简体   繁体   中英

How can I select a specific function overload?

I want to pass an overloaded operator to a function, which can't figure out which of the two overloads it should use.

// This works, not overloaded:
chai.add(chaiscript::fun(&HttpRequest::operator+), "+");
// This does not, overloaded:
chai.add(chaiscript::fun(&(
    std::map<std::string,std::string>::operator[]), "[]");

The chaiscript::fun expects a generic parameter, but it can't figure out which overload to use.
I need to specify the overload, but I don't know the syntax. I tried things like:

chai.add(chaiscript::fun(&(
    std::map<std::string,std::string>::operator[]<foo>), "[]");

but this doesn't work.

How is the syntax to specify the overload?

The following may help you to choose one overload of map::operator[]

static_cast<
    std::string& (std::map<std::string, std::string>::*)(const std::string&)>(
        &std::map<std::string, std::string>::operator []);

Or with a typedef :

using MyMap = std::map<std::string, std::string>;

static_cast<std::string& (MyMap::*)(const std::string&)>(&MyMap::operator []);
//          ReturnType    Class     params

Firstly, operators are just functions with a special syntax. So, overloaded operators are just overloaded functions. With that in mind, you question boils down to "How do I select one of an overloaded set of functions?". The simple answer to that is the use of a static_cast.

Alternatively, I think you can also use it in a context where the context already dictates which of the overloads to select. In your case, you are using a template function (I think) and it doesn't work there because it needs the type to instantiate the template and it needs the template instatiation to select the type and thus the overload. Using a temporary variable is one way to resolve this, another is to explicitly specify the template function (like eg max<float>(0, x) where otherwise 0 is considered of integer type).

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM