How does the scanf function work in C?

Question

Why do you require ampersand (&) in the scanf function. What will the output or type of error (compile or runtime) be in the following C code?

Answer 1

In C, all function arguments are passed by value; any changes to the function's formal parameter are not reflected in the actual parameter. For example:

void foo(int bar)
{
  bar = bar + 1;
}

int main(void)
{
  int x = 0;
  printf("x before foo = %d\n", x);
  foo(x);
  printf("x after foo = %d\n", x);
  return 0;
}

The output of the program will be

x before foo = 0
x after foo = 0

because bar receives the value of x (0), not a reference to x itself. Changing bar has no effect on x.

In C, the way around this is to pass a pointer to a variable:

void foo(int *bar)
{
  *bar = *bar + 1;
}

int main(void)
{
  int x = 0;
  printf("x before foo = %d\n", x);
  foo(&x);
  printf("x after foo = %d\n", x);
  return 0;
}

Now the output of the program is

x before foo = 0
x after foo = 1

This time, the formal parameter bar is not an int, but a pointer to int, and it receives the address of x (given by the expression &x in the call to foo), not the value contained in x. The expression *bar means "get the value in the location bar points to", so *bar = *bar + 1 corresponds to x = x + 1.

Since scanf() needs to write to its arguments, it expects those arguments to typed as pointers. The "%d" conversion specifier expects the corresponding argument to be a pointer to int (int *), the "%u" conversion specifier expects a pointer to unsigned int (unsigned *), "%s" expects a pointer to char (char *), "%f" expects a pointer to float (float *), etc. In your example, since a is typed int, you need to use the expression &a to get a pointer.

Note that if a were already a pointer type, you would not need to use the & operator in the call to scanf():

int main(void)
{
  int a, *pa;      // declare pa as a pointer to int
  ...
  pa = &a;         // assign address of a to pa
  scanf("%d", pa); // scanf() will write to a through pa
  ...
}

Note also that when passing an array to a function (such as when using the "%s" conversion specifier to read a string), you don't need to use the & operator; the array expression will implicitly be converted to a pointer type:

int main(void)
{
  char name[20];
  ...
  scanf("%19s", name); // name implicitly converted from "char [20]" to "char *"
  ...
}