Using scanf for input - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to read an integer from the user using scanf.

C

#include <stdio.h>

int main() {
    int number;
    scanf([1], &number);
    printf("You entered: %d\n", number);
    return 0;
}

Drag options to blanks, or click blank then click option'

A"%d"

B"%f"

C"%c"

D"%s"

Attempts:

3 left

💡 Hint

Common Mistakes

Using %f or %c instead of %d for integer input.

Forgetting the quotes around the format specifier.

The format specifier %d is used to read an integer with scanf.

2fill in blank

medium

Complete the code to read a float value from the user using scanf.

C

#include <stdio.h>

int main() {
    float value;
    scanf([1], &value);
    printf("You entered: %f\n", value);
    return 0;
}

Drag options to blanks, or click blank then click option'

A"%d"

B"%c"

C"%f"

D"%s"

Attempts:

3 left

💡 Hint

Common Mistakes

Using %d instead of %f for float input.

Forgetting the ampersand (&) before the variable.

The format specifier %f is used to read a float with scanf.

3fill in blank

hard

Fix the error in the scanf statement to correctly read a character.

C

#include <stdio.h>

int main() {
    char ch;
    scanf([1], &ch);
    printf("You entered: %c\n", ch);
    return 0;
}

Drag options to blanks, or click blank then click option'

A"%f"

B"%d"

C"%s"

D"%c"

Attempts:

3 left

💡 Hint

Common Mistakes

Using %d, %f or %s instead of %c for character input.

The format specifier %c is used to read a single character with scanf.

4fill in blank

hard

Complete the code to read a string safely using scanf.

C

#include <stdio.h>

int main() {
    char name[20];
    scanf([1], name);
    printf("Hello, %s!\n", name);
    return 0;
}

Drag options to blanks, or click blank then click option'

A"%19s"

B"%s"

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using %s without size limit, risking buffer overflow.

Using & before the array name when reading strings.

To safely read a string into an array, use %19s to limit input size. When reading strings into arrays, do not use the ampersand (&) because the array name is already a pointer.

5fill in blank

hard

Fill all three blanks to read two integers and one float from the user.

C

#include <stdio.h>

int main() {
    int a, b;
    float c;
    scanf("[1] [2] [3]", &a, &b, &c);
    printf("Integers: %d, %d and Float: %f\n", a, b, c);
    return 0;
}

Drag options to blanks, or click blank then click option'

A"%d %d %f"

B"%d %f %d"

C"%f"

D"%d"

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up the order of format specifiers.

Using wrong format specifiers for variables.

The format string should match the variables: "%d %d %f" to read two integers and one float. Fill "%d" for the first two blanks and "%f" for the third.

Practice

1. What is the purpose of the scanf function in C?

easy

A. To read input from the user during program execution

B. To print output to the screen

C. To declare variables

D. To perform mathematical calculations

Solution

1. Step 1: Understand the role of scanf

   scanf is used to get input from the user while the program runs.

2. Step 2: Differentiate from other functions

   Functions like printf print output, but scanf reads input.

3. Final Answer:

   To read input from the user during program execution -> Option A

4. Quick Check:

   scanf reads input [OK]

Hint: scanf reads input, printf prints output [OK]

Common Mistakes:

• Confusing scanf with printf
• Thinking scanf declares variables
• Assuming scanf performs calculations

2. Which of the following is the correct way to read an integer variable num using scanf?

easy

A. scanf("%s", &num);

B. scanf("%d", num);

C. scanf("%f", &num);

D. scanf("%d", &num);

Solution

1. Step 1: Identify the correct format specifier for integer

   Use %d to read an integer.

2. Step 2: Use the address operator for variables

   Use &num to pass the address of the variable to scanf.

3. Final Answer:

   scanf("%d", &num); -> Option D

4. Quick Check:

   Integer input needs %d and &variable [OK]

Hint: Use & before variable except arrays [OK]

Common Mistakes:

• Omitting & before variable
• Using wrong format specifier like %f or %s
• Passing variable instead of its address

3. What will be the output of the following code if the user inputs 25?

int age;
scanf("%d", &age);
printf("Age: %d", age);

medium

A. Age: 0

B. Age: 25

C. Age: %d

D. Compilation error

Solution

1. Step 1: Understand input reading

   scanf("%d", &age); reads the integer 25 and stores it in age.

2. Step 2: Print the stored value

   printf("Age: %d", age); prints the string with the value of age, which is 25.

3. Final Answer:

   Age: 25 -> Option B

4. Quick Check:

   scanf stores input, printf prints it [OK]

Hint: scanf stores input, printf prints variable [OK]

Common Mistakes:

• Forgetting & in scanf causing runtime error
• Using wrong format specifier in printf
• Expecting scanf to print output

4. Identify the error in the following code snippet:

int x;
printf("Enter a number: ");
scanf("%d", x);
printf("You entered %d", x);

medium

A. Missing & before variable x in scanf

B. Wrong format specifier in scanf

C. Missing semicolon after printf

D. Variable x not declared

Solution

1. Step 1: Check scanf usage

   scanf("%d", x); is incorrect because it should use &x to pass the address.

2. Step 2: Confirm other parts are correct

   Format specifier is correct, semicolons are present, and x is declared.

3. Final Answer:

   Missing & before variable x in scanf -> Option A

4. Quick Check:

   scanf needs &variable for input [OK]

Hint: Always use & before variables in scanf except arrays [OK]

Common Mistakes:

• Omitting & causing segmentation fault
• Using wrong format specifier
• Assuming scanf prints output

5. You want to read a string (word) from the user into a character array name[20]. Which of the following scanf statements is correct and safe to avoid buffer overflow?

hard

A. scanf("%s", name);

B. scanf("%s", &name);

C. scanf("%19s", name);

D. scanf("%20s", &name);

Solution

1. Step 1: Understand string input with scanf

   To read a string into name, use %s without & because arrays decay to pointers.

2. Step 2: Prevent buffer overflow

   Limit input length to 19 characters using %19s to leave space for the null terminator.

3. Final Answer:

   scanf("%19s", name); -> Option C

4. Quick Check:

   Limit string input length with %19s [OK]

Hint: Use %19s for 20-char array to avoid overflow [OK]

Common Mistakes:

• Using & with arrays in scanf
• Not limiting input length causing overflow
• Using wrong format specifier for strings