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Author: disha2sinha

Methods/VarargsMethodOverloading.java

@@ -9,7 +9,7 @@ void varargs(int...a)
         }
     }
 
-    void varargs(boolean...a) {
+    /*void varargs(boolean...a) {
         System.out.println("METHOD:varargs(boolean...)\tNUMBER OF ARGUMENTS:" + a.length + "\tITEMS:");
         for(boolean i:a)
         {
@@ -23,7 +23,7 @@ void varargs(String... a) {
         {
             System.out.println(i+" ");
         }
-    }
+    }*/
 
     void varargs(float... a) {
         System.out.println("METHOD:varargs(float...)\tNUMBER OF ARGUMENTS:" + a.length + "\tITEMS:");
@@ -46,8 +46,18 @@ public static void main(String[] args) {
         Test obj=new Test();
         obj.varargs("Hello World",10,30,80,90);
         obj.varargs(10.9f,8.0f);
-        obj.varargs(true,false);
-        obj.varargs("Disha","Argha","World");
+        //obj.varargs(true,false);
+        //obj.varargs("Disha","Argha","World");
+        obj.varargs();/*int is more specific than double. If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run-time method dispatch. The Java programming language uses the rule that the most specific method is chosen according to type promotion. The following rules define the direct supertype relation among the primitive types in this case:
+
+    double > float
+    float > long
+    long > int
+    int > char
+    int > short
+    short > byte*/
+
+
     }
 
 }