There is a pattern when dealing with arrays and functions; it's just a little hard to see at first.

When dealing with arrays, it's useful to remember the following: when an array expression appears in most contexts, the type of the expression is implicitly converted from "N-element array of T" to "pointer to T", and its value is set to point to the first element in the array. The exceptions to this rule are when the array expression appears as an operand of either the & or sizeof operators, or when it is a string literal being used as an initializer in a declaration.

Thus, when you call a function with an array expression as an argument, the function will receive a pointer, not an array:

int arr[10];
...
foo(arr);
...

void foo(int *arr) { ... }

This is why you don't use the & operator for arguments corresponding to "%s" in scanf():

char str[STRING_LENGTH];
...
scanf("%s", str);

Because of the implicit conversion, scanf() receives a char * value that points to the beginning of the str array. This holds true for any function called with an array expression as an argument (just about any of the str* functions, *scanf and *printf functions, etc.).

In practice, you will probably never call a function with an array expression using the & operator, as in:

int arr[N];
...
foo(&arr);

void foo(int (*p)[N]) {...}

Such code is not very common; you have to know the size of the array in the function declaration, and the function only works with pointers to arrays of specific sizes (a pointer to a 10-element array of T is a different type than a pointer to a 11-element array of T).

When an array expression appears as an operand to the & operator, the type of the resulting expression is "pointer to N-element array of T", or T (*)[N], which is different from an array of pointers (T *[N]) and a pointer to the base type (T *).

When dealing with functions and pointers, the rule to remember is: if you want to change the value of an argument and have it reflected in the calling code, you must pass a pointer to the thing you want to modify. Again, arrays throw a bit of a monkey wrench into the works, but we'll deal with the normal cases first.

Remember that C passes all function arguments by value; the formal parameter receives a copy of the value in the actual parameter, and any changes to the formal parameter are not reflected in the actual parameter. The common example is a swap function:

void swap(int x, int y) { int tmp = x; x = y; y = tmp; }
...
int a = 1, b = 2;
printf("before swap: a = %d, b = %d\n", a, b);
swap(a, b);
printf("after swap: a = %d, b = %d\n", a, b);

You'll get the following output:

before swap: a = 1, b = 2
after swap: a = 1, b = 2

The formal parameters x and y are distinct objects from a and b, so changes to x and y are not reflected in a and b. Since we want to modify the values of a and b, we must pass pointers to them to the swap function:

void swap(int *x, int *y) {int tmp = *x; *x = *y; *y = tmp; }
...
int a = 1, b = 2;
printf("before swap: a = %d, b = %d\n", a, b);
swap(&a, &b);
printf("after swap: a = %d, b = %d\n", a, b);

Now your output will be

before swap: a = 1, b = 2
after swap: a = 2, b = 1

Note that, in the swap function, we don't change the values of x and y, but the values of what x and y point to. Writing to *x is different from writing to x; we're not updating the value in x itself, we get a location from x and update the value in that location.

This is equally true if we want to modify a pointer value; if we write

int myFopen(FILE *stream) {stream = fopen("myfile.dat", "r"); }
...
FILE *in;
myFopen(in);

then we're modifying the value of the input parameter stream, not what stream points to, so changing stream has no effect on the value of in; in order for this to work, we must pass in a pointer to the pointer:

int myFopen(FILE **stream) {*stream = fopen("myFile.dat", "r"); }
...
FILE *in;
myFopen(&in);

Again, arrays throw a bit of a monkey wrench into the works. When you pass an array expression to a function, what the function receives is a pointer. Because of how array subscripting is defined, you can use a subscript operator on a pointer the same way you can use it on an array:

int arr[N];
init(arr, N);
...
void init(int *arr, int N) {size_t i; for (i = 0; i < N; i++) arr[i] = i*i;}

Note that array objects may not be assigned; i.e., you can't do something like

int a[10], b[10];
...
a = b;

so you want to be careful when you're dealing with pointers to arrays; something like

void (int (*foo)[N])
{
  ...
  *foo = ...;
}

won't work.

I was asked by a student if & and * were chosen because they were next to each other on the keyboard (something I had never noticed before). Much googling led me to B and BCPL documentation, and this thread. However, I couldn't find much at all. It seemed like there were lots of reasons for * in B, but I couldn't find anything for &.

So following @MichaelT's suggestion, I asked Ken Thompson:

From: Ken Thompson < [email protected] >

near on the keyboard: no.
c copied from b so & and * are same there.
b got * from earlier languages - some assembly,
bcpl and i think pl/1.
i think that i used & because the name (ampersand)
sounds like "address." b was designed to be run with
a teletype model 33 teletype. (5 bit baud-o code)
so the use of symbols was restricted.

If you look at it another way, *myVariable is of type int, which makes some sense.