[ad_1]
\n<\/p>\n
Two methods with the same name but different signatures is called overloaded method. At compile time Compiler will identify the method based on the method signature even though the name of the method is same.<\/p>\n
void Add(int x, int y)\n{\n Console.WriteLine(\"Add int\");\n}\nvoid Add(double x, double y)\n{\n Console.WriteLine(\"Add double\");\n} \n<\/code><\/pre>\nHere the Add(double x, double y) overlaods the first fuction Add(int x, int y) same name different signatures.
\nIf we call Add(4,2) it will call the Add(int x, int y) function
\nand if we call Add(4.0,2.0) it will call the Add(double x, double y) function.<\/p>\n
Criteria for method overloading.<\/h2>\n\n- Same name different signature.<\/li>\n
- \n
Different number of arguments or Different types of arguments or Different order of arguments.<\/p>\n<\/li>\n
- \n
Can’t overload method by Changing the return type of the method. Because return value alone is not sufficient for the compiler to figure out which method to call. Also the return type of a method is not considered to be part of a method’s signature.<\/p>\n<\/li>\n<\/ol>\n
Also note, the out<\/code> and ref<\/code> parameter modifies can<\/em> be used to differinate to overloads, like so:<\/p>\nvoid method1(int x);\nvoid method1(out int x);\n\nvoid method2(char c);\nvoid method2(ref char c);\n<\/code><\/pre>\nHowever, you cannot do this:<\/p>\n
void method(out int x);\nvoid method(ref int x);\n<\/code><\/pre>\nAs ref<\/code> and out<\/code> do not differentiate from one another, in an overload sense.
\nAnd be careful with overloads like this:<\/p>\nvoid method(int x) => Console.Writeline(x);\nvoid method(int x, int y = 0) => Console.Writeline(x * y);\n<\/code><\/pre>\nThis will compile and work, but the compiler will default to the first method that has one parameter, until you specify the second parameter, then it will call the second method.<\/p>\n<\/p><\/div>\n