hmwrk 5

# hmwrk 5 - University of Colorado Introduction to...

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University of Colorado Introduction to Engineering Computing – GEEN 1300 Spring 2008 Homework 5 Solutions Problem 2 on page 367 of Volume II of the textbook. Begin by re-reading section 3 of Chapter 12, especially pp 326-329, and reviewing slides 20-23 of Lecture 10. One key point in this solution is that Sub Switch thinks that its first argument is named e, its second argument is named b, and its third argument is named d. When you call Switch with some other variables – say Switch(x, y, z) – it “hooks” those variables that you called it with (x,y,z) to its own internal names (e,b,d). This acts as a pipeline out to those other variables: if you refer to e inside Switch, it'll reach out and access the variable x. This got confusing in this problem because the textbook authors had variables in different places with the same names. They did this in order to make you think very carefully about which variable gets accessed by each reference, but it created more confusion than thinking! Another key point in this problem is the control flow: Sub Calc calls Sub Switch, which means that the execution looks like this: That means that the four points in the problem where you're asked for the values of a, b, etc. come up in different order in the computer than they do on the textbook pages.

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The first line of Calc (a = -1: and so on) happens before the branch to Switch. It sets the values of three of Switch’s internal variables (b, c, and d), leaves the other one (e) alone, and also sets the value of the “global” variable a. After this line of code is executed, the values of a, b, c, d, and e are -1, 2, 8, 4, 0. This is the
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hmwrk 5 - University of Colorado Introduction to...

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