String-Double not working for negative numbers

Question

public class Convert{
    public static void main(String args[]){
        String num="-12.0";
        int g=0,fnl=0;
        int exp=(num.indexOf(".")-1);
        String num2=num;
        if(num.charAt(0)==('-')){
            num=num.substring(1);
            System.out.println(num);
        }
        String numb=num;
        System.out.println(numb);
        int d=0;
        while(numb.charAt(d)!='.'){
                g=numb.charAt(d)-48;
                System.out.println(g);
                int k = 1;
                for(int f=0;f<exp;f++){
                k=(k*10);
                }
                fnl+=(k*g);
                d++;
                exp--;
                System.out.println(fnl);

        }

        num=(num.substring(d) );

         double num1= Double.parseDouble(num); 
         double finalnum=(fnl+num1);
         System.out.println(fnl);
         if(num2.charAt(0)=='-'){
             finalnum*=-1;
         }

    }
}

I created my own method to convert the integer part of the string and it's only working for positive numbers. :( It should work for both positive and negative numbers. and I used parse double to convert the numbers after the decimal point. I cant use parse double to convert the whole string to double because we're not allowed to do so.

you probably need to account for the fact that the position of the exponent that you expect will be different when a negative sign is prepended to the number — ealfonso, Sep 14 '13 at 08:28
@user2768260 Is your requirement is just to convert String or what? — SpringLearner, Sep 14 '13 at 08:33
Is this a homework problem?? You said you were not allowed to do so. What does that mean?? — prasanth, Sep 14 '13 at 08:35
@prasanth : even if it is a homework he has atleast come up with this much of code. so there is nothing wrong in helping him. I hope you would agree. :) — Thirumalai Parthasarathi, Sep 14 '13 at 08:56
@prasanth : no offense meant, usually the comments like "is this your homework?" always ends in "this is not a code-writing service, bla bla" as most OP would post requirements and not the code they have written so far.. you know what i mean :) — Thirumalai Parthasarathi, Sep 14 '13 at 10:02
@user2768260 : Please mark an answer that has solved your problem — Thirumalai Parthasarathi, Sep 14 '13 at 16:48

progrenhard · Answer 1 · 2013-09-14T09:15:23.047

public static double toDouble(String a){
     int sign = 1;
     int start = 0;
     if(a.charAt(0) == '-'){
         start = 1;
         sign = -1;
     }
     double value = 0;
     boolean decimal = false;
     int exp = 0;

     for(int i = start; i < a.length(); i++){
         if(a.charAt(i) == '.'){
             if(decimal) return 0.0;
             decimal = true;
         }else{
             value += (a.charAt(i) - 48) * Math.pow(10,a.length() - i - 1);
         }

         if(decimal) exp++;
     }

     value =  value / Math.pow(10,exp);
     value *= sign;
      System.out.println(value);
     return value;

 }

I sort of rewrote it the way i would do this problem. If you remember i posted psuedo code on your other question this is my "programmed" version.

This is how my code works.

First it checks the sign at the first character [0] then increments the start position to 1 instead of zero to start the loop past the - sign.

It loops through the array and checks to see if there is a . if it is it sets the decimal variable to true instead of false. If it reads more than one decimal point in the string it will return a 0.0 which i denoted as a fail value. However, if it doesn't pick up anymore decimals it increments the exponent value to see how many spaces it will need to jump.

value += (a.charAt(i) - 48) * Math.pow(10,a.length() - i - 1);

this line of code is very interesting. a.charAt(i) - 48 converts the character to an integer value. then i multiply it by 10^(what power of ten the value is at then subtract one to compensate for 1.0). Then i add it value.

Everything else is pretty self explanatory.

NOTE:

I tested this code myself, however one important thing is that i haven't added checks for invalid strings with other characters.

may i know the use of this statement `if(decimal) return 0.0`? — Thirumalai Parthasarathi, Sep 14 '13 at 09:58
@BlackPanther It's a error check essentially. If it detects more than one `.` it would be an invalid string `3.333.3` is not a decimal so it returns a `null` in this case i represented it as `0.0` — progrenhard, Sep 14 '13 at 16:29
oh nice.. :) but if this `int count = a.length() - a.replace(".", "").length();` is implemented and a check is made on count before the for loop, saves time because if `3.333.3` is entered we wouldn't be wasting time by parsing 5 characters unnecessarily :) — Thirumalai Parthasarathi, Sep 14 '13 at 16:41
@BlackPanther `.replace` is a linear operation we would be adding more complexity. The complexity right now (not including pow function) would be `0(n)` if we included that it would be `0(n+n)` <-- Even though that just boils down to O(n) haha but you get my point. Also, i don't believe in we should "fix" there string that they imputed.. — progrenhard, Sep 14 '13 at 17:18
dude for a minute there i was like "Say what?" with all those big O and stuffs... But yeah i get your point.. thx for the info.. +1 for that :) — Thirumalai Parthasarathi, Sep 14 '13 at 17:28

Thirumalai Parthasarathi · Answer 2 · 2013-09-14T09:31:57.543

1

int g=0,fnl=0;
String input  = "-12.0";
String temp = input;
String forDecimalPart = input;
if(input.charAt(0)=='-') {
    temp=input.substring(1);
}

int exp = temp.indexOf(".");
System.out.println(exp);
int d=0;
while(temp.charAt(d)!='.') {
    g=temp.charAt(d)-48;
    int k = 1;
    for(int f=1;f<exp;f++) {
      k*=10;
    }
    fnl+=(k*g);
    d++;
    exp--;
    System.out.println(fnl);
}

forDecimalPart=temp.substring(d) ;

double num1= Double.parseDouble(forDecimalPart); 
double finalnum=(fnl+num1);
if(input.charAt(0)=='-'){
finalnum*=-1;
}
System.out.println(finalnum);

i have given you the answer as you have put the effort to come up with atleast this code. good luck :) I have scripted the code resembling your question without adding any Extra methods from other classes like Math.pow()

edited Sep 14 '13 at 09:31

answered Sep 14 '13 at 08:55

Thirumalai Parthasarathi

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Are you sure declaring all variables in a single place is a convention? I think it was in "Effective Java" where the opposite was recommended. – Igor Rodriguez Sep 14 '13 at 09:23
@IgorRodriguez : i stand corrected... im sorry for the misinformation. – Thirumalai Parthasarathi Sep 14 '13 at 09:26
1

@IgorRodriguez : i believe the convention is to declare the variable before the immediate use of it. or at the start of the block in which it is going to be used. Thanks for the tip, it made me look it up :) +1 for that – Thirumalai Parthasarathi Sep 14 '13 at 09:33

ealfonso · Answer 3 · 2013-09-14T17:10:07.170

import java.util.regex.*;

public class DoubleParse{

public static void main(String[] args)
{

    String[] doubles = "12.3 4.08 5 -5 -5.4 4.0 0 0.0 5.800 asda".split(" ");
    for (String num: doubles)
    {
        System.out.println(parse_double(num));
    }
}


final static Pattern double_format = Pattern.compile("^(\\-?)[0]*([0-9]+)(?:\\.([0-9]+?)[0]*)?$");


public static double parse_double(String num)
{

    Matcher m = double_format.matcher(num);
    if (!m.matches())
        throw new IllegalArgumentException(num+" is not in proper format!");

    int sign = num.charAt(0)=='-'? -1 : 1;
    int point = num.indexOf('.');
    int whole = parse_int(num.substring(sign==-1? 1 : 0, (point!=-1)? point : num.length()));
    double fraction=0;
    if (point != -1)
    {
        String fractionS = num.substring(point+1);
        fraction = parse_int( fractionS)/Math.pow(10, fractionS.length());
    }
    return (whole+fraction)*sign;
}

public static int parse_int(String numS)
{
    int num=0;
    for (int i=0;i<numS.length();i++)
    {
        num *= 10;
        num += (int) (numS.charAt(i)-'0');
    }
    return num;
}
}

you doing essentially three 0(n) operations in your question http://stackoverflow.com/questions/4378455/what-is-the-complexity-of-regular-expression — progrenhard, Sep 15 '13 at 03:12
lol. I do check that the input is in valid format (unlike your solution). However, this does not affect the asymptotic time complexity of the code. — ealfonso, Sep 15 '13 at 03:26

String-Double not working for negative numbers

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