I am a student taking a class in Turbo Pascal (the 5th Ed. book).
I'm terrible at programming and missed the last class. I haven't
been able to find anyone that could help me understand the last work
we were given . I was hoping someone might be able
to explain this to me. I won't receive a grade in this until I complete
all my work and I don't seem to be able to get past this.

self-check pg. 439 #1: (on arrays) Turbo Pascal
The following sequence of statements changes the contents of array X.
Describe what each statement does to the array, and show the final
contents of array X after all statements execute.

X = array
I = index range

I :=3;
X[I] := X[I] + 10.0;
X[I - 1] := X[2 * 1 - 1];
X[I + 1] := X[2 * 1] + X[2 * I + 1);
for I :=3 downto 1 do
X[I + 1] := X[I]

I would appreciate any help I can get with this.
Serina

: I am a student taking a class in Turbo Pascal (the 5th Ed. book).
: I'm terrible at programming and missed the last class. I haven't
: been able to find anyone that could help me understand the last work
: we were given . I was hoping someone might be able
: to explain this to me. I won't receive a grade in this until I complete
: all my work and I don't seem to be able to get past this.
:
: self-check pg. 439 #1: (on arrays) Turbo Pascal
: The following sequence of statements changes the contents of array X.
: Describe what each statement does to the array, and show the final
: contents of array X after all statements execute.
:
: X = array
: I = index range
:
: I :=3;
: X[I] := X[I] + 10.0;
: X[I - 1] := X[2 * 1 - 1];
: X[I + 1] := X[2 * 1] + X[2 * I + 1);
: for I :=3 downto 1 do
: X[I + 1] := X[I]
:
: I would appreciate any help I can get with this.
: Serina

Hey Serina,

First off, you need to understand what an array is. Think of it like a row of boxes on graph paper. In this case, we'll use 10 boxes in a row and number them from 1..10 You can change put tick marks in any one box, but it will only effect the one.

For the example above:
I :=3;

so just change all the 'I' to '3'. The second line then becomes:
X[3] := X[3] + 10.0;

So box #3 gets 10.0 ticks added to it.

X[3 - 1] := X[2 * 1 - 1];

So box #2 (3-1) will get the same number of ticks that are in box #1 (2*1-1). So if Box[1] (Box #1) has 6 tick marks, then you would make it so box[2] has 6 tick marks as well.

Hope this helps a bit. Monday mornings are not a good time to catch me...

Phat Nat

: : I am a student taking a class in Turbo Pascal (the 5th Ed. book).
: : I'm terrible at programming and missed the last class. I haven't
: : been able to find anyone that could help me understand the last work
: : we were given . I was hoping someone might be able
: : to explain this to me. I won't receive a grade in this until I complete
: : all my work and I don't seem to be able to get past this.
: :
: : self-check pg. 439 #1: (on arrays) Turbo Pascal
: : The following sequence of statements changes the contents of array X.
: : Describe what each statement does to the array, and show the final
: : contents of array X after all statements execute.
: :
: : X = array
: : I = index range
: :
: : I :=3;
: : X[I] := X[I] + 10.0;
: : X[I - 1] := X[2 * 1 - 1];
: : X[I + 1] := X[2 * 1] + X[2 * I + 1);
: : for I :=3 downto 1 do
: : X[I + 1] := X[I]
: :
: : I would appreciate any help I can get with this.
: : Serina
:
:
: Thanx Nat,
I appreciate your time. I think I understand it better.
Is this right?
1. I :=3; {index = 3}

