Function overloading by return type?

Question

Why don\'t more mainstream statically typed languages support function/method overloading by return type? I can\'t think of any that do. It seems no less useful or reasona

Answer 1

This one is slightly different for C++; I don't know if it would be considered overloading by return type directly. It is more of a template specialization that acts in the manner of.

util.h

#ifndef UTIL_H
#define UTIL_H

#include <string>
#include <sstream>
#include <algorithm>

class util {
public: 
    static int      convertToInt( const std::string& str );
    static unsigned convertToUnsigned( const std::string& str );
    static float    convertToFloat( const std::string& str );
    static double   convertToDouble( const std::string& str );

private:
    util();
    util( const util& c );
    util& operator=( const util& c );

    template<typename T>
    static bool stringToValue( const std::string& str, T* pVal, unsigned numValues );

    template<typename T>
    static T getValue( const std::string& str, std::size_t& remainder );
};

#include "util.inl"

#endif UTIL_H

util.inl

template<typename T>
static bool util::stringToValue( const std::string& str, T* pValue, unsigned numValues ) {
    int numCommas = std::count(str.begin(), str.end(), ',');
    if (numCommas != numValues - 1) {
        return false;
    }

    std::size_t remainder;
    pValue[0] = getValue<T>(str, remainder);

    if (numValues == 1) {
        if (str.size() != remainder) {
            return false;
        }
    }
    else {
        std::size_t offset = remainder;
        if (str.at(offset) != ',') {
            return false;
        }

        unsigned lastIdx = numValues - 1;
        for (unsigned u = 1; u < numValues; ++u) {
            pValue[u] = getValue<T>(str.substr(++offset), remainder);
            offset += remainder;
            if ((u < lastIdx && str.at(offset) != ',') ||
                (u == lastIdx && offset != str.size()))
            {
                return false;
            }
        }
    }
    return true;    
}

util.cpp

#include "util.h"

template<>
int util::getValue( const std::string& str, std::size_t& remainder ) {
    return std::stoi( str, &remainder );
} 

template<>
unsigned util::getValue( const std::string& str, std::size_t& remainder ) {
    return std::stoul( str, &remainder );
}

template<>
float util::getValue( const std::string& str, std::size_t& remainder ) {
    return std::stof( str, &remainder );
}     

template<>   
double util::getValue( const std::string& str, std::size_t& remainder ) {
    return std::stod( str, &remainder );
}

int util::convertToInt( const std::string& str ) {
    int i = 0;
    if ( !stringToValue( str, &i, 1 ) ) {
        std::ostringstream strStream;
        strStream << __FUNCTION__ << " Bad conversion of [" << str << "] to int";
        throw strStream.str();
    }
    return i;
}

unsigned util::convertToUnsigned( const std::string& str ) {
    unsigned u = 0;
    if ( !stringToValue( str, &u, 1 ) ) {
        std::ostringstream strStream;
        strStream << __FUNCTION__ << " Bad conversion of [" << str << "] to unsigned";
        throw strStream.str();
    }
    return u;
}     

float util::convertToFloat(const std::string& str) {
    float f = 0;
    if (!stringToValue(str, &f, 1)) {
        std::ostringstream strStream;
        strStream << __FUNCTION__ << " Bad conversion of [" << str << "] to float";
        throw strStream.str();
    }
    return f;
}

double util::convertToDouble(const std::string& str) {
    float d = 0;
    if (!stringToValue(str, &d, 1)) {
        std::ostringstream strStream;
        strStream << __FUNCTION__ << " Bad conversion of [" << str << "] to double";
        throw strStream.str();
    }
    return d;
}

This example is not exactly using function overload resolution by return type, however this c++ non object class is using template specialization to simulate function overload resolution by return type with a private static method.

Each of the convertToType functions are calling the function template stringToValue() and if you look at the implementation details or algorithm of this function template it is calling getValue<T>( param, param ) and it is returning back a type T and storing it into a T* that is passed into the stringToValue() function template as one of its parameters.

Other than something like this; C++ does not really have a mechanism to have function overloading resolution by return type. There may be other constructs or mechanisms that I'm not aware of that could simulate resolution by return type.

Answer 2

In .NET, sometimes we use one parameter to indicate the desired output from a generic result, and then made a conversion to get what we expect.

