Input with scanf()

scanf(), declared in <stdio.h>, reads formatted input from standard input (typically the keyboard). Like printf, it takes a format string describing what to read, followed by pointers to the variables that should receive the parsed values.

#include <stdio.h>

int main(void) {
    int age;
    double price;

    printf("Enter your age: ");
    scanf("%d", &age);

    printf("Enter a price: ");
    scanf("%lf", &price);

    printf("Age: %d, Price: %.2f\n", age, price);
    return 0;
}

Never Forget the & for Scalar Variables

Missing & is one of THE most common C beginner bugs

scanf needs the address of each variable so it can write the parsed value directly into memory — that's what the & (address-of) operator provides. Forgetting it for a scalar variable like int or double passes the variable's current value instead of its address. scanf then interprets that value as a memory address and tries to write there, which typically causes a segfault or other undefined behavior. (The one common exception is a char array used as a string buffer, which already decays to an address — no & needed there.)

int age;
/* scanf("%d", age); */  // BUG: missing & — passes a garbage value as an address

scanf("%d", &age);       // CORRECT — passes the address of age

char name[50];
scanf("%s", name);       // CORRECT — arrays decay to a pointer automatically, no & needed

Always Check scanf's Return Value

Ignoring the return value hides malformed input

scanf returns the number of input items successfully matched and assigned — which can be less than the number of conversion specifiers if the input doesn't match the expected format (or reaches end-of-file). Code that ignores this return value will silently continue with uninitialized or stale variable values when the user types something unexpected.

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

if (matched != 1) {
    printf("Invalid input — expected a number\n");
} else {
    printf("You entered: %d\n", age);
}

The Trailing Newline Gotcha

When a user types a number and presses Enter, scanf("%d", ...) consumes the digits but leaves the newline character sitting in the input buffer. If the next input call is getchar() or fgets(), that leftover newline is read immediately, often making it look like the following input was skipped entirely.

#include <stdio.h>

int main(void) {
    int age;
    char name[50];

    printf("Enter your age: ");
    scanf("%d", &age);

    printf("Enter your name: ");
    /* BUG: fgets immediately reads the leftover '\n' from the previous
       scanf call and returns an empty string, before the user can type. */
    fgets(name, sizeof(name), stdin);

    printf("Age: %d, Name: %s\n", age, name);
    return 0;
}

The fix: consume the leftover newline explicitly

After a numeric scanf call that will be followed by line-based input, drain the rest of the current line first — either with a small loop calling getchar() until a newline or EOF, or by adding a leading space in the next scanf format string (a space in a scanf format matches and discards any amount of leading whitespace, including newlines).

int age;
char name[50];

scanf("%d", &age);
while (getchar() != '\n') {
    /* discard the rest of the line, including the leftover newline */
}

printf("Enter your name: ");
fgets(name, sizeof(name), stdin); // now reads the real input correctly

scanf writes into the address you give it — always pass &variable for scalars
Always check the return value of scanf (number of items matched) before trusting the result
scanf("%d", ...) leaves the trailing newline in the input buffer — drain it before calling getchar/fgets