Using scanf for input

Introduction

We use scanf to get information from the user while the program is running. It helps the program read what the user types.

When you want to ask the user for their age and use it in the program.

When you need to get a number from the user to do math calculations.

When you want to read a word or sentence typed by the user.

When you want to make a simple quiz that asks questions and reads answers.

When you want to get multiple pieces of information at once, like name and score.

Syntax

C

scanf("%format_specifier", &variable);

The format_specifier tells scanf what type of data to expect, like %d for integers or %f for decimals.

The & symbol before the variable means 'address of' and is needed so scanf can store the input in the right place.

Examples

This reads an integer number from the user and saves it in the variable age.

C

int age;
scanf("%d", &age);

This reads a decimal number (float) from the user and saves it in price.

C

float price;
scanf("%f", &price);

This reads a word (string) from the user and saves it in the array name. No & needed for arrays.

C

char name[20];
scanf("%s", name);

Sample Program

This program asks the user to enter their age, reads it using scanf, and then prints it back.

C

#include <stdio.h>

int main() {
    int age;
    printf("Enter your age: ");
    scanf("%d", &age);
    printf("You are %d years old.\n", age);
    return 0;
}

Important Notes

Always use the correct format specifier for the type of data you want to read.

For strings (arrays), do not use & before the variable name.

scanf stops reading a string at the first space, so it reads only one word.

Summary

scanf reads user input during program execution.

Use format specifiers like %d, %f, and %s to tell scanf what to expect.

Remember to use & before variables except for arrays when using scanf.

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