Static vs dynamic type checking in C

I'm curious in the differences between static and dynamic type checking.

Jul 1, 2022 in C++ by Nicholas
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Cases of static and dynamic binding in C++

When an object's static type is used to associate it with a member function, this is known as static binding (understand the type of its class). When a pointer or reference is associated with a member function based on the dynamic type of the object, this is known as dynamic binding (understand the instance of the variable at runtime). Before continuing, keep in mind that dynamic binding only works with pointers, references, and virtual functions for the base class. Because everything needed to call the function is known at compile time, the first call is a static binding (also known as early binding). Derived1 d1(1, 10); d1.display_data(); You already know that the d1 instance is a Derived1 automatic variable, and that it will call the Derived1::display data method (). The first condition is incorrect: d1 is neither a pointer nor a reference. The second condition isn't acceptable: There is no virtual Derived1::display data. The second call is for ...READ MORE

answered Jun 7, 2022 in C++ by Damon
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C++ - Overloading vs Overriding in Inheritance

In C++, a derived class's method only overrides the base class's method if their declarations match (I say "match," but I'm not sure what the formal term is). That is, all arguments must be of the same type, with the same const qualification. If there are any mismatches, the derived class's method hides all methods with the same name rather than overriding. This is what the "ERROR" in your image is attempting to convey. So, in that image, / overrides in a comment is incorrect and misleading. Yes, many C++ instructors are unaware of these somewhat esoteric details. Furthermore, if you want to override, your base class's method must be virtual; otherwise, polymorphism will not work. . We could also say that the derived-class method hides the base-class method if it wasn't virtual. The part about hiding, on the other hand, has almost no meaning here; what this term really means is that you're not in charge. Furthermore, overloading is the presence of multiple methods with the same name but different signatures, as you may have noticed. To be useful, they must all be present in the derived class; otherwise, they will be hidden if the derived class only has one method, fa1, and the other fa1 are in the base. There is, however, a syntax sugar that "copies" all fa1 from base to derived, removing all the hidden semantics: class A { public: void fa1(); ...READ MORE

answered Jun 7, 2022 in C++ by Damon
• 4,960 points • 827 views

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C++ - Overloading vs Overriding in Inheritance

In C++, a derived class's method only overrides the base class's method if their declarations match (I say "match," but I'm not sure what the formal term is). That is, all arguments must be of the same type, with the same const qualification. If there are any mismatches, the derived class's method hides all methods with the same name rather than overriding. This is what the "ERROR" in your image is attempting to convey. So, in that image, / overrides in a comment is incorrect and misleading. Yes, many C++ instructors are unaware of these somewhat esoteric details. Furthermore, if you want to override, your base class's method must be virtual; otherwise, polymorphism will not work. We could also say that the derived-class method hides the base-class method if it wasn't virtual. The part about hiding, on the other hand, has almost no meaning here; what this term really means is that you're not in charge. Furthermore, overloading is the presence of multiple methods with the same name but different signatures, as you may have noticed. To be useful, they must all be present in the derived class; otherwise, they will be hidden if the derived class only has one method, fa1, and the other fa1 are in the base. There is, however, a syntax sugar that "copies" all fa1 from the base to the derived. class A { public: void fa1(); ...READ MORE

answered Jun 7, 2022 in C++ by Damon
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The static keyword and its various uses in C++

Static variables exist during the "lifetime" of the translation unit in which they are declared, and: It cannot be accessible from any other translation unit if it is in a namespace scope (i.e. outside of functions and classes). This is referred to as "internal linking" or "static storage lifetime." (Except for constexpr, do not do this in headers; otherwise, you would wind up with a different variable in each translation unit, which is really confusing.) If it is a variable in a function, it, like any other local variable, cannot be accessed from outside the function. (This is the mentioned local) Class members have no limited scope owing to static, but they may be referenced from both the class and an instance (like std::string::npos). locations as code: static std::string namespaceScope = "Hello"; void ...READ MORE

answered Jun 27, 2022 in C++ by Damon
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*this vs this in C++

This is a pointer, and *this is a pointer that has been dereferenced. If you had a function that returned this, it would be a pointer to the current object, but a function that returned *this would be a "clone" of the current object, created on the stack unless you defined the method's return type to be a reference. A small application that demonstrates the difference between working with copies and working with references: #include <iostream> class Foo { public: ...READ MORE

answered Jun 28, 2022 in C++ by Damon
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+1 vote

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Using push_back vs at in a vector in C++

I'm not sure how to use vectors in C++. It has to do with the vector's push back technique. I used push back to insert entries into the vector in the first programme. I used at() to insert entries into the vector in the second application. #include <iostream> #include <vector> #include <string> using namespace std; int main ...READ MORE