2. X[3] := X[3] +10; {stores value of 10 in index 3}

3. X[3 - 1] := X[2 * 3 - 1]; {stores same value in index [2] as in index[5]}

4. X[3 + 1] := X[2 * 3] + X[2 * 3 + 1]; {index 4 = sum of index 6 and index 7}

5. for I := 5 to 7 do {initilizes I to 5}

6. X[I] := X[I + 1]; {adds 1 to I}
{index 5 := 5 + 1}
{index 6 := 6 + 1}
{index 7 := 7 +1}

7. for I := 3 down to 1 do {initializes I to 3}

8. X[3 + 1] := X[3] {adds 1 to I}
{X[2 + 1] := X[2]}
{X[1 + 1] := X[1]}

Does this look ok or am I still way off?
Thanx,
Serina

: : : I am a student taking a class in Turbo Pascal (the 5th Ed. book).
: : : I'm terrible at programming and missed the last class. I haven't
: : : been able to find anyone that could help me understand the last work
: : : we were given . I was hoping someone might be able
: : : to explain this to me. I won't receive a grade in this until I complete
: : : all my work and I don't seem to be able to get past this.
: : :
: : : self-check pg. 439 #1: (on arrays) Turbo Pascal
: : : The following sequence of statements changes the contents of array X.
: : : Describe what each statement does to the array, and show the final
: : : contents of array X after all statements execute.
: : :
: : : X = array
: : : I = index range
: : :
: : : I :=3;
: : : X[I] := X[I] + 10.0;
: : : X[I - 1] := X[2 * 1 - 1];
: : : X[I + 1] := X[2 * 1] + X[2 * I + 1);
: : : for I :=3 downto 1 do
: : : X[I + 1] := X[I]
: : :
: : : I would appreciate any help I can get with this.
: : : Serina
: :
: :
: : Thanx Nat,
: I appreciate your time. I think I understand it better.
: Is this right?
: 1. I :=3; {index = 3}
:
: 2. X[3] := X[3] +10; {stores value of 10 in index 3}
:
: 3. X[3 - 1] := X[2 * 3 - 1]; {stores same value in index [2] as in index[5]}
:
: 4. X[3 + 1] := X[2 * 3] + X[2 * 3 + 1]; {index 4 = sum of index 6 and index 7}
:
: 5. for I := 5 to 7 do {initilizes I to 5}
:
: 6. X[I] := X[I + 1]; {adds 1 to I}
: {index 5 := 5 + 1}
: {index 6 := 6 + 1}
: {index 7 := 7 +1}
:
: 7. for I := 3 down to 1 do {initializes I to 3}
:
: 8. X[3 + 1] := X[3] {adds 1 to I}
: {X[2 + 1] := X[2]}
: {X[1 + 1] := X[1]}
:
: Does this look ok or am I still way off?
: Thanx,
: Serina
:
A number of your explanations is still a little off. These are 2 and 5 thru 8. 5 & 6 are actually 2 parts of the same statement:

for I := 5 to 7 do  
  X[I] := X[I + 1]; { <== statement ends here } 

Just remember that Pascal uses a semicolon as statement separator. The for-do loop does a little more than initializing I to 5. It also checks if I <= 7 and increments I with each step. The same goes for 7 & 8.

: A number of your explanations is still a little off. These are 2 and 5 thru 8. 5 & 6 are actually 2 parts of the same statement:
:

: for I := 5 to 7 do  
:   X[I] := X[I + 1]; { <== statement ends here } 
: 

: Just remember that Pascal uses a semicolon as statement separator. The for-do loop does a little more than initializing I to 5. It also checks if I <= 7 and increments I with each step. The same goes for 7 & 8.


:Thanks zibadian,
That helped, I'm working on it.
Serina

: Thanx Nat,
: I appreciate your time. I think I understand it better.
: Is this right?
: 1. I :=3; {index = 3}
:
: 2. X[3] := X[3] +10; {stores value of 10 in index 3}
:
: 3. X[3 - 1] := X[2 * 3 - 1]; {stores same value in index [2] as in index[5]}
:
: 4. X[3 + 1] := X[2 * 3] + X[2 * 3 + 1]; {index 4 = sum of index 6 and index 7}
:
: 5. for I := 5 to 7 do {initilizes I to 5}
:
: 6. X[I] := X[I + 1]; {adds 1 to I}
: {index 5 := 5 + 1}
: {index 6 := 6 + 1}
: {index 7 := 7 +1}
:
: 7. for I := 3 down to 1 do {initializes I to 3}
:
: 8. X[3 + 1] := X[3] {adds 1 to I}
: {X[2 + 1] := X[2]}
: {X[1 + 1] := X[1]}
:
: Does this look ok or am I still way off?
: Thanx,
: Serina

Pretty close. There are just a few that are a little off:

X = array
I = index range

1. I :=3;
1. I :=3; {index = 3}

2. X[I] := X[I] + 10.0;
2. X[3] := X[3] +10; {stores value of 10 in index 3}
This actually adds 10 to the existing value. If X[3] was set to 5, then this would make it 15

