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C++: How to overload pow for user type?


C++: How to overload pow for user type?

By : Dias
Date : November 26 2020, 01:01 AM
it should still fix some issue You are not allowed to overloaded function into the std namespace. A typical solution for this is to declare your function in the same namespace as the custom type like
code :
namespace my_code
{
    struct foo {};

    foo pow(foo base, int power)
    {
        // code here
    }
}
void bar()
{
    using std::pow;
    //...
    auto a = pow(10,3);
    //...
    auto b = pow(foo{}, 3);

}


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Child Controls are null when loading User Control programmatically via LoadControl(Type, object[]) overload

Child Controls are null when loading User Control programmatically via LoadControl(Type, object[]) overload


By : Daniel Stephenson
Date : March 29 2020, 07:55 AM
Hope that helps This is by design. An .ascx file actually inherits from the code-behind class, so the .ascx is a derived type of the code-behind class.
This means that when you load the parent code-behind class using the LoadControl(Type, object[]) method, it's instantiating the parent class defined in the code-behind, and not the derived .ascx class which contains the child controls.
How to correctly overload operator< for use in std::map with my user-defined type?

How to correctly overload operator< for use in std::map with my user-defined type?


By : audiologic
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , You should change your inner loop comparison akin to your length conparison:
code :
    for (int i = 1; i < s1.arr[0]+1; i++)
    {
        if (s1.arr[i] < s2.arr[i])
            return true;
        else if (s1.arr[i] > s2.arr[i])
            return false;
    }
Why is it is impossible to overload return type in C++, but possible to overload parameters?

Why is it is impossible to overload return type in C++, but possible to overload parameters?


By : Andry Smirnov
Date : March 29 2020, 07:55 AM
Does that help It is a technical limitation, if you want to call it that:
C++ types are inferred bottom-up: the type of an expression only depends on its sub-expressions, and not of the context expression(s) it appears in. So the type of the arguments of an overloaded method can be determined in order to pick which version to call, but it would be impossible to tell which method to call if there was overloading on the return type.
Is it well-formed to overload an operator for a standard-defined type in global namespace that doesn't depend on a user-

Is it well-formed to overload an operator for a standard-defined type in global namespace that doesn't depend on a user-


By : jleusjr
Date : March 29 2020, 07:55 AM
it fixes the issue I don't see anything in the section named Constraints on programs that forbids defining a function like yours.
I say, it is perfectly valid to do so.
User-defined constructor overload not being parameter-matched to the overload with the argument's superclass

User-defined constructor overload not being parameter-matched to the overload with the argument's superclass


By : user3077670
Date : December 25 2020, 06:01 PM
I wish this helpful for you Following the code-snippet below, I attempt to pass a Boo instance through the Boo::Boo(Foo const &) constructor overload, which I cannot manage to do. , Use a delegating constructor:
code :
template <typename T>
struct Boo : public Foo
{
    Boo(Boo<T> const & arg) : Boo(static_cast<Foo const&>(arg)) {};

    Boo()            { std::cout << "Beep " << sizeof(T) << std::endl; }
    Boo(Foo const &) { std::cout << "Boop " << sizeof(T) << std::endl; }
};
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