Base conversion in Java

Question

int a=012;
int b=13;
System.out.println(a+b);

The result of the print is 23, so one of them must have been declared in another base format. Which one is this and why?

E. Armand · Answer 1 · 2018-12-09T11:49:09.503

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That's because your variable a is in base-8 , so a is 0*8^2+1*8^1+2*8^0 (^ means power) , so a=10 in decimal base and that's why your answer is 23.

If you want it to be in base 10 just delete the 0 in a=012.

edited Dec 09 '18 at 11:49

answered Dec 09 '18 at 11:44

E. Armand

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score 1 · Answer 2 · answered Dec 09 '18 at 11:58

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Well, you can easily test it whats the base 10 of that number by simply:

System.out.println(Integer.toString(a,10)); // print in the console

All the rest is explained well by E. Armand, ie. how to convert base8 number to base10.

answered Dec 09 '18 at 11:58

MS90

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Wouldn't `System.out.println(a);` be a much shorter and more familiar way of showing a base-10 number? – Erwin Bolwidt Dec 09 '18 at 12:09
Sure, even from his Q. it could be denoted, but I thought it would be more concise to him doing it the way I did. He could later go and explore what's binary presentation of that octal by calling Integer.toString(012,2). @ErwinBolwidt – MS90 Dec 09 '18 at 12:17

score 0 · Answer 3 · answered Dec 09 '18 at 12:02

int a = 012;  // first '0' means that this number is octal (=10 in decimal)
int b = 13;
System.out.println(a+b);  // 012 + 13 = 10 + 13 = 23.

To change output, do use Integer.toString(int i, int radix):

System.out.println("binary: 0b" + Integer.toString(a + b, 2)); // binary: 0b10111
System.out.println("octal: 0" + Integer.toString(a + b, 8));   // octal: 027
System.out.println("dec: " + Integer.toString(a + b, 10));     // dec: 23
System.out.println("hex: 0x" + Integer.toString(a + b, 16));   // hex: 0x17

Base conversion in Java

3 Answers3