3. X[I-1] := X[2*I-1];
3. X[3-1] := X[2*3-1]; {stores same value in index [2] as in index[5]}

4. X[3+1] := X[2*3] + X[2*3+1]; {index 4 = sum of index 6 and index 7}

5. for I := 5 to 7 do {initilizes I to 5}
this is a loop; usually used with BEGIN and END. This will set I=5, then add one until I=7. Example:

For I := 5 to 7 Do
Begin
     WriteLn(I);
End;

If you ran this, You would see:

5
6
7

6. X[I] := X[I + 1]; {adds 1 to I}
{index 5 := 5 + 1}
{index 6 := 6 + 1}
{index 7 := 7 + 1}

{index 5 := index 6 (5 + 1)}
{index 6 := index 7 (6 + 1)}
{index 7 := index 8 (7 + 1)}


7. for I :=3 downto 1 do
X[I + 1] := X[I]

7. For I := 3 down to 1 do {initializes I to 3}
X[3 + 1] := X[3] {adds 1 to I}
{X[2 + 1] := X[2]}
{X[1 + 1] := X[1]}
For I := 3 down to 1 do {initializes I to 3}
X[3 + 1] := X[3] {index 4 (3+1) = index[3]}
{X[2 + 1] := X[2]}
{X[1 + 1] := X[1]}
When DOWNTO is used in place of TO, I counts down, instead of up

Hope this clears up any problems. If you need more help, just leave a message.

Phat Nat

This message was edited by sweet_serina at 2003-6-24 20:5:16

---

:
: Thanx Phat Nat,

This is making more sense to me. This is what I have now:


: X = array
I = index range

1. I :=3; {index = 3}
:
: 2. X[3] := X[3] +10; {adds value of 10 in index 3}
:
: 3. X[3 - 1] := X[2 * 3 - 1]; { same value in index [2] as in
index[5]}
:
: 4. X[3 + 1] := X[2 * 3] + X[2 * 3 + 1];
{index 4 = sum of index 6 and index 7}
:
: 5. for I := 5 to 7 do {initilizes I to 5}
: X[I] := X[I + 1]; {checks if I is <=7 and adds 1 to I}
: {loop}
{index 5 := index 6 (5 + 1)}
: {index 6 := index 7 (6 + 1)}
: {index 7 := index 8 (7 + 1)}
:
: 6. for I := 3 down to 1 do {initializes I to 3}
: X[3 + 1] := X[3] {checks if I is <= 1 and adds 1 to I}
{loop}
: {index [4]= index [3]}
X[2 + 1] := X[2]
{index [3] = index [2]}
X[1 + 1] := X[1]
{index [2] = index [1]}

Am I closer? I'm supposed to show the final contents of array X after all statements execute. I just figured out that I am supposed to substitute code from one program with this code and then tell what it does. My mind doesn't even want to mess with this anymore. I have one program to finish writing and this assignment here to finish and I will earn an ass. degree in computer science, and that is what I feel like, an ass. I have until the 7th to finish but I am beginning to wonder. I think if I get a good nights sleep I'll do better tomorrow. Thanx for looking at this for me and all your help.
Serina

: 6. for I := 3 down to 1 do {initializes I to 3}
: X[3 + 1] := X[3] {checks if I is <= 1 and adds 1 to I}
: {loop}
: {index [4]= index [3]}
: X[2 + 1] := X[2]
: {index [3] = index [2]}
: X[1 + 1] := X[1]
: {index [2] = index [1]}
:
: Am I closer? I'm supposed to show the final contents of array X after all statements execute. I just figured out that I am supposed to substitute code from one program with this code and then tell what it does. My mind doesn't even want to mess with this anymore. I have one program to finish writing and this assignment here to finish and I will earn an ass. degree in computer science, and that is what I feel like, an ass. I have until the 7th to finish but I am beginning to wonder. I think if I get a good nights sleep I'll do better tomorrow. Thanx for looking at this for me and all your help.
: Serina


