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How to prevent constructor from casting argument types in c++?


How to prevent constructor from casting argument types in c++?

By : analogchops
Date : November 22 2020, 09:00 AM
like below fixes the issue I have a class to store big unsigned numbers and I'd like to allow user to create object using types such as long long, int, unsigned int and so on and also from string . I created a constructor BigNumber(const unsigned long long) and BigNumber(const std::string) but I want to disallow user to use constructions like: BigNumber('a') or BigNumber(true). I heard about explicit so I decided to write in my class definition following lines: , You can achieve this using a template:
code :
template<typename T,
         typename = typename std::enable_if<
            std::is_integral<T>::value
            && !std::is_same<char, T>::value
            && !std::is_same<bool, T>::value>::type>
explicit BigNumber(T const n);


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Does std::function's copy-constructor require the template type's argument types to be complete types?

Does std::function's copy-constructor require the template type's argument types to be complete types?


By : Deepak Vanmelil
Date : March 29 2020, 07:55 AM
will help you Edit: Apperently, this issue is now fixed, so the below text can be seen as history. :)
code :
#include <functional>

void f(int* p){}

int main(){
  std::function<void(int)> fun(f);
}
template <class F> function(F f);
template <class F, class A> function(allocator_arg_t, const A& a, F f);
template<bool Copy, class F>
struct lazy_callable{
  static bool const value = callable<F>::value;
};

template<class F>
struct lazy_callable<true, F>{
  static bool const value = false;
};

template<class F>
function(F f, typename enable_if<lazy_callable<!std::is_same<F,function>::value>::type* = 0);
Issue with type casting the argument of the constructor

Issue with type casting the argument of the constructor


By : Carolatta Watson
Date : March 29 2020, 07:55 AM
Does that help Foo f(int(x));
It is not a type cast, it's a function declaration - function f that takes an int called x and returns a Foo.
code :
Foo f(int x);
Foo f( int (((x))) );
Foo f((int (x)));
//    ^       ^
Up-casting template argument types

Up-casting template argument types


By : user4178091
Date : March 29 2020, 07:55 AM
like below fixes the issue No need to cast. The problem here is that Arrays.asList(new Derived()) naturally tries to create a List, and then calling .iterator() on a List naturally gives an Iterator, which is not a sub-type of Iterator, so you get a compilation error.
You can specify that you want a List, using Arrays.asList. This works, because you can certainly put a Derived instance into a List, and then calling .iterator() on a List naturally gives an Iterator.
code :
    class Example implements Iterable<Base> {
        @Override
        public Iterator<Base> iterator() {
            return Arrays.<Base>asList(new Derived()).iterator();
        }
    }
C++ no instance of constructor matches the argument list argument types are: (std::string, double)

C++ no instance of constructor matches the argument list argument types are: (std::string, double)


By : Aamir
Date : March 29 2020, 07:55 AM
around this issue The error message given to you by the compiler is precise and correct. There are two possible fixes here:
define the constructor use brace initialization
code :
class variable {
public:
    variable(string n = {}, double v = 0.0) : name{n}, value{v} {}
    string name;
    double value;
};
    var_table.push_back( {var, val} );
no instance of constructor matches the argument list -- argument types are:

no instance of constructor matches the argument list -- argument types are:


By : Sadan Singh Rana
Date : March 29 2020, 07:55 AM
Hope that helps The problem is that MyClass and MyClass are different types, and you only have a copy constructor that takes objects of the same type.
To get your code to work, I had to change two things:
code :
template<typename u>
MyClass(const MyClass<u>& obj){
    x=(t) obj.x;
    z=(t)obj.z;
}
prog.cpp:14:19: error: ‘int MyClass<int>::x’ is private within this context
         x=(t) obj.x;
template<typename> friend class MyClass;
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