Very useful. All the things you have explained in this video are very important to keep in mind. Thank you so much.
@PortfolioCourses2 жыл бұрын
You're welcome Eduardo! :-)
@Klusio192 жыл бұрын
I have a question. In 3:46 You said, that we have to follow the same order when initializing member list as the member variables were declared in the class. As with most things, I went and check that out. But I couldn't find any diffrence when initialazing in order or not. In both ways it worked. Take a look at my example class: class TestClass{ public: int number; string name; TestClass(string name_set, int number_set) : name(name_set), number(number_set) { cout
@Klusio192 жыл бұрын
Ok, I did a little bit more KZbin research and I think I found out what is going on. First of all my example is a little bit unoftunate to show how the initialization works. If I understand correctly, by using member initializer list, the variables will be initialized in order like they have been declared in the class, and compiler will not care in what order I entered them. So in the class below: class Student{ public: int age; int IQ; Student(int a) : age(a), IQ(5 * age) {} }; everything would be ok if I wanted to print IQ in the main function. But when I would flip the order and declare "IQ" first, and then "age" like that: class Student{ public: int IQ; int age; Student(int a) : age(a), IQ(5 * age) {} }; and then I would like to print out the IQ in the main function , it would be 0, because first, the IQ is initialized, which will be 5 times 0, because age is not initalized at all, and after that age will be initialized. Am I correct now?
@PortfolioCourses2 жыл бұрын
The first bullet point here explains it, basically if we do things out of order like that it's not garaunteed to work as we could end up using uninitialized values: medium.com/pranayaggarwal25/using-modern-class-members-and-initializations-c11e931c3ba. :-)
@Klusio192 жыл бұрын
@@PortfolioCourses thank you for fast answer! So it turns out, that I noticed that correctly. Thank you again! 😊
@PortfolioCourses2 жыл бұрын
@@Klusio19 Yes, sounds like you've got it now. 🙂
@itsmaxim0111 ай бұрын
21:10 just pointing out that long int (which is the same as long) is the exact same 4 bytes on x64 systems, which are almost every system nowadays. On such systems long and int are identical, therefore there will be no type-narrowing.
@SIVAJI_33310 ай бұрын
7:24 So,the Major class has two constructors,the constructor is invoked based arguments we passed,then its compile time polymorphism i.e function overloading,Am I right?
@jaishreejain80302 жыл бұрын
Amazing😀 .Very helpful Thank you Sir
@PortfolioCourses2 жыл бұрын
You're welcome Jai! 😀 I'm glad you enjoyed the video.
@yuwownly86302 жыл бұрын
I'm just wondering. What is the difference between major(Major(major)) and major(major) as the member initializer list?
@PortfolioCourses2 жыл бұрын
Great question Yuw! 😀 It looks to me like in the first example, an object Major is being instantiated using major as an argument, a bit similar to this code here in terms of the general idea: stackoverflow.com/a/12927220
@sallaklamhayyen98762 жыл бұрын
thank you
@PortfolioCourses2 жыл бұрын
You’re welcome! :-)
@shafayet0198 Жыл бұрын
What will happen if I don't add Major(the class name) in major{Major{major}}
@PortfolioCourses Жыл бұрын
Wow, great question! :-) If we don't put it in, then the constructor for Major will be called and it will be provided with "major" as an argument, which is the same thing that will happen as the code is now. But let's say I have a class SpecialMajor that is a derived class of Major, and let's say it has a constructor that also accepts a string "major" as an argument. Now I *might* want to have the major member variable (with type Major) reference a SpecialMajor object, because it also *is* a Major, it's a dervied type / subtype of Major. But if I want to do that, then I *need* to include the classname like this: major(SpecialMajor(major)) in order for a SpecialMajor object to be instantiated here. Hopefully that helps! :-)