Accept all types as argument in function

Question

How can I in C++ make a function accept every Object, so I can give it numbers, String or other Objects. I am not very well in C++, I hope it\'s not a totally stupid questio

Answer 1

You can overload your function for different types, i.e.

size_t func(int);
size_t func(std::string);

Alternatively and/or additionally, you can provide a function template, which is a way to tell the compiler how to generate your function for any particular type, for example

template<typename T>
size_t func(T const&) { return sizeof(T); }

You may use more advanced techniques such as SFINAE to effectively overload those template functions, i.e. to use different templates for different kind of types T (i.e. integral types, pointer, built-in types, pod, etc). The compiler will then pick the best-fitting func() (if any) for any function call it encounters and, if this is a template, generate an appropriate function. This requires no re-coding.

A completely different approach is to use a generic erasure type, such as boost::any, when the function will need to resolve the expected types at coding-time (as opposed to compile-time):

size_t func(boost::any const&x)
{
  auto i = boost::any_cast<const int*>(x);
  if(i) return func(*i);
  // etc for other types, but this must be done at coding time!
}

Answer 2

You can use templates for this purpose:

template <typename T>
void foo(T const & value)
{
    // value is of some type T, which can be any type at all.
}

What you can actually do with the value may be rather limited without knowing its type -- it depends on the goal of your function. (If someone attempts to call the function with an argument type that causes that function specialization to be ill-formed then your template function will fail to instantiate and it will be a compile-time error.)

Answer 3

If I understood you correctly you might wanna try using templates http://en.cppreference.com/w/cpp/language/function_template

Answer 4

You are probably looking for templates.
I suggest you read this.

Answer 5

I'm not sure what you're trying to accomplish, but you can pass a void pointer as a parameter.

void foo(void* bar);