Spring 2001 CS100M Solutions to Exercise E2

Remarks Recall the steps to perform an assignment lhs=rhs in Matlab:

+ Evaluate $lhs$ to determine a location.
+ Evaluate $rhs$ to determine a value
  (the value of a variable $v$ is its current value.)
+ If $lhs$ does not exist, then declare it as a new variable.
  (Declare means create a new named box to hold values.)
+ Place the value into the location,
  replacing whatever value used to be in the location.

Observe that according to these rules, if $x$ and $y$ are different variables, then after $x=y$ is performed, modifying $y$ has NO effect on the value of $x$.

NOTE: the 2nd and 3rd steps above were originally in the wrong order here and in 1/25 lecture notes.

~~Note: This sentence shows how crossing out looks for numbers: it should appear crossed out, or at least italicized.~~ Crossed out vectors and structs are shown with a gray background.

w 1 8 64 512

x 3 1 4 6

y 2 7 2 7 2 7 2 7

z 1 3 1 8 24 8 64 192 64 512 1536 512 4096

genny

age 1314

friends 7

age 11

toys 8

entire struct crossed out

rosie

age 11

toys 8

junie

age 13

friends 7

sister

age 11

toys 8

y

13

8 9

2 3

9 10

entire vector crossed out

i
3 5 4 10 4 11

j
1 2 3 4 5

Here are all the updates, listed in order (not required of you; provided in case you find it helpful):

    x = 5 1 4 8 2 6 3 9 7
    i = 3
    j = 1
    i = 5
    x(5) = 3
    j = 2
    i = 4
    x(4) = 9
    j = 3
    i = 10
    x = 5 1 4 9 3 6 3 9 7 3 <-- entire vector gets replaced
    j = 4
    i = 4
    x(4) = 10
    j = 5
    i = 11
    x = 5 1 4 10 3 6 3 9 7 3 5 <-- entire vector gets replaced


[2/02] TIPS

+ draw a variable as the name next to the box for its value
+ draw values inside or nearby the box for a variable
  + all fields of a struct should be inside the box,
    but you are allowed to extend the box, e.g. $x.a=5$
    might look like this:
    +------+                                +---+ 
  x | a[5] |                    but not x.a | 5 | 
    +------+                                +---+ 

    and then after also executing $x.b=7$, you could have

    +------+          +------+              +---+      +---+
  x | a[5] |  or    x | a[5] |  but not x.a | 5 |  x.b | 7 |
    +------+          |      |              +---+      +---+
    | b[7] |          | b[7] |
    +------+          +------+
    
+ cross out, not scribble out, old values
  + for structs and vectors, if the whole thing is replaced,
    then the whole value gets crossed out.
    e.g. continuing the example above,
    if we next execute $x=0$, then you could have:

  +-------+               +---------+   
x | a[5/] |             x | a[5/]   |   
  |   /   |               |   /   0 |   
  | b/[7] | 0      or     | b/[7]   |
  | /     |               | /       |   
  +-------+               +---------+