c++@ overloaded functions

For a detailed analysis reference for overloading:

Http://www.cnblogs.com/skynet/archive/2010/09/05/1818636.html

The internal mechanism involves how overloaded functions resolve naming conflicts and invoke matching problems.

Criteria for resolving overloads:

• exact match : parameter matching without conversion, or just trivial conversions, such as array name to pointer, function name to pointer to function, t to const t;
• Promotion matches : integer promotion (such as bool to int, char to int, short to int), float to double
• use standard conversion matching : such as int to double, double to int, double to long double, derived* to base*, t* to void*, int to unsigned int;
• use user-defined matches ;
• use ellipsis to match : similar to the ellipsis parameter in printf

If multiple matching functions are found at the highest level, the call is rejected (because of ambiguity, ambiguous). Look at the following example:

void print (int), void print (const char*), void print (double), void print (long), void print (char), void H (char c,int i,short s , float f) {     print (c);//exact match, call print (char)     print (i);//exact match, call print (int)     print (s);//integer promotion, call print (int)     print (f);//float to double lift, call print (double)     print (' a ');//exact match, call print (char)     print (49);//Exact Match, Call print (int)     print (0);//exact match, call print (int)     print ("a");//exact match, call print (const char*)}

Defining too few or too many overloaded functions can lead to ambiguity and look at one of the following examples:

void F1 (char), void F1 (long), void F2 (char*), void F2 (int*), void K (int i) {       F1 (i);//Call F1 (char)? F1 (Long)?       F2 (0);//Call F2 (char*)? F2 (int*)? }

At this point, the compiler will error, throw the errors to the user to handle: by the display type conversion to call and so on (such as F2 (Static_cast<int *> (0), of course, it is ugly, and you want to invoke other methods when you use to convert). The above example is just a case of a parameter, let's look at a two-parameter scenario:

int pow (int, int);d ouble pow (double,double), void G () {       double D=pow (2.0,2)//Call POW (int (2.0), 2)? Pow (2.0,double (2) )?}

/* * Overloading: Two or more functions have the same function name, different number of parameters. * When the number of arguments is the same, the parameter type is different, and the overload will call the function of the type parameter first, * If not, cast the type. */#include <iostream>using namespace Std;int Ave (int n1, int n2);d ouble Ave (double n1, double n2);d ouble Ave (double N1, double N2, double n3); int main () {cout << "the average of 2.0 2.5 3.0 is:" << Ave (2.0, 2.5, 3.0) <<e  Ndl;cout << "The average of 4.5 5.5 is:" << Ave (4.5, 5.5) << endl;cout << "The average of 5 8 is:" << Ave (8, 5) << Endl;return 0;} int Ave (int n1, int n2) {return (N1-N2);} Double Ave (double n1, double n2) {return ((N1+N2)/2.0);} Double Ave (double n1, double N2, double n3) {return ((N1+N2+N3)/3.0);}

Operation Result:
The average of 2.0 2.5 3.0 Is:2.5the average of 4.5 5.5 is:5the average of 5 8 is:3

　　

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