Cracking the Coding Interview 5.2

Given a(decimal -e.g. 3.72)number that is passed in as a string, print the binary representation. If the number can not be represented accurately in binary, print "ERROR"

整数部分：

对2取余，然后向右移动一位，重复直到整数部分变为0

小数部分：

乘以2，看结果是否大于1，大于1则2^-1位上位1，否则为0。如果大于1，将结果减1再乘以2，否则直接乘以2，继续判断，直到小数部分超过32位，返回ERROR

例如：0.75d = 0.11b，乘以2实际上相当于左移，结果大于1，则说明2^-1位上是1，然后减1，继续乘以2，结果等于1，说明2^-2上是1，且后面没有小数了。

#include<iostream>
#include<string>
#include<stdlib.h>
using namespace std;

string func(const string &str)
{
string strInt;
string strDou;

string::size_type idx = str.find('.');
if(idx == string::npos)
{
strInt = str;
}
else
{
strInt = str.substr(0,idx);
strDou = str.substr(idx);
}

int intPart = atoi(strInt.c_str());
double douPart;
if(!strDou.empty())
{
douPart = atof(strDou.c_str());
}
else
{
douPart = 0.0;
}

string strIntB,strDouB;
while(intPart!=0)
{
if((intPart&1)>0)
{
strIntB = '1'+strIntB;
}
else
{
strIntB = '0'+strIntB;
}
intPart=intPart>>1;
}
while(!(douPart>-10e-15 && douPart<10e-15))
{
if(douPart*2>1)
{
strDouB = '1'+strDouB;
douPart = douPart*2-1;
}

else if((douPart*2-1)>-10e-15 && (douPart*2-1)<10e-15)
{
strDouB = '1'+strDouB;
break;
}

else
{
strDouB = '0'+strDouB;
}
if(strDouB.size()>32)
{
return "ERROR";
}
}

if(strDouB.empty())
{
return strIntB;
}
else
{
return (strIntB+'.'+strDouB);
}
}

int main()
{
string str("3.75");
cout<<func(str)<<endl;
return 0;
}