conversion from long to float

[tony kanvas]

see this code

class q20 { 
public static void main(String[] args) { 
System.out.println("Hello World!"); 
q20 qq = new q20(); 
qq.callme(7l,7l,7l); 
} 
void callme(float d, float f, long i){ 
System.out.println("One"); 
} 
void callme(float ff, double dd , long ll){ 
System.out.println("Two"); 
} 
void callme(int ff, int dd , int ll){ 
System.out.println("Three"); 
} 
  

i need some explanation about the conversion from long to float and why convert to double as we know double is A defult and long 64 and float32

[Igor Zeta]

Originally posted by tony kanvas:

class q20 { 
public static void main(String[] args) { 
System.out.println("Hello World!"); 
q20 qq = new q20(); 
qq.callme(7l,7l,7l); 
} 
void callme(float d, float f, long i){ 
System.out.println("One"); 
} 
void callme(float ff, double dd , long ll){ 
System.out.println("Two"); 
} 
void callme(int ff, int dd , int ll){ 
System.out.println("Three"); 
} 
  

In this code i see there's a call to an overloaded method,i think it will be called the triple int parameters method because best fits the call,
so why the thought of converting?
I apologize for my poor english...!

[Rehana Shaik]

Conversion rule's in java are as follows
byte-->short--> int -->long--> float -->double

[Thomas Paul]

Originally posted by Rehana Shaik:
Conversion rule's in java are as follows
byte-->short--> int -->long--> float -->double

Anything moving in the diredction of the arrows does not require a cast. Anything moving against the arrows requires a cast.

[Sridhar Srikanthan]

Originally posted by Thomas Paul:

Anything moving in the diredction of the arrows does not require a cast. Anything moving against the arrows requires a cast.

If the order given above is to be taken into account, how could one explain the result of
the above program as
Hello World!
One
instead of
Hello World!
Two
because the middle varibale of the method which prints Two is a double.That would be even more precise...or am I missing something here?
Thanks in advance
Sri

[Rehana Shaik]

Method call conversions always look for the exact data type or a wider one in the method signatures. They will not do narrowing conversions to resolve methods, instead we will get a compile error. so it will go for float other than double.

[John Paverd]

From the Java Language Specification, Section 15.12:

The Java programming language uses the rule that the most specific method is chosen.
The informal intuition is that one method declaration is more specific than another if any invocation handled by the first method could be passed on to the other one without a compile-time type error.

The method "Two" could not "pass on" the call to the method "One" because it would require an explicit cast of the 2nd parameter from double to float.
However, method "One" could "pass on" the call to method "Two", because the compiler can implicitly widen the 2nd parameter from float to a double.
Therefore method "One" is more specific than method "Two" and is chosen by the compiler.
And it's difficult to see in my browser, but the arguments in the method call are 7L, not 71.

[Sridhar Srikanthan]

Oops,
John,
Even i thought them to be 71 instead of 7L. my mistake
Thanks for the answer
Sri