Looks good other than the part above. You got the numbers right, but the check is :
6. for I := 3 down to 1 do {initializes I to 3}
X[3 + 1] := X[3] {checks if I is >= 1 and adds 1 to I}

because it starts at 3 and goes down until 1

Phat Nat

Thanx Phat Nat,
Now I am supposed to use numbers from another program and say what it does and what the final outcome is.
(value of X[I] from other program)
X[1]= 16
X[2]= 12
X[3]=6
X[4]=8
X[5]=2.5
X[6]=12
X[7]=14
X[8]=-54.5
=======================================
I :=3;
X[I] := X[I] + 10;
X[I - 1] := X[2*I-1];
X[I+1] := X[2*I] + X[2*I+1];
for I := 5 to 7 do
X[I] := X[I + 1];
for I := 3 down to 1 do
X[I + 1] := X[I]
=========================================
Is this right?

I = index range
X = array

I := 3; {I = 3}
X[I] := X[I] + 10; { X[3] stores 6(from other program) + 10 = 16}
X[I - 1] := X[2*I-1]; {X[2] = X[5], X[5] = 2.5 }
X[I+1] := X[2*I] + X[2*I+1]; {X[4] = X[6] + X[7], X[4] = X[4]=26}
for I := 5 to 7 do {initializes I to 5 and checks if I>= 7
X[I] := X[I + 1];and increments with each loop}
{X[5]:= X[6], X[5] and X[6] = 12}
{X[6] := X[7], X[6]= 14}
{X[7] := X[8], X[7]= -54.5}

for I := 3 down to 1 do {initializes I to 3}
X[I + 1] := X[I]
{X[4] := X[3], X[4]= 16}
{X[3] := X[2], X[3]= 2.5}
{X[2] ;= X[1], X[2]= 16}

show final contents of array X:
X[1] = 16
X[2] = 16
X[3] = 2.5
X[4] = 16
X[5] = 12
X[6] = 14
X[7] = -54.5
X[8] = -54.5

I think I still did it wrong. I would appreciate it if you can check this for me.
Thanx, Serina

: Thanx Phat Nat,
: Now I am supposed to use numbers from another program and say what it does and what the final outcome is.
: (value of X[I] from other program)
: X[1]= 16
: X[2]= 12
: X[3]=6
: X[4]=8
: X[5]=2.5
: X[6]=12
: X[7]=14
: X[8]=-54.5
: =======================================
: I :=3;
: X[I] := X[I] + 10;
: X[I - 1] := X[2*I-1];
: X[I+1] := X[2*I] + X[2*I+1];
: for I := 5 to 7 do
: X[I] := X[I + 1];
: for I := 3 down to 1 do
: X[I + 1] := X[I]
: =========================================
: Is this right?
:
: I = index range
: X = array
:
: I := 3; {I = 3}
: X[I] := X[I] + 10; { X[3] stores 6(from other program) + 10 = 16}
: X[I - 1] := X[2*I-1]; {X[2] = X[5], X[5] = 2.5 }
: X[I+1] := X[2*I] + X[2*I+1]; {X[4] = X[6] + X[7], X[4] = X[4]=26}
: for I := 5 to 7 do {initializes I to 5 and checks if I>= 7
: X[I] := X[I + 1];and increments with each loop}
: {X[5]:= X[6], X[5] and X[6] = 12}
: {X[6] := X[7], X[6]= 14}
: {X[7] := X[8], X[7]= -54.5}
:
: for I := 3 down to 1 do {initializes I to 3}
: X[I + 1] := X[I]
: {X[4] := X[3], X[4]= 16}
: {X[3] := X[2], X[3]= 2.5}
: {X[2] ;= X[1], X[2]= 16}
:
: show final contents of array X:
: X[1] = 16
: X[2] = 16
: X[3] = 2.5
: X[4] = 16
: X[5] = 12
: X[6] = 14
: X[7] = -54.5
: X[8] = -54.5
:
: I think I still did it wrong. I would appreciate it if you can check this for me.
: Thanx, Serina

Looks right, but I'm in a rush so I may have counted wrong too. But you have the idea. I'll check it again later and make another post if it is wrong.
Phat Nat

: Phat Nat,
I wanted to thank you for all your help. I finished my work and turned it in on time. I couldn't have done it without understanding this last assignment. I hope alot of good things come your way. Thanks again.
Sweet Serina
:
:
: