<0
    fprintf(' %f is not a valid number\n',n) 
end
  while n~=0
    strtemp=strcat(num2str(mod(n,2)),strtemp);
    n=floor(n/2);
  end
  y=strtemp;
方案二：
function Mydec2bin(x)
%x----------the decimal number 
%
%Recod of version:
% Date Programmer Description of change
% ======= ========= ================= 
% 9/26/2009 Iptest Original code
flag=x>0;T=abs(x);T1=fix(T);T2=T-T1;flag1=T1>0;flag2=T2>0;
i=1;j=1;
%for positive number
%integer part origin code
if flag1==1;
flag3=1;
while T1>0
N0(i)=mod(T1,2);T1=fix(T1/2);
i=i+1;
end
n=length(N0);
for ii=1:n
N(ii)=N0(n-ii+1);
end
else
N=[];flag3=0;
end
%point part origin code
if flag2==1;
flag4=1;
while T2~=0 %maybe come out no ending situation
M(j)=(T2*2>1)*fix(T2*2)+(T2*2==1);
T2=(T2*2>1)*(T2*2-1)+(T2*2<1)*T2*2;
j=j+1;
end
else
M=[];flag4=0;
end
%for real situation
if flag==1%for positive number
    if flag3==1;
        disp('integer part origin code');
        disp(N);
        disp('integer part reverse code:');
        disp(N);
    else
        disp('Not have integer part !');
    end
    if flag4==0;
        disp('Not have point part !');
    else
        disp('point part origin code:');
        disp(M);
        disp('point part reverse code:');
        disp(M);
    end
else %for negative number
N=[1 N];
if flag1==1
    if flag3==1;
        disp('integer part origin code');
        disp(N);
        disp('integer part reverse code:');
        N1=[N(1) ~N(2:end)];
        disp(N1);
    else
        disp('Not have integer part !');
    end
    if flag4==0;
        disp('Not have point part !');
    else
        disp('point part origin code:');
        disp(M);
        disp('point part reverse code:');
        M1=[~M];
        disp(M1);
    end
else
disp('integer part reverse code for symbol code:');
disp(N); 
if flag4==0;
    disp('Not have point part !');
else
    disp('point part origin code:');
    disp(M);
    disp('point part reverse code:');
    M1=[~M];
    disp(M1);
end
end
end
disp （'if there have some problem for the code please turn it out,thanks!')      

转载于:https://www.cnblogs.com/xiaoxuesheng993/p/7444355.html