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Lab 2: Formatted and Unformatted Input/output in C – C Programming Lab Report
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Lab 2: Formatted and Unformatted Input/output in C

A C Programming Lab Report covering Formatted and Unformatted Input/output in C

Lab Report Information

Lab No.: 2

Title: Formatted and Unformatted Input/output in C

Course: Computer Programming (CT 101), I Year I Part

Credit: Important Notes

Objective: To become familiar with data types, operators, expressions, and formatted input/output functions.

C Programming Lab Report: Theory

This lab report focuses on variables, operators, expressions, and the core concepts of Formatted and Unformatted Input/output in C.

Datatypes and Operators

C uses various data types (like `int`, `float`, `char`) to define variables. Operators are symbols used to perform operations. These include arithmetic (`+`, `-`), relational (`==`, `>`), logical (`&&`, `||`), and assignment (`=`, `+=`) operators.

Type Casting and Promotion

Type casting is the manual conversion of one data type to another, e.g., `(float)my_integer;`. Type promotion is the automatic conversion done by the compiler in expressions to prevent data loss.

Format Specifiers

Format specifiers (like `%d`, `%f`, `%s`) are used in `printf()` and `scanf()` to define the type and format of data. They can be modified with flags (`-`, `+`, `0`), width specifiers, and precision specifiers (`.2f`).

Lab 2 Programs: Formatted and Unformatted Input/output in C

1. Program to Illustrate the Increment Operator

Question:

WAP to illustrate the increment operator.

Program:

#include <stdio.h>
#include <conio.h>

void main() {
    clrscr();
    int a = 10, b = 10, x, y;
    x = ++a * ++a - ++a;
    y = ++b * b++ - b++;
    printf("Value of a = %d and value of x = %d\n", a, x);
    printf("Value of b = %d and value of y = %d", b, y);
    getch();
}

Output:

Value of a = 13 and value of x = 143
Value of b = 13 and value of y = 119

2. Program for Seconds to Minutes Conversion

Question:

WAP to input seconds and convert them into minutes.

Program:

#include <stdio.h>
#include <conio.h>

void main() {
    clrscr();
    int s;
    float m;
    printf("Enter seconds: ");
    scanf("%d", &s);
    m = s / 60.0;
    printf("\n%d sec is = %.2f minutes", s, m);
    getch();
}

Output:

Enter seconds: 150
150 sec is = 2.50 minutes

Algorithm:

  1. Start
  2. Declare integer `s` and float `m`.
  3. Read the value for seconds `s`.
  4. Calculate minutes `m = s / 60.0`.
  5. Display the result `m`.
  6. Stop.

Flowchart:

Start
↓
Declare int s, float m
↓
Input s
↓
m = s / 60.0
↓
Display m
↓
Stop

3. Program for Equivalent Resistance

Question:

WAP to compute the equivalent resistance of two resistors R1 and R2 when they are connected in series & parallel connection.

Program:

#include <stdio.h>
#include <conio.h>

void main() {
    clrscr();
    float r1, r2, series, parallel;
    printf("\nEnter resistance of R1: ");
    scanf("%f", &r1);
    printf("\nEnter resistance of R2: ");
    scanf("%f", &r2);
    series = r1 + r2;
    parallel = (r1 * r2) / (r1 + r2);
    printf("\nThe equivalent resistance of %.2f and %.2f\n", r1, r2);
    printf("While connecting in series = %.2f\n", series);
    printf("While connecting in parallel = %.2f", parallel);
    getch();
}

Output:

Enter resistance of R1: 10
Enter resistance of R2: 20
The equivalent resistance of 10.00 and 20.00
While connecting in series = 30.00
While connecting in parallel = 6.67

Algorithm:

  1. Start
  2. Declare float variables r1, r2, series, parallel.
  3. Read values for r1 and r2.
  4. Calculate series = r1 + r2.
  5. Calculate parallel = (r1 * r2) / (r1 + r2).
  6. Display series and parallel resistances.
  7. Stop.

Flowchart:

Start
↓
Input r1, r2
↓
series = r1 + r2
parallel = (r1*r2)/(r1+r2)
↓
Display series, parallel
↓
Stop

4. Program for Expression Evaluation

Question:

WAP to evaluate the expression v = ((x+y)/(3a+b)) * (p+q) + (x/y)^3.5

Program:

#include <stdio.h>
#include <conio.h>
#include <math.h>

void main() {
    clrscr();
    float v, x, y, a, b, p, q;
    printf("\nEnter x, y, a, b, p, q respectively: ");
    scanf("%f %f %f %f %f %f", &x, &y, &a, &b, &p, &q);
    v = ((x + y) / (3 * a + b) * (p + q)) + pow((x / y), 3.5);
    printf("\nThe result is %f", v);
    getch();
}

Output:

Enter x, y, a, b, p, q respectively: 10 5 2 3 4 5
The result is 26.313709

Algorithm:

  1. Start
  2. Declare float variables v, x, y, a, b, p, q.
  3. Read values for x, y, a, b, p, q.
  4. Calculate v using the given formula.
  5. Display the value of v.
  6. Stop.

