Why "539" and not "1"?

[Robert Ploch]

hi
i have to to create a method which is able to read a string, check, if it is a correct number (e.g. +10, -50, 10, but not +5w8) and return that number as an int
i coded this so far:

/**
 * @param inputString: initialized String
 * @param index: position in array
 */
 
public int makeNumber(String inputString, int index)
{
    char[] input = inputString.toCharArray();
    int number=input[index];
    
    /*stringCheck is an extern method to check, if the string was a correct number */
    
    if (stringCheck(inputString,index))
    {
      while (index < input.length)
      {
      number = number*10+input[index++];
      }
      if (i < 0)
      {
        number = number * (-1);
      }
    }
    return number;
}

when i enter "1" as a string --> number = 539, which shouldnt be!!
any idea someone?
thx
[ June 05, 2002: Message edited by: Dirk Schreckmann ]

[Jamie Robertson]

**
 * @param inputString: initialized String
 * @param index: position in array
 */
public int makeNumber(String inputString)
{
   //exception handling in the method is optional
   return Integer.parseInt( inputString.trim() );
}

Am I missing something? what is the index variable for?
Jamie

[Robert Ploch]

the index should be the position in the array, for example: if index=1, and inputString=123, the number will be generated out of 2 and 3, if index=0, the whole number will be taken into account

[Jamie Robertson]

I thought that might be the case. try this instead

**
 * @param inputString: initialized String
 * @param index: position in array
 */
public int makeNumber(String inputString, int offset)
{
   //exception handling in the method is optional
   return Integer.parseInt( inputString.substring( offset ) );
}

you might have to deal with the NumberFormatException in case of a conversion error.
Jamie
[ June 05, 2002: Message edited by: Jamie Robertson ]

[Robert Ploch]

uff, i`m a newbie and dont know exactly, what you mean with Numberformatexception etc.
anyway, if i replace "return number" with
"return Integer.parseInt( inputString.substring( index ) )", an error occurs

[Jim Yingst]

Well, what does the error message say? And what is the value of the inputString and index you're using when it occurs?

[Jim Yingst]

Also Bulla, please change your display name to include a first and last name, as discussed in our display name policy. Thanks.

[Robert Ploch]

ok
the following error occurs, if inputString=12345 and index=0
java test
please type in a string: 12345
java.lang.NumberFormatException:
at java.lang.Integer.parseInt(Compiled Code)
at java.lang.Integer.parseInt(Integer.java:375)
at MakeNumberAutomat.makeNumber(Compiled Code)
at test.main(test.java:12)

the goal is: check, if a string is a real number, and calculate a number out of a string using that formula:
((((1*10+2)*10+3)*10+4)*10+5)*10..etc.
that number should be returned.
if the inputString is <0, the number should be mulitiplicated with (-1) and also be returned
the inputString might also be "+12345" or "-12345" and even " +12345"

[Jim Yingst]

Thanks, Robert. Any relation to the author of "Psycho"?
The NumberFormatException indicates that Integer.parseInt() doesn't like the format of the input string - it doesn't look like a valid number. But this isn't making much sense, as it seems that input should parse without a problem. Perhaps the input isn't being passed to the Integer.parseInt() properly? I suggest showing the somplete code you currently have for this. Also, try inserting a few lines:

    System.out.println("Parsing: \"" + inputString 
                       + "\" at index " + index);
    String substr = inputString.substring(index);
    System.out.println("Substring: \"" + substr + "\"");
    int value = Integer.parseInt(substr);
    System.out.println("Value: " + value);
    return value;

This way you can see the results of each step, and know exactly what's going into the Integer.parseInt() just before the error occurs. You can remove the print statements later when you're happy with how everything works.

[Dirk Schreckmann]

Robert, note that you may soon discover that
Integer.parseInt( "+123" );
would throw a NumberFormatException.
You may wish to take a look at this old conversation in which I stumbled through learning about the location of the standard API source files and the basics of using jar files after coming across the above problem unexpectedly.

[Ilja Preuss]

Originally posted by Bulla:
when i enter "1" as a string --> number = 539, which shouldnt be!!
any idea someone?

It's because (int)'1' doesn't give you the integer value 1 but the unicode value of the character '1'.

[Jamie Robertson]

mabe one of these can help you accomplish your task...or give you some ideas

public static int numberFormat( String s )
{
char c[] = s.trim().toCharArray();
int sum = 0;
for (int i = 0; i < c.length ; i ++)
{
if ( c[i] == '0' )
{
sum *= 10;
}
else if ( c[i] == '1' )
{
sum *= 10;
sum += 1;
}
else if ( c[i] == '2' )
{
sum *= 10;
sum += 2;
}
else if ( c[i] == '3' )
{
sum *= 10;
sum += 3;
}
else if ( c[i] == '4' )
{
sum *= 10;
sum += 4;
}
else if ( c[i] == '5' )
{
sum *= 10;
sum += 5;
}
else if ( c[i] == '6' )
{
sum *= 10;
sum += 6;
}
else if ( c[i] == '7' )
{
sum *= 10;
sum += 7;
}
else if ( c[i] == '8' )
{
sum *= 10;
sum += 8;
}
else if ( c[i] == '9' )
{
sum *= 10;
sum += 9;
}
else
{
//invalid character routine
}

}
if ( c[0] == '-' )
{
sum *= (-1);
}

return sum;
}
public static int numberFormat2( String s )
{
//assuming the first digit is always +/- sign
int value = Integer.parseInt( s.substring(1) );
if ( s.substring( 0, 1 ).equals("-") )
{
value *= (-1);
}
return value;
}
public static int numberFormat3( String s )
{
int value = 0;
if ( s.trim().substring( 0, 1 ).equals("-") )
{
value = Integer.parseInt( s.substring(1) );
value *= (-1);
}
else if ( s.trim().substring( 0, 1 ).equals( "+") )
{
value = Integer.parseInt( s.substring(1) );
}
else
{
value = Integer.parseInt( s.trim() );
}
return value;
}

Jamie

[Robert Ploch]

thx jamie, numberformat1 was the right one, i had to modify it a littlebit cause an introducing '+' would return a wrong result, seems to be a problem in java

[David Weitzman]

Note that the massive switch is not really necessary. You can use:
char c = ...;
if (c >= '0' && c <= '9') {
int currentDigit = c - '0';
} else {
// it isn't a digit. What could it be...
}