overloaded function with no contextual type information

Compiler does not know, which overload of cb to chose.
2) Even if it did know, it would not convert from void* to, say void(*)(int), without any explicit conversion.

However, it can deduce it if you give compiler enough information:

void cb(int x)
{
    std::cout <<"print inside integer callback : " << x << "\n" ;
}

void cb(float x)
{
    std::cout <<"print inside float callback :" << x <<  "\n" ;
}

void cb(const std::string& x)
{
    std::cout <<"print inside string callback : " << x << "\n" ;
}

int main()
{

    void(*CallbackInt)(int);
    void(*CallbackFloat)(float);
    void(*CallbackString)(const std::string&);

    CallbackInt = cb;
    CallbackInt(5);

    CallbackFloat = cb;
    CallbackFloat(6.3);

    CallbackString = cb;
    CallbackString("John");

    return 0;

}