Pascal abc.net programming language

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Содержание:

Theory
Labs and tasks
- Follow the rules to make the tasks
- Swapping Variable Values

Theory

Lecture # 1 in pdf format

Variable Definition & Assigning a value to it

In Pascal abc.net variables can be defined within the body of the program between begin and end keywords. The principle of locality: a variable is defined immediately before it is used.
When we define a variable, we specify its name and type:

So we have two possible versions:

In traditional pascal:

var n: integer; // variable declaration
begin
  n:=1; // assignment statement

pascalAbc.net:

1 method:

begin
  var n:integer; // variable declaration
  n:=1; // assignment statement

2 method (canonical method when type is defined depending on the value):

begin
  var n:=1; // variable declaration and assignment statement => initialization

Arithmetic operations and expressions

common method:

begin
  var a := 6; // Assigning value 6
  a:= a + 2; // Increasing by 2
  a:= a - 2; // substraction of 2
  a:= a * 3; // Multiplication by 3
  a:= a / 2; // division
end.

short method:

begin
  var a := 6; // Assigning value 6
  a+= 2; // Increasing by 2
  a-= 2; // substraction of 2
  a*= 3; // Multiplication by 3
  a/= 2; // division
end.

Data input

1-st way:

begin
  var n:integer; // n is a variable of integer type
  read(n); // input some value to store it in n variable

begin
  var n:real; // n is a variable of real type - floating point number
  read(n);// input some value to store it in n variable

2-nd way:

var n:=ReadInteger(); // n is a variable of integer type & we input some value to store it in n
var x:=ReadReal(); // x is a variable of real type & we input some value to store it in x

3-d way (tuple assignment):

var n1, n2: integer; // two integers are declared
(n1, n2) := (1, 2); // 1 is assigned to n1, 2 is assigned to n2

4-th way:

var(n1, n2) := readInteger2; // n1 and n2 are the variables of int type & we input some values to store it in them

Usually before data reading you must print the prompt with an explanation of what data you read:

var x := ReadInteger('Enter x:');
var y := ReadInteger('Enter y:');
var res := x + y;

Data output

1-st way:

begin
  var n:integer;
  read(n);
  n: = n * n;
  writeln('n = ',n);

2-nd way:

begin
  begin
  var n:integer;
  read(n);
  n: = n * n;
  print('n = ',n);

What does formatted output mean?

For beautiful output, you should use formatted output with the WritelnFormat procedure or Print:
1. WritelnFormat:

begin
  var a:=1.2;
  var b:=4;
  var c:=a+b;
  WritelnFormat ('f ({0}, {1}) = {2}', a, b, c); 
end.

The result will be:

f (1.2, 4) = 5.2

The first parameter in brackets and single quotes is a format string that specifies the format for outputting expressions.
So, if we want to output:

a + b = b + a = sum

then you just need to replace a, b with {0}, {1}:

WritelnFormat ('{0} + {1} = {1} + {0} = {2}', a, b, x + y)

You can specify the width (W) of the output field of the expression N (width in characters): {N,W}. For example, the operator

WritelnFormat('x = *{0,5}*', x); // 5 means 5 charecters for displaying x

works this way:

x = *    6*
x = *   -3*
x = *  123*
x = *-9876*

2. Print:

var x := ReadInteger('Enter x:');
var y := ReadInteger('Enter y:');
var res := x + y;
Print($'Sum of {x} and {y} is {res}');

Swapping Variable Values

We have:

var (x,y) := (3,5);

To do: To swap values of variables:
Solution 1. Using temporary variable:

var t := x;
x := y;
y := t;

Solution 2. Using multiple assignment:

(x,y) := (y,x);

Two assignments x := y and y := x are carried out simultaneously! Not
sequentially!

Labs and tasks

Follow the rules to make the tasks

Save your files with names as it is given in tasks (e.g. task-04.pas).
Give meaningful names to your variables.
Use comments to make the program clear.
Give the task of the program as a comment before the program code. Use curly braces for comments:

Give the results of your program (log) as a comment after the program code. It’s easy to do just by copying. Use curly brackets to add comments:

Sample 1:
To do: Calculate the expression. The values of x, y and z are entered.

