Java overloading int to long

Question

Referring to Java overloading - long and float, which mentions rules in JLS #15

The following rules define the direct supertype relation among the primitive types:

double >1 float

float >1 long

long >1 int

int >1 char

int >1 short

short >1 byte

where "S >1 T" means "T is a direct subtype of S", as per JLS #4.10 immediately above this section.

Why is the following code prints float?

int q = 2;
a(q);

void b(long a) {
        System.out.println("long");
}

void a(float a) {
        System.out.println("float");
}

did you intend both of the methods to have the same name? – Eran Dec 23 '18 at 12:19 — Eran, Dec 23 '18 at 12:19

Ori Marko · Accepted Answer · 2018-12-23T13:33:08.397

0

There's only 1 method named a which accept float parameter,

Java support relation (direct supertype relation) from float to int (using long in the middle):

float > long - > int

If you rename method b to a, you will execute method with long as you expected

void a(long a) {
        System.out.println("long");
}

edited Dec 23 '18 at 13:33

answered Dec 23 '18 at 12:15

Ori Marko

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Why conversion didn't stop at long? and if there were static void c(double c) { System.out.println("double"); }, the result is not "double" but still "float" – Sajin Surendran Dec 23 '18 at 12:19

Java overloading int to long

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