What is the radix parameter in Java, and how does

问题:

I understand that radix for the function Integer.parseInt() is the base to convert the string into. Shouldn't 11 base 10 converted with a radix/base 16 be a B instead of 17?

The following code prints 17 according to the textbook:

public class Test {
  public static void main(String[] args) {
    System.out.println( Integer.parseInt("11", 16) );
  }
}

回答1:

When you perform the ParseInt operation with the radix, the 11 base 16 is parsed as 17, which is a simple value. It is then printed as radix 10.

You want:

System.out.println(Integer.toString(11, 16));

This takes the decimal value 11(not having a base at the moment, like having "eleven" watermelons(one more than the number of fingers a person has)) and prints it with radix 16, resulting in B.

When we take an int value it's stored as base 2 within the computer's physical memory (in nearly all cases) but this is irrelevant since the parse and tostring conversions work with an arbitrary radix (10 by default).

回答2:

It's actually taking 11 in hex and converting it to decimal. So for example if you had the same code but with "A" in the string, it would output 10.

回答3:

Here,

public class Test {
      public static void main(String[] args) {
      System.out.println(Integer.parseInt("11", 16));
    }
}

11 is 16 based number and should be converted at 10 i.e decimal.

 So, integer of (11)16 = 1*16^1 +1*16^0 = 16+1 = 17

回答4:

The function act backwards as you think. You convert "11" in base 16 to base 10, so the result is 17.

回答5:

To convert from base 10 to base 16 use

System.out.println(Integer.toString(11, 16));

Output will be b.