Using scanf for input - Step-by-Step Execution

Concept Flow - Using scanf for input

Start Program

↓

Declare variables

↓

Call scanf

↓

Wait for user input

↓

Store input in variables

↓

Use input in program

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End Program

The program declares variables, calls scanf to get user input, waits for input, stores it, then uses it.

Execution Sample

C

#include <stdio.h>
int main() {
  int age;
  scanf("%d", &age);
  printf("Age: %d\n", age);
  return 0;
}

Reads an integer from the user and prints it.

Execution Table

Step	Action	Input Provided	Variable 'age' Value	Output
1	Program starts, variable 'age' declared	N/A	undefined	N/A
2	scanf called, waiting for input	N/A	undefined	N/A
3	User inputs '25' and presses Enter	25	25	N/A
4	printf prints the value of 'age'	N/A	25	Age: 25
5	Program ends	N/A	25	N/A

💡 Program ends after printing the input value.

Variable Tracker

Variable	Start	After scanf input	Final
age	undefined	25	25

Key Moments - 2 Insights

Why do we use &age in scanf instead of just age?

scanf needs the address of the variable to store the input there. Using &age gives scanf the memory location to write the value, as shown in step 3 of the execution_table.

What happens if the user inputs a non-integer value?

scanf expects an integer for %d. If the input is not an integer, scanf will fail to store a valid value, and 'age' may remain unchanged or undefined. This is why input must match the format specifier.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'age' immediately after scanf reads input?

A 25

B undefined

C 0

D Input string

💡 Hint

Check the 'Variable age Value' column at Step 3 in the execution_table.

At which step does the program wait for user input?

A Step 1

B Step 2

C Step 4

D Step 5

💡 Hint

Look for the step where scanf is called and waiting for input in the execution_table.

If the user enters '30' instead of '25', how does the variable_tracker change?

A The output changes but 'age' stays 25

B The variable 'age' remains undefined

C The 'After scanf input' and 'Final' values for 'age' become 30

D The program crashes

💡 Hint

Variable_tracker shows the value stored after scanf input; changing input changes stored value.

Concept Snapshot

scanf("%d", &var) reads an integer from user input.
Use &var to pass variable address.
Input must match format specifier.
scanf waits for input during execution.
Value is stored in variable for later use.

Full Transcript

This program shows how scanf reads input from the user. First, it declares an integer variable 'age'. Then, scanf is called with "%d" and &age, which means it waits for the user to type an integer and press Enter. The input is stored in 'age'. Finally, printf prints the stored value. The execution table shows each step: starting the program, waiting for input, storing input, printing output, and ending. The variable tracker shows how 'age' changes from undefined to the input value. Remember, scanf needs the address of the variable to store the input. If input does not match the expected type, scanf may fail to store correctly.