regarding long and float

[saxena neera]

i was developing the below code....

     static String m(float i)
    		{return "float";
    		}
    		static String m(double i) 
    		{return "double";
    		}		
    		public static void main (String[] args) {
    		int a1 = 1;
    		long b1 = 2;
    		System.out.print(m(a1)+","+ m(b1));
    	}

Both results in output float, float , what the reason behind that please advise and how can I call double please advise thanks a lot..!

[Winston Gutkowski]

saxena neera wrote:Both results in output float, float , what the reason behind that please advise and how can I call double please advise thanks a lot..!

I suggest you look here.

Winston

[Vinod Vijay]

saxena neera wrote:i was developing the below code....

     static String m(float i)
    		{return "float";
    		}
    		static String m(double i) 
    		{return "double";
    		}		
    		public static void main (String[] args) {
    		int a1 = 1;
    		long b1 = 2;
    		System.out.print(m(a1)+","+ m(b1));
    	}

Both results in output float, float , what the reason behind that please advise and how can I call double please advise thanks a lot..!

Both are returning different string objects because you are returning string only. Method parameters have nothing to do here in this case. they are not doing any job. Period

[Ulf Dittmer]

Vinod Vijay wrote:Method parameters have nothing to do here in this case. they are not doing any job.

I think you misunderstood the issue at hand. It is entirely about method parameters, and in particular, their type. It hinges on the fact that the parameter that fits most closely is chosen, and for a long, that is float, not double.

One way to achieve this would to call the method like this: m((double) b1)