Why does scanf require &?

Question

I want to read a number from stdin. I don't understand why scanf requires the use of & before the name of my variable:

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

Why does scanf need the address of the variable?

Answer 1

It needs to change the variable. Since all arguments in C are passed by value you need to pass a pointer if you want a function to be able to change a parameter.

Here's a super-simple example showing it:

void nochange(int var) {
    // Here, var is a copy of the original number. &var != &value
    var = 1337;
}
void change(int *var) {
    // Here, var is a pointer to the original number. var == &value
    // Writing to `*var` modifies the variable the pointer points to
    *var = 1337;
}
int main() {
    int value = 42;
    nochange(value);
    change(&value);
    return 0;
}

Answer 2

C function parameters are always "pass-by-value", which means that the function scanf only sees a copy of the current value of whatever you specify as the argument expression.

In this case &i is a pointer value that refers to the variable i. scanf can use this to modify i. If you passed i, then it would only see an uninitialized value, which (a) is UB, (b) is not sufficient information for scanf to know how to modify i.

Answer 3

It's not needed.

char s[1234];

scanf("%s", s); 

Works just fine without a single & anywhere. What scanf and company need are pointers. To let it modify a particular variable, you pass the address of that variable. For a few types that happens by default. For others, you use & to take the address (get a pointer to that variable).