What is wrong in this statement? scanf("%d",whatnumber);

Question

What is wrong in this statement?  scanf("%d",whatnumber);

Accepted Answer

C Programming Language

What Is Wrong in This Statement?
`scanf("%d", whatnumber);`

This is a frequently asked C programming interview question. The mistake is related to passing the memory address of a variable to the scanf() function.

Answer: The statement

scanf("%d", whatnumber);

is incorrect because the variable whatnumber is passed directly instead of passing its memory address.

The correct statement is:


scanf("%d", &whatnumber);

The & (address-of) operator tells scanf() where the entered value should be stored.

Why Is It Wrong?

The scanf() function stores the user input into a variable. Therefore, it requires the memory address of the variable, not its current value.

Statement	Correct?	Reason
`scanf("%d", whatnumber);`	No ❌	Variable value is passed instead of its address.
`scanf("%d", &whatnumber);`	Yes ✔	The address of the variable is passed.

Incorrect Program


#include 

int main()
{
    int whatnumber;

    scanf("%d", whatnumber);

    printf("%d", whatnumber);

    return 0;
}

Problem

scanf() receives an integer value instead of its address.
The program may produce a compilation warning or undefined behavior.
The entered value cannot be stored correctly.

Correct Program


#include 

int main()
{
    int whatnumber;

    scanf("%d", &whatnumber);

    printf("%d", whatnumber);

    return 0;
}

Sample Output


Enter a number: 25
25

How `scanf()` Works


User enters 25
       │
       ▼
scanf("%d", &whatnumber)
       │
       ▼
Address of whatnumber is passed
       │
       ▼
25 is stored in memory
       │
       ▼
printf() prints 25

Why Is the `&` Operator Needed?

Expression	Meaning
`whatnumber`	The value stored in the variable.
`&whatnumber`	The memory address of the variable.
`scanf()`	Requires the variable's memory address to store user input.

Important Note: Most data types used with scanf() require the address-of operator (&), such as int, float, double, and char.

Exception: Character arrays (strings) do not use the & operator because the array name itself represents the address of its first element.


char name[20];

scanf("%s", name);

Common Data Types with `scanf()`

Data Type	Format Specifier	Example
int	`%d`	`scanf("%d", &num);`
float	`%f`	`scanf("%f", &price);`
double	`%lf`	`scanf("%lf", &value);`
char	`%c`	`scanf("%c", &ch);`
String	`%s`	`scanf("%s", name);`

Common Mistakes

Avoid These Mistakes

Forgetting to use the & operator with numeric variables.
Using the wrong format specifier for the data type.
Using & with a string (character array).
Passing an uninitialized pointer to scanf().

Best Practices

Always pass the variable's address for numeric input.
Use the correct format specifier for each data type.
Read compiler warnings carefully.
Test your program with different input values.

Prerequisites

Before Learning This Topic

Basic understanding of variables and data types.
Knowledge of the scanf() function.
Basic understanding of memory addresses.
Familiarity with the address-of (&) operator.

Interview Questions

Why does scanf() require the address-of (&) operator?
What is wrong with scanf("%d", number);?
Why is & not used with %s in scanf()?
What is the difference between number and &number?
Which data types require the & operator in scanf()?

Key Takeaway

The statement scanf("%d", whatnumber); is incorrect because scanf() needs the memory address of the variable to store the user's input. The correct statement is scanf("%d", &whatnumber);, where the & operator provides the address of the variable.