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"

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整數部分：

? ? 對2取余，然后向右移動一位，重復直到整數部分變為0

小數部分：

? ? 乘以2，看結果是否大于1，大于1則2^-1位上位1，否則為0。如果大于1，將結果減1再乘以2，否則直接乘以2，繼續判斷，直到小數部分超過32位，返回ERROR

? ? 例如：0.75d = 0.11b，乘以2實際上相當于左移，結果大于1，則說明2^-1位上是1，然后減1，繼續乘以2，結果等于1，說明2^-2上是1，且后面沒有小數了。

#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;
}

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