public class TestParseInt  {
    public static void main(String[ ] args) {
//        int n=Integer.parseInt("12");     //A  no error because 12 is string
//        int n=Integer.parseInt("12", 2);  //B  has error because ?      
//        int n=Integer.parseInt("345", 8); //C  no error because ?
//        int n=Integer.parseInt("100");    //D  no error because 100 is string
//        int n=Integer.parseInt(100);      //E  has error because 100 is not string
//        int n=Integer.parseInt(100, 16);  //F  has error because 100 is not string

in the front of each line, I have written my understanding of why there is or is not a compile error.

Please comment and enhance these statements.

Thank you.
LVL 34
Mike EghtebasDatabase and Application DeveloperAsked:
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B has errior and second paramemetr is radix - if your radix is 2 - you cannot have digit 2 in binary number
C has no error as 3,4,5 - a vialid digits in radix 8
Threst of thinsg are obvious - you cannot have int when you need String and vice versa - ther are just no such methiods with parseInt(int, int)
or parseInt(int)
There are jsut two methods parseInt in Integer class:

static int       parseInt(String s)
          Parses the string argument as a signed decimal integer.
static int       parseInt(String s, int radix)
          Parses the string argument as a signed integer in the radix specified by the second argument.

Both have first argument of type String
So any parseInt which have forst argument not a String will fail

The second argument is optional - it is int and has a meaning of radix
So obviously the string should conform to the radix, so strings continaing digit 2 with radix 2
cannot be interpreed as int - so paarseInt ("12",2) - is not legal, as binary numbers should have only 0 and 1
"345",8 on the contrary is fine, as any digits less than 8will be fine with octal numbers

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