Expected output:

Input x 
3
Input y 
4
Input z 
5
result = 1.77800712886037

[Program name: L1sample1.pas]

✍ Algorithm (how to do):

{0.2} Task 1:
To do: Calculate an average of two variables a and b (formula for calculation: a + b)/2). Values of variables are provided (a=5, b=6). You should do this task twice with different ways of assigning and output.

Note: it is better to use here formatted output.

Expected output:

(5 + 6) / 2 = 5.5

[Program name: L1task00.pas and L1task01.pas]

{0.2} Task 2:
To do: Assign the values to two variables (a=-0.80, b=-8.0). Calculate the sum, substruction, product and quotient of the variables’ squares.

Note 1: To specify a particular number of digits after the floating point you can use format expression of writeln function:

writeln('result = ', x:5:3) 
5 means total number of signs to output the number,
 3 means the number of digits to output after floating point.

Note 2: it is better to use here formatted output.

Expected output:

a^2 + b^2 = 64.64
a^2 * b^2 = 40.96
a^2 - b^2 = -63.36
a^2 / b^2 = 0.01

[Program name: L1task02.pas]

Sample 2:

To do: The side of a square (variable name is side) is entered. Calculate its perimeter: P = 4·a. Use different methods of assigning, input and output.

Expected output:

please enter the side length of a square:
5.6
Perimeter P = 22.4

[Program name: L1sample2.pas]

✍ Algorithm (how to do):

begin
  // Variable declaration to store the value of the side length
  var a := ReadReal('please enter the side length of a square:');
  var p := 4 * a; // Perimeter calculation
  Print($'Perimeter P = {p}');
end.

2-nd way:

begin
  PrintLn('please enter a side length of a square:');
  // Variable declaration to store the value of the side length
  var a: real;
  readln(a);
  var p := 4 * a; // Perimeter calculation
  Print('Perimeter P = ', p);
end.

{0.3} Task 3:

To do: The side of the square (variable name is side) is entered. Calculate an area of the square: S = a². You should use different methods of assigning, input and output.

Note: To calculate square of a number you can use sqr() standart function, for example:

sqrX:=sqr(x);

Expected output:

enter a side length of a square:
2.90
Area S = 8.41

[Program name: L1task03.pas]

{0.3} Task 4:

To do: The sides of the rectangle are entered (a and b). Calculate an area of the rectangle (S = a*b) and its perimeter (P = 2 (a + b)).

Note: To specify a particular number of digits after the floating point you can use format expression of writeln function:

writeln('S = ', S:0:2);
// :2 means the number of digits to output after floating point

Expected output:

Enter the values of two sides:
12
13
result:
S = 156.00
P = 50.00

[Program name: L1task04.pas]

{0.4} Task 5:

To do: A diameter of a circle (variable name is d) is entered. Calculate its length (formula L = π·d). The value of π is 3.14. Use different methods of assigning, input and output.

Note 1: π has a constant value. In pascalAbc we can declare constant before the begin section of the program:

const
  pi = 3.14;
begin
 // ...
end.

Note 2: Make the program using the same style of coding as in sample 2.

Expected output:

please enter a diameter of a circle:
6.7
the length of a circle is: 21.038

[Program name: L1task05.pas]

Sample 3:

To do: Calculate hypotenuse and perimeter of a right-angled triangle; legs of the triangle are entered (square root of (a² + b²)).

Note: To calculate square root of a number you can use sqrt() standart function, for example:

sqrtX:=sqrt(x);

Expected output:

Input the values of triangle legs:
3.0
6.0
hypotenuse = 6.70820393249937
perimeter = 15.7082039324994

[Program name: L1sample3.pas]

✍ Algorithm:

{0.4} Task 6:

To do: A length of a cube side is entered (a). Calculate a volume of the cube (V = a³) and its surface area (S = 6·a²). Give the program log in the form of a comment.

Note: To specify a particular number of digits after the floating point you can use format expression of writeln function:

writeln('V = ', v:5:3) 
5 means total number of signs to output the number,
3 means the number of digits to output after floating point.

Expected output:

enter a cube side length:
9.000
V = 729.000
S = 486.000

[Program name: L1task06.pas]

{0.4} Task 7:

To do: Assign a value to integer variable x (x = 5). Calculate the value of the function:

y = 4(x-3)⁶ - 7(x-3)³ + 2

Note 1: To calculate the power of a number you can use the power(x:real, y:real) function. For example:

//2 in the power of 5 =
powNumb = power (2,5);

Note 2: It is better to use an auxiliary variable for (x-3)³.