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();
    }

Maybe this example could help too:

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;
}

Answer 3

Contrary to what others are saying, overloading by return type is possible and is done by some modern languages. The usual objection is that in code like

int func();
string func();
int main() { func(); }

you can't tell which func() is being called. This can be resolved in a few ways:

1. Have a predictable method to determine which function is called in such a situation.
2. Whenever such a situation occurs, it's a compile-time error. However, have a syntax that allows the programmer to disambiguate, e.g. int main() { (string)func(); }.
3. Don't have side effects. If you don't have side effects and you never use the return value of a function, then the compiler can avoid ever calling the function in the first place.

Two of the languages I regularly (ab)use overload by return type: Perl and Haskell. Let me describe what they do.

In Perl, there is a fundamental distinction between scalar and list context (and others, but we'll pretend there are two). Every built-in function in Perl can do different things depending on the context in which it is called. For example, the join operator forces list context (on the thing being joined) while the scalar operator forces scalar context, so compare:

print join " ", localtime(); # printed "58 11 2 14 0 109 3 13 0" for me right now
print scalar localtime(); # printed "Wed Jan 14 02:12:44 2009" for me right now.

Every operator in Perl does something in scalar context and something in list context, and they may be different, as illustrated. (This isn't just for random operators like localtime. If you use an array @a in list context, it returns the array, while in scalar context, it returns the number of elements. So for example print @a prints out the elements, while print 0+@a prints the size.) Furthermore, every operator can force a context, e.g. addition + forces scalar context. Every entry in man perlfunc documents this. For example, here is part of the entry for glob EXPR:

In list context, returns a (possibly empty) list of filename expansions on the value of EXPR such as the standard Unix shell /bin/csh would do. In scalar context, glob iterates through such filename expansions, returning undef when the list is exhausted.

Now, what's the relation between list and scalar context? Well, man perlfunc says

Remember the following important rule: There is no rule that relates the behavior of an expression in list context to its behavior in scalar context, or vice versa. It might do two totally different things. Each operator and function decides which sort of value it would be most appropriate to return in scalar context. Some operators return the length of the list that would have been returned in list context. Some operators return the first value in the list. Some operators return the last value in the list. Some operators return a count of successful operations. In general, they do what you want, unless you want consistency.

so it's not a simple matter of having a single function, and then you do simple conversion at the end. In fact, I chose the localtime example for that reason.

It's not just the built-ins that have this behavior. Any user can define such a function using wantarray, which allows you to distinguish between list, scalar, and void context. So, for example, you can decide to do nothing if you're being called in void context.

Now, you may complain that this isn't true overloading by return value because you only have one function, which is told the context it's called in and then acts on that information. However, this is clearly equivalent (and analogous to how Perl doesn't allow usual overloading literally, but a function can just examine its arguments). Moreover, it nicely resolves the ambiguous situation mentioned at the beginning of this response. Perl doesn't complain that it doesn't know which method to call; it just calls it. All it has to do is figure out what context the function was called in, which is always possible:

sub func {
    if( not defined wantarray ) {
        print "void\n";
    } elsif( wantarray ) {
        print "list\n";
    } else {
        print "scalar\n";
    }
}

func(); # prints "void"
() = func(); # prints "list"
0+func(); # prints "scalar"

(Note: I may sometimes say Perl operator when I mean function. This is not crucial to this discussion.)

Haskell takes the other approach, namely to not have side effects. It also has a strong type system, and so you can write code like the following:

main = do n <- readLn
          print (sqrt n) -- note that this is aligned below the n, if you care to run this

This code reads a floating point number from standard input, and prints its square root. But what is surprising about this? Well, the type of readLn is readLn :: Read a => IO a. What this means is that for any type that can be Read (formally, every type that is an instance of the Read type class), readLn can read it. How did Haskell know that I wanted to read a floating point number? Well, the type of sqrt is sqrt :: Floating a => a -> a, which essentially means that sqrt can only accept floating point numbers as inputs, and so Haskell inferred what I wanted.

What happens when Haskell can't infer what I want? Well, there a few possibilities. If I don't use the return value at all, Haskell simply won't call the function in the first place. However, if I do use the return value, then Haskell will complain that it can't infer the type:

main = do n <- readLn
          print n
-- this program results in a compile-time error "Unresolved top-level overloading"

I can resolve the ambiguity by specifying the type I want:

main = do n <- readLn
          print (n::Int)
-- this compiles (and does what I want)

Anyway, what this whole discussion means is that overloading by return value is possible and is done, which answers part of your question.

The other part of your question is why more languages don't do it. I'll let others answer that. However, a few comments: the principle reason is probably that the opportunity for confusion is truly greater here than in overloading by argument type. You can also look at rationales from individual languages:

Ada: "It might appear that the simplest overload resolution rule is to use everything - all information from as wide a context as possible - to resolve the overloaded reference. This rule may be simple, but it is not helpful. It requires the human reader to scan arbitrarily large pieces of text, and to make arbitrarily complex inferences (such as (g) above). We believe that a better rule is one that makes explicit the task a human reader or a compiler must perform, and that makes this task as natural for the human reader as possible."

C++ (subsection 7.4.1of Bjarne Stroustrup's "The C++ Programming Language"): "Return types are not considered in overload resolution. The reason is to keep resolution for an individual operator or function call context-independent. Consider:

float sqrt(float);
double sqrt(double);

void f(double da, float fla)
{
    float fl = sqrt(da);     // call sqrt(double)
    double d = sqrt(da); // call sqrt(double)
    fl = sqrt(fla);            // call sqrt(float)
    d = sqrt(fla);             // call sqrt(float)
}

If the return type were taken into account, it would no longer be possible to look at a call of sqrt() in isolation and determine which function was called." (Note, for comparison, that in Haskell there are no implicit conversions.)

Java (Java Language Specification 9.4.1): "One of the inherited methods must be return-type-substitutable for every other inherited method, or else a compile-time error occurs." (Yes, I know this doesn't give a rationale. I'm sure the rationale is given by Gosling in "the Java Programming Language". Maybe someone has a copy? I bet it's the "principle of least surprise" in essence.) However, fun fact about Java: the JVM allows overloading by return value! This is used, for example, in Scala, and can be accessed directly through Java as well by playing around with internals.

PS. As a final note, it is actually possible to overload by return value in C++ with a trick. Witness:

struct func {
    operator string() { return "1";}
    operator int() { return 2; }
};

int main( ) {
    int x    = func(); // calls int version
    string y = func(); // calls string version
    double d = func(); // calls int version
    cout << func() << endl; // calls int version
    func(); // calls neither
}

Answer 4

If functions were overloaded by the return type and you had these two overloads

int func();
string func();

there is no way the compiler could figure out which of those two functions to call upon seeing a call like this

void main() 
{
    func();
}

For this reason, language designers often disallow return-value overloading.

Some languages (such as MSIL), however, do allow overloading by return type. They too face the above difficulty of course, but they have workarounds, for which you'll have to consult their documentation.

Answer 5

In haskell it's possible even though it doesn't have function overloading. Haskell uses type classes. In a program you could see:

class Example a where
    example :: Integer -> a

instance Example Integer where  -- example is now implemented for Integer
    example :: Integer -> Integer
    example i = i * 10

Function overloading itself is not so popular. Mostly languages I've seen with it are C++, perhaps java and/or C#. In all dynamic languages it's a shorthand for:

define example:i
  ↑i type route:
    Integer = [↑i & 0xff]
    String = [↑i upper]


def example(i):
    if isinstance(i, int):
        return i & 0xff
    elif isinstance(i, str):
        return i.upper()

Therefore there's no much point in it. Most people aren't interested whether language can help you drop a single line per where ever you use it.

Pattern matching is somewhat similar to function overloading, and I guess sometimes work similarly. It's not common though because it is useful only for few programs and is tricky to implement on most of languages.

You see there's infinitely many other better easier-to-implement features to implement into the language, including:

• Dynamic typing
• Internal support for lists, dictionaries and unicode strings
• Optimizations (JIT, type inferencing, compiling)
• Integrated deployment tools
• Library support
• Community support and gathering places
• Rich standard libraries
• Good syntax
• Read eval print loop
• Support for reflective programming

Answer 6

As already shown - ambiguous calls of a function that differs only by return type introduces ambiguity. Ambiguity induces defective code. Defective code must be avoided.

The complexity driven by the attempt to ambiguity shows that this is not a good hack. Apart from an intellectual exercise - why not use procedures with reference parameters.

procedure(reference string){};
procedure(reference int){};
string blah;
procedure(blah)