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;
}