CareerCup

class Float2String
{
public static void main(String[] args)
{
float f=7.64f;
String result;
result=""+f;
System.out.println(result);
}
}

int i;
i = (int) f;
//convert this integer into string ... and put a decimal ...
.
.
f = f - i; //now f contains the fractional part ...
f *=10;
i = (int) f; //this is first digit to the right of decimal ...
.
.
f = f-i;
f *= 10;
i = (int) f; // this is second digit to the right of decimal...
// repeate above steps four more times ...

is use of overloaded operator allowed ? if so in Java

String result = "" + <number value i.e 7.64>;
but it can not be this simple :)

I solved it in c++...Please let me know any improvements..

#include<iostream>
#include<fstream>
#include<string>
#include<cstring>

#define MAXLEN 350

using namespace std;

void reverse(char * str , int len){
    int i=0,j=len-1;

    char tmp;

    while(i<j){
        tmp=str[j];
        str[j]=str[i];
        str[i]=tmp;
        j--;i++;
    }

}

char * floatToString(double f ,char * str, int fractionPartSize)
{
    ofstream fout1 ("float1.out");

    int sign=f<0?1:0;
    int pos =0;
    int intPart = (int)f;
    double fractionPart= f-intPart;

    //one char at a time , first int part then fraction part

    if(intPart==0)
    {
        str[pos++]='0';
    }
    else if(sign==1)
    {
        str[pos++]='-';
        intPart=-intPart;
    }
        while(intPart>0){
            str[pos++]='0'+intPart%10;
            intPart/=10;
        }

        int intPartSize=pos-sign;
        reverse(str+sign,intPartSize);
        //int i=0;
        str[pos++]='.';
        while(pos<fractionPartSize+intPartSize+1){
            f=f-(int)f;
            f*=10;
            str[pos++]='0'+(int)f;
        }
         str[pos]='\0';
         fout1<<"string"<<str;


}

int main()
{
    ifstream fin ("float.in");
    ofstream fout("float.out");

    char mainStr[MAXLEN];

    double f;

    fin>>f;

    int fractionPartSize=6;

    floatToString(f,mainStr,fractionPartSize);

    return 0;

}

#include<iostream.h>
#include<string.h>

string reverse (string src)
{
int i = 0, j = src.length() - 1;
while(i < j)
{
char temp = src.at(i);
src.at(i) = src.at(j);
src.at(j) = temp;
i++;
j--;
}
return src;
}

string intToString(int src)
{
string result;
do
{
int res = (int)'0' + src%10;
src = src/10;
result.push_back(res);
}
while(src != 0);
return reverse(result);
}


int main(int argc, char* argv[])
{
string str;
float f = 7.65;
int intPart = (int)f;

int decPart = 0;
float temp = f - int(f);
while(temp - (int)temp != 0.0)
{
temp = temp*10;
}
decPart = temp;

str = intToString(intPart)+ '.'+ intToString(decPart);
cout << str;
cin.get();
return 0;
}

float x = 7.98;
	char a[5] ;
	for (int i =0; i<4; i++){
		int m = (int)x;
		if(i==1)
			a[i] = '.';
		else 
			a[i] = m+48;
		x = x-m;
		x = x*10;
	}

	a[4] = '\0';
	string str = a;
	cout<<str<<endl;

#include<stdio.h>
#include<string.h>
main()
{
  float x = 7.98;
  char a[5] ;
  int i; 
	for (i =0; i<=4; i++){
		int m = (int)x;
		if(i==1)
			{a[i] = '.';a[i+1]=m+48;i++;}
		else 
			a[i] = m+48;
		x = x-m;
		x = x*10;
	}

	a[4] = '\0';
	
        puts(a);
}

int convert_f_t_s(float num, int decimal_num)
{
int i = (int) num;
char *str = NULL;
float decimal = num - i;
int i_num = 0;
int tmp = 0;
int j = 0;
while (i) {
i = i / 10;
i_num++;
}

str = (char *)malloc(i_num + decimal_num + 2);
memset (str, 0, i_num + decimal_num + 2);
i = (int) num;
decimal = num - i;
for (j = 0; j < i_num; j++) {
tmp = i % 10;
i = i / 10;
str[i_num - j -1] = '0' + tmp;
}
str[i_num] = '.';
for (j = 0; j < decimal_num; j++) {
decimal = decimal * 10;
tmp = ((int)decimal % 10);

str[i_num + j + 1] = '0' + tmp;
}

printf("orl str is %s\n", str);
}
int main(int argc, char **argv)
{
float num = 34.234;

convert_f_t_s(num, 3);
}

I guess questions means not using anything to convert from number to characters.
And thus the answer should be independent of the language used.
My solution : use a trie containing 0-9. and then traverse in the trie with different digits from the number as input.

In JAVA

public static void main(){

float val = 7.6f ;
String str = " "+val;  //  This same statement can also be use in C# to convert float variable    	//into string
System.out.print(str);
}

What they basically asked for is the implementation of itoa function.
Not the use of it to get the work done.

stringstream ss;
    string s;
    double f=1.94;
    ss<<f;
    ss>>s;
    cout<<s;

geekyjumps.blogspot.in/2013/09/given-float-number-convert-it-into.html

Here is a really good answer from stackoverflow:

1. Use the log-function to find out the magnitude m of your number. If the magnitude is negative print "0." and an appropriate amount of zeros.
2. Consecutively divide by 10^m and cast the result to int to get the decimal digits. m-- for the next digit.
3. If you came accross m==0, don't forget to print the decimal point ".".
4. Break off after a couple of digits. If m>0 when you break of, don't forget to print "E" and itoa(m). (Here I meant your own implementation of itoa, which is easy to implement)

Instead of the log-function you can also directly extract the exponent by bitshifting and correcting for the exponent's offset (see IEEE 754, where 1st bit is sign, next 8 bits is exponent and next 23 bits significand). For instance, Java has a floatToLongBits etc. functions to get the binary representation and than you can create the string using bit manipulation.