3.1 C++ Classes & Object Orientation

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来自 University of California, Santa Cruz 的课程

C++ For C Programmers, Part A

442 个评分

University of California, Santa Cruz

C++ For C Programmers, Part A

442 个评分

本课程是面向有C语言经验的，并想要学习C++编程的程序员们。涉及到的例子和练习需要有对基础算法和面向对象软件的理解。

从本节课中

Module 3

Point: Default constructor and initializing syntax. Conversion Constructors. Copy Constructor. List and dynamic memory allocation. Deep Copy.

3.1 C++ Classes & Object Orientation7:19

3.2 Point and its Constructor12:42

3.3 More Constructors18:22

3.4 Using Dijkstra's Algorithm14:22

3.5 The is-connected Algorithm12:05

3.6 Complex Aggregates - Lists6:53

3.8 Print() Chaining6:20

3.9 More Elaborate List3:43

3.10 Deep vs. Shallow Copy8:48

3.11 Destructor4:49

3.12 Dynamic Data Structures in STL7:50

与讲师见面

Ira Pohl
Professor
Computer Science

[MUSIC]

Today, we're going to get deeper into the object oriented programming style.

So we're going to look at more C++ classes and

how that builds on in an object oriented manner.

We're going to look again, at constructors and

destructors which are these critical methods that both build and

deallocate the new extended data types.

And then we're going to discuss our next significant homework assignment and

that's a homework assignment based on graph theory.

So again, all of you came in having a background in

computer science roughly at a sophomore level at least and

you may well have already seen standard graph theory algorithm.

So this may be familiar to you.

In any case, I will review all that material.

So even if you have forgotten it or

haven't in the past had an adequate background,

we're going to really understand Dijkstra's shortest path algorithm and

that's going to be the basis for our second homework assignment.

Objects.

Objects could be anything.

So here in this little example, we're thinking of shapes.

Shapes could be boxes, polygons, text shapes.

There can be operations like delete on

them operations that involve counting the shapes.

And so if you wanted to build a windowing environment,

what would be our set of widgets?

What would be the operations?

What would the user expect, the user of this package of routines?

How would we properly design things and keep the internal details hidden away.

All of that is what you have to think about when you're

an object oriented C++ programmer.

Again, a best guideline to doing this thought

process is to know what happens in the language for native types.

So with native types, we have to fully understand what happens

when we declare something, even a simple object like an int.

What does it mean to say int i, j = 3?

Well, we typically do that inside block scope.

Block maybe inside the main function or

it maybe in the auxiliary file scope functions.

So we're declaring here, we're allocating.

We're initializing.

Now for a simple native type, all this happens automatically on the stack.

We have to replicate that for our user defined types for our objects.

And again, as we started with last time,

we saw we had a special method for doing this called a constructor.

So just for review purposes in the C language or in basic C++,

what is supposed to happen here where we have nested blocks?

So here's a a main.

We have two declarations, i is initialized the 9.

J has initialized the 3.

We're going to print what goes to screen.

So that's what we're expecting you to do.

This is going to be very similar to the kinds of exams that I give.

This is a question that I could readily put on are in our final exam.

And then secondarily, we have an internal block is not,

here's the internal block then the nested internal block.

Now, I actually would have preferred an indented style.

This is not, you should reformat that if you're going to try the code.

So that the internal block is indented.

So that you can visually.

That's again, a convention.

But it makes it more readable, you can visually see the inner block.

And so in the inner block, there's a redeclaration int i and

we're asking what that prints versus at the outer block.

And the answer is the inner block,

the outer block has 9 and 3.

But the inner block has what you might call i inner,

which was recalculated to be 5 and j was left.

J just got passed in, because there was no inner block j.

So that's really j outer.

And then over here, we have outer again.

And when we go to the outer block and

we print it, we see that value 9 was retained.

So that should all be in your knowledge base already.

From an allocation point of view, i and

j are placed on the stack.

Inner block.

I inner is on the stack.

I outer is hidden.

Then we have block exit.

I inner Is destroyed.

And then we return to the outer block and we have again,

i9 and j3 which are the outer block variables that are still on the stack.

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