Flowchart:

Start
↓
Input x, y, a, b, p, q
↓
v = ((x+y)/(3a+b)) * (p+q) + pow(x/y, 3.5)
↓
Display v
↓
Stop

5. Program for Swapping Values

Question:

WAP to swap the values of two variables x and y.

Program:

#include <stdio.h>
#include <conio.h>

void main() {
    clrscr();
    int x, y, temp;
    printf("Enter values for x and y: ");
    scanf("%d %d", &x, &y);
    printf("\nBefore Swapping:\n");
    printf("x=%d, y=%d\n", x, y);
    temp = x;
    x = y;
    y = temp;
    printf("\nAfter Swapping:\n");
    printf("x=%d, y=%d", x, y);
    getch();
}

Output:

Enter values for x and y: 10 20
Before Swapping:
x=10, y=20
After Swapping:
x=20, y=10

Algorithm:

  1. Start
  2. Declare integer variables x, y, temp.
  3. Read values for x and y.
  4. Store value of x in temp.
  5. Assign value of y to x.
  6. Assign value of temp to y.
  7. Display the new values of x and y.
  8. Stop.

Flowchart:

Start
↓
Input x, y
↓
temp = x
x = y
y = temp
↓
Display x, y
↓
Stop

6. Program for Format Specification Observation

Question:

Run the following programs and observe the output regarding format specification.

i. Integer Formatting

#include <stdio.h>
#include <conio.h>
void main() {
    int x = 1234;
    printf("Case 1: %d\n", x);
    printf("Case 2: %6d\n", x);
    printf("Case 3: %2d\n", x);
    printf("Case 4: %-6d\n", x);
    printf("Case 5: %06d\n", x);
    printf("Case 6: %-+6d\n", x);
    printf("Case 7: %-2d", x);
    getch();
}

Output:

Case 1: 1234
Case 2: 1234
Case 3: 1234
Case 4: 1234
Case 5: 001234
Case 6: +1234
Case 7: 1234

ii. Float Formatting

#include <stdio.h>
#include <conio.h>
void main() {
    float x = 123.8765;
    printf("Case 1: %f\n", x);
    printf("Case 2: %e\n", x);
    printf("Case 3: %.2f\n", x);
    printf("Case 4: %12.4f\n", x);
    printf("Case 5: %-12.2f\n", x);
    printf("Case 6: %012.2f\n", x);
    printf("Case 7: %g\n", x);
    printf("Case 8: %.0f", x);
    getch();
}

Output:

Case 1: 123.876500
Case 2: 1.238765e+02
Case 3: 123.88
Case 4: 123.8765
Case 5: 123.88
Case 6: 000000123.88
Case 7: 123.877
Case 8: 124

iii. Character Formatting

#include <stdio.h>
#include <conio.h>
void main() {
    char ch = 'b';
    printf("Case 1: %c\n", ch);
    printf("Case 2: %8c\n", ch);
    printf("Case 3: %-8c", ch);
    getch();
}

Output:

Case 1: b
Case 2: b
Case 3: b

iv. String Formatting

#include <stdio.h>
#include <conio.h>
void main() {
    char str[20] = "My Name is Diwakar.";
    printf("Case 1: %s\n", str);
    printf("Case 2: %-19s\n", str);
    printf("Case 3: %-19.13s\n", str);
    printf("Case 4: %19.5s\n", str);
    printf("Case 5: %-19.7s\n", str);
    printf("Case 6: %-1.7s\n", str);
    printf("Case 7: %.9s", str);
    getch();
}

Output:

Case 1: My Name is Diwakar.
Case 2: My Name is Diwakar.
Case 3: My Name is Diw
Case 4: My Nam
Case 5: My Name
Case 6: My Name
Case 7: My Name i

v. Unformatted Input with %[^\n]

#include <stdio.h>
#include <conio.h>
void main() {
    char str[50];
    printf("Enter a string: ");
    scanf("%[^\n]", str);
    printf("Read string is: %s", str);
    getch();
}

Output:

Enter a string: my name is diwakar bhandari
Read string is: my name is diwakar bhandari

vi. Unformatted Input with Space Delimiter

#include <stdio.h>
#include <conio.h>
void main() {
    char str[20];
    printf("Enter string: ");
    scanf("%s", str);
    printf("Read String is: %s", str);
    getch();
}

Output:

Enter string: my name
Read String is: my

vii. Character Set Input

#include <stdio.h>
#include <conio.h>
void main() {
    char str[20];
    printf("Enter String: ");
    scanf("%[a-z]", str);
    printf("Read String is: %s", str);
    getch();
}

Output:

Enter String: diwakar123
Read String is: diwakar

Discussion and Conclusion

Through this C Programming Lab Report on Formatted and Unformatted Input/output in C, I learned a lot. In this lab, I explored different ways to use operators, especially the pre-increment and post-increment functions which can yield different results depending on compiler behavior. After researching it, I conclude that the output result of complex increment/decrement expressions is often compiler-dependent and should be used with caution. The exercises on formatted output (`printf`) demonstrated the power of format specifiers, flags, width, and precision for controlling the appearance of data, which is essential for creating clean and readable user interfaces.

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