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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) ); } } 
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The parseInt function converts its first argument to a string, parses that string, then returns an integer or NaN . If not NaN , the return value will be the integer that is the first argument taken as a number in the specified radix .
The parseInt method parses a value as a string and returns the first integer. A radix parameter specifies the number system to use: 2 = binary, 8 = octal, 10 = decimal, 16 = hexadecimal. If radix is omitted, JavaScript assumes radix 10.
Java Integer parseInt (String s) Method It returns the integer value which is represented by the argument in a decimal integer.
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).
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