为什么主流的静态类型语言不支持按返回类型重载函数/方法?我想不出有什么能做到。这似乎并不比支持按参数类型重载更有用或更合理。为什么它不那么受欢迎呢?
当前回答
在. net中，有时我们使用一个参数来指示一个通用结果的期望输出，然后进行转换以得到我们期望的结果。
C#
public enum FooReturnType{
IntType,
StringType,
WeaType
}
class Wea {
public override string ToString()
{
return "Wea class";
}
}
public static object Foo(FooReturnType type){
object result = null;
if (type == FooReturnType.IntType)
{
/*Int related actions*/
result = 1;
}
else if (type == FooReturnType.StringType)
{
/*String related actions*/
result = "Some important text";
}
else if (type == FooReturnType.WeaType)
{
/*Wea related actions*/
result = new Wea();
}
return result;
}
static void Main(string[] args)
{
Console.WriteLine("Expecting Int from Foo: " + Foo(FooReturnType.IntType));
Console.WriteLine("Expecting String from Foo: " + Foo(FooReturnType.StringType));
Console.WriteLine("Expecting Wea from Foo: " + Foo(FooReturnType.WeaType));
Console.Read();
}
也许这个例子也有帮助:
C++
#include <iostream>
enum class FooReturnType{ //Only C++11
IntType,
StringType,
WeaType
}_FooReturnType;
class Wea{
public:
const char* ToString(){
return "Wea class";
}
};
void* Foo(FooReturnType type){
void* result = 0;
if (type == FooReturnType::IntType) //Only C++11
{
/*Int related actions*/
result = (void*)1;
}
else if (type == FooReturnType::StringType) //Only C++11
{
/*String related actions*/
result = (void*)"Some important text";
}
else if (type == FooReturnType::WeaType) //Only C++11
{
/*Wea related actions*/
result = (void*)new Wea();
}
return result;
}
int main(int argc, char* argv[])
{
int intReturn = (int)Foo(FooReturnType::IntType);
const char* stringReturn = (const char*)Foo(FooReturnType::StringType);
Wea *someWea = static_cast<Wea*>(Foo(FooReturnType::WeaType));
std::cout << "Expecting Int from Foo: " << intReturn << std::endl;
std::cout << "Expecting String from Foo: " << stringReturn << std::endl;
std::cout << "Expecting Wea from Foo: " << someWea->ToString() << std::endl;
delete someWea; // Don't leak oil!
return 0;
}
其他回答
在. net中，有时我们使用一个参数来指示一个通用结果的期望输出，然后进行转换以得到我们期望的结果。
C#
public enum FooReturnType{
IntType,
StringType,
WeaType
}
class Wea {
public override string ToString()
{
return "Wea class";
}
}
public static object Foo(FooReturnType type){
object result = null;
if (type == FooReturnType.IntType)
{
/*Int related actions*/
result = 1;
}
else if (type == FooReturnType.StringType)
{
/*String related actions*/
result = "Some important text";
}
else if (type == FooReturnType.WeaType)
{
/*Wea related actions*/
result = new Wea();
}
return result;
}
static void Main(string[] args)
{
Console.WriteLine("Expecting Int from Foo: " + Foo(FooReturnType.IntType));
Console.WriteLine("Expecting String from Foo: " + Foo(FooReturnType.StringType));
Console.WriteLine("Expecting Wea from Foo: " + Foo(FooReturnType.WeaType));
Console.Read();
}
也许这个例子也有帮助:
C++
#include <iostream>
enum class FooReturnType{ //Only C++11
IntType,
StringType,
WeaType
}_FooReturnType;
class Wea{
public:
const char* ToString(){
return "Wea class";
}
};
void* Foo(FooReturnType type){
void* result = 0;
if (type == FooReturnType::IntType) //Only C++11
{
/*Int related actions*/
result = (void*)1;
}
else if (type == FooReturnType::StringType) //Only C++11
{
/*String related actions*/
result = (void*)"Some important text";
}
else if (type == FooReturnType::WeaType) //Only C++11
{
/*Wea related actions*/
result = (void*)new Wea();
}
return result;
}
int main(int argc, char* argv[])
{
int intReturn = (int)Foo(FooReturnType::IntType);
const char* stringReturn = (const char*)Foo(FooReturnType::StringType);
Wea *someWea = static_cast<Wea*>(Foo(FooReturnType::WeaType));
std::cout << "Expecting Int from Foo: " << intReturn << std::endl;
std::cout << "Expecting String from Foo: " << stringReturn << std::endl;
std::cout << "Expecting Wea from Foo: " << someWea->ToString() << std::endl;
delete someWea; // Don't leak oil!
return 0;
}
大多数静态语言现在也支持泛型，这将解决您的问题。如前所述，如果没有参数差异，就无法知道调用哪一个。如果你想这样做，使用泛型就可以了。
如果函数被返回类型重载并且有这两个重载
int func();
string func();
在看到这样的调用时，编译器无法确定调用这两个函数中的哪一个
void main()
{
func();
}
由于这个原因，语言设计者通常不允许返回值重载。
然而，有些语言(如MSIL)允许按返回类型重载。当然，它们也面临上述困难，但它们有变通办法，为此您必须查阅它们的文档。
如前所述，对仅因返回类型不同而不同的函数的模糊调用会引入模糊。 模糊性会导致有缺陷的代码。 必须避免有缺陷的代码。
试图模糊化所带来的复杂性表明这不是一个好的hack。 除了智力练习之外，为什么不使用带有引用参数的过程呢?
procedure(reference string){};
procedure(reference int){};
string blah;
procedure(blah)
如果你稍微换个角度来看，这个重载特性并不难管理。考虑以下几点:
public Integer | String f(int choice){
if(choice==1){
return new string();
}else{
return new Integer();
}}
如果一种语言确实返回重载，它将允许参数重载，但不允许重复。 这将解决以下问题:
main (){
f(x)
}
因为只有一个f(int)选项可供选择。
