Using scanf for input - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Reading an integer with scanf

What is the output of this C program when the user inputs 42?

C

#include <stdio.h>

int main() {
    int num;
    printf("Enter a number: ");
    scanf("%d", &num);
    printf("You entered: %d\n", num);
    return 0;
}

A You entered: 42

B Compilation error

C You entered: %d

D You entered: 0

💡 Hint

Remember scanf needs the address of the variable to store the input.

The scanf function reads the integer input 42 and stores it in the variable num. Then printf prints it correctly.

❓ Predict Output

intermediate

Reading multiple inputs with scanf

What will be printed if the user inputs 10 20?

C

#include <stdio.h>

int main() {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("Sum: %d\n", a + b);
    return 0;
}

A Sum: 30

B Sum: 1020

C Sum: 10

D Runtime error

💡 Hint

scanf reads each integer separately when given multiple format specifiers.

scanf reads two integers 10 and 20, stores them in a and b, then prints their sum 30.

❓ Predict Output

advanced

Using scanf with wrong format specifier

What happens when this program runs and the user inputs 3.14?

C

#include <stdio.h>

int main() {
    int x;
    scanf("%d", &x);
    printf("Value: %d\n", x);
    return 0;
}

A Runtime error

B Value: 0

C Value: 3.14

D Value: 3

💡 Hint

scanf with %d reads integer part and stops at decimal point.

scanf with %d reads integer part 3 from input 3.14 and stores it in x. The decimal part is ignored.

❓ Predict Output

advanced

Reading a string with scanf

What will be the output if the user inputs HelloWorld?

C

#include <stdio.h>

int main() {
    char str[20];
    scanf("%s", str);
    printf("You typed: %s\n", str);
    return 0;
}

A You typed: Hello World

B You typed: HelloWorld

C You typed: HelloWorld\n

D Buffer overflow error

💡 Hint

scanf with %s reads input until first whitespace.

scanf reads the string HelloWorld into str and prints it exactly.

❓ Predict Output

expert

Using scanf with character input and newline

What will be printed if the user inputs A and presses Enter?

C

#include <stdio.h>

int main() {
    char c1, c2;
    scanf("%c", &c1);
    scanf("%c", &c2);
    printf("Chars: %c and %c\n", c1, c2);
    return 0;
}

A Chars: A and (space)

B Chars: A and A

C Chars: A and \n

D Runtime error

💡 Hint

The second scanf reads the leftover newline character from input buffer.

First scanf reads 'A', second scanf reads the newline character '\n' left in the input buffer.

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