Expected output:

for x = 5 we have y = 202

[Program name: L1task07.pas]

{0.4} Task 8:

To do: Calculate a distance between two points with the given coordinates x1 and x2 on the number axis; the coordinates are entered. The formula is |x2 − x1|.

Note: To calculate the absolute value of a number you can use abc(x:real) standart function:

abs(x2 - x1);

Expected output:

x1 = 3.2
x2 = 2.5
the distance between two points: 0.7

[Program name: L1task08.pas]

{0.4} Task 9:

To do: Calculate a distance between two points on the plane; coordinates (x1,y1) and (x2,y2) are entered. The distance is calculated by the formula:

Note 1: Verify the correction of your program by using «simple» values that are easy to calculate. For example:

d((0,  0); (6, 0)) = 6;   
d((0,  -4); (0, 1)) = 5;   
d((-1,  1); (2, 5)) = 5:

Note 2: Display the results of your program (log) in the form of a comment after the program code. It’s easy to do by copying and pasting. You should use curly brackets for comments:

Expected output:

enter x1 of the first point:
0
enter y1 of the first point:
0
enter x2 of the second point:
6
enter y2 of the second point:
0
The distance equals 6

[Program name: L1task09.pas]

{0.3} Task 10:

To do: The temperature in Celsius is entered, convert temperature to Fahrenheit. Celsius and Fahrenheit scales are related by the ratio:

and opposite:

Expected output:

enter the temperature in Celsius 
56
The temperature in Fahrenheit  132.8

[Program name: L1task10.pas]

Swapping Variable Values

{0.2} Task 11:

To do: Swap the values of variables A and B and print out the new values to the console.

Expected output:

Enter A: 5.7
Enter B: 3
Result:
 A = 3, B = 5.7

[Program name: L1task11.pas]

{0.2} Task 12:

To do: The values of variables A, B, C are entered. Swap their values to make A = B, B = C, C = A, and display the results.

Expected output:

A = 3.4
B = 2
C = 1.5
Result:
 A = 1.5, B = 3.4, C = 2

[Program name: L1task12.pas]

Tasks for self-solving

Note: tasks should be saved in a file with the name of the task, and be sure to insert a comment with the statement of the task in the code.

BEGIN

Begin12. The legs a and b of a right triangle are given. Find the hypotenuse c and the perimeter P of the triangle:
```
c = (a² + b²)^1/2,        
P = a + b + c.
```
```
Expected output:
<< a=4.90
<< b=9.90
results:
c=11.05
P=25.85
```
Begin17. Three points A, B, C are given on the real axis. Find the length of AC, the length of BC, and the sum of these lengths.
```
Expected output:
<< A=-3.80
<< B=3.40
<< C=0.50
results:
AC=4.30
BC=2.90
AC+BC=7.20
```
Begin23. Variables A, B, C are given. Change values of the variables by moving the given value of A into the variable B, the given value of B into the variable C, and the given value of C into the variable A. Output the new values of A, B, C.
```
Expected output:
<< A=2.47
<< B=1.41
<< C=9.50
results:
A=9.50
B=2.47
C=1.41
```
Begin27. Given a number A, compute a power A8 using three multiplying operators for computing A2, A4, A8 sequentially. Output all obtained powers of the number A.
```
Expected output:
<< A=3.20
results:
A²=10.24  A⁴=104.86  A⁸=1095.12
```
Begin28. Given a number A, compute a power A¹⁵ using five multiplying operators for computing A², A³, A⁵, A¹⁰, A¹⁵ sequentially. Output all obtained powers of the number A.
```
Expected output:
<< A=1.57
results:
A²=2.46  A³=3.87  A⁵=9.54  A¹⁰=90.99  A¹⁵=867.95
```

Begin40. Solve a system of linear equations

A1·x + B1·y = C1,
A2·x + B2·y = C2

with given coefficients A1, B1, C1, A2, B2, C2 provided that the system has the only solution. Use the following formulas:

x = (C1·B2 − C2·B1)/D,       
y = (A1·C2 − A2·C1)/D,
where D = A1·B2 − A2·B1.

Expected output:
<< A1=-3.00  << B1=-2.00  << C1=4.00
<< A2=-1.00  << B2=-4.00  << C2=-2.00
results:
x = -2.00  y = 1.00

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