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Execution of Switch/Case

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Programming Fundamentals

Université Duke

4.7 (1,358 notes) | 40K étudiants inscrits

Cours 1 de 4 dans Introduction to Programming in C Spécialisation

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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Compétences que vous apprendrez

Algorithms, Programming Language Concepts, Problem Solving, C Programming

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4.7 (1,358 notes)

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SS

Apr 21, 2019

I love the focus of this first course in the specialization is about problem solving and developing a mindset for programming. The course is well thought out and concepts are clearly explained!

AM

Apr 08, 2019

Really great course. Took this as a refresher. I feel that the content of this course was significantly higher quality than the programming tuition I received in first semester of university.

À partir de la leçon

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

Enseigné par

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

Transcription

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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