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Execution of Switch/Case

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Programming Fundamentals

Университет Дьюка

4.7 (оценок: 1,049) | Зарегистрировано учащихся: 34K

Курс 1 из 4 — Introduction to Programming in C Специализация

Programming is an increasingly important skill, whether you aspire to a career in software development, or in other fields. This course is the first in the specialization Introduction to Programming in C, but its lessons extend to any language you might want to learn. This is because programming is fundamentally about figuring out how to solve a class of problems and writing the algorithm, a clear set of steps to solve any problem in its class. This course will introduce you to a powerful problem-solving process—the Seven Steps—which you can use to solve any programming problem. In this course, you will learn how to develop an algorithm, then progress to reading code and understanding how programming concepts relate to algorithms.

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Получаемые навыки

Algorithms, Programming Language Concepts, Problem Solving, C Programming

Рецензии

4.7 (оценок: 1,049)

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MM

Dec 27, 2017

Well worth it! Hated it in the beginning, couldn't wrap my head around the algorithms, but the course studies helped, kept reading, watching the videos, FINALLY got it!\n\n100% can recommend!

VV

Nov 11, 2018

Best Course I've taken on programming. Special thanks to the mentors, very clear and patient to explain each thing elaborately! Course is in great detail and made easy for Beginners. Thanks!

Из урока

Reading Code

In this module, you will learn to read code—this means you will be able to execute a piece of code by hand, and clearly illustrate what each statement does and what the state of the program is. Understanding how to read code is the only way to be sure you can write correct code. By the end of this module, you will be able to read and understand code with functions, conditional statements, iteration, and other fundamental techniques.

Execution of If/Else3:17

Execution of Switch/Case3:05

Преподаватели

Andrew D. Hilton
Associate Professor of the Practice
Genevieve M. Lipp
Assistant Professor of the Practice
Anne Bracy
Senior Lecturer

Текст видео

In this video, we're going to execute code with a function g that contains a switch

case statement.

As always, we begin with the execution arrow at the start of main for

the statement a = g(10, 4);.

We make a box per a since we're declaring it, then we create a frame for

the function g, and populate it with the values 10 for n and 4 for x.

We make a note of where we are and we'll return to when the function is complete,

and enter function g.

We have the switch statement x + n in this case is 4 + 10 which is 14.

So we find the matching case label, case 14 and

transition our execution arrow inside that case and begin executing statements there.

The first statement we encounter is the return statement n- x, 10- 4 = 6.

As always when we encounter a return statement,

we're going to make a note of the value to return and leave the function that we're

in going back to the calling function and destroying the frame.

We finished the assignment statement with a = 6.

Now we reach int b = g(a, 2).

So we create a box for b, draw a frame for g, passing in the arguments a,

which is 6 and 2.

We note our position and enter function g.

Here, our selection expression is 2 + 6, which is 8, so

we're going to go into the case for 8 and begin executing statements there.

The first statement we encounter is x = x + 1, so

we update the box for x to have 3 instead of 2.

Now we fall through to the next case.

There is no break statement here, and

we don't worry about the fact that there is another case label.

We just keep executing statements until we reach a break.

The next statement is n = n- 1.

So we're going to update the value of n to be 5.

And now we do reach a break statement.

This break statement is going to take us out of the innermost

enclosing switch statement, the boundaries which are here.

As we'll see later, break could take us out of other constructs.

But in this case, it'll take us out of the switch statement that encloses it and

we begin executing code after the switch.

The next statement we encounter says to return x * n, which is 3 * 5, which is 15.

So we return 15, return to main, and

destroy the frame for g and assign 15 to b.

Now we have int c = g(9, b).

So we create a box for c, a frame for g with arguments 9 and

b which is 15 and note where we are and enter function g.

Evaluating this expression gives us 24.

Looking at our case labels, we have 8, 0, and 14.

None of those match 24, so we use the default which matches anything that's not

named explicitly by another case label.

We'll jump into the default case and begin executing statements there.

We do x = n, assigning 9 to x which brings us to a break

which takes us out of the inside of the switch statement.

We now return x * n which is 81.

Back to main, we finish our assignment and return zero from main exiting the program.

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