Everything you need to know about pointers in C

Version 1.3. Copyright 2005–2010 Peter Hosey.

This work is licensed under a Creative Commons Attribution 2.5 License.

This document comes with a companion example program, available as one file or as multiple files (zipped).

Style used in this document This is regular text. This is a variable , some code , and some sample output . This is a line of code. This is output you'd see on your screen.

Definition of a pointer A pointer is a memory address. (Mmm, short paragraphs.)

Starting off Say you declare a variable named foo . int foo ; This variable occupies some memory. On current mainstream Intel processors, it occupies four bytes of memory (because an int is four bytes wide). Now let's declare another variable. int * foo_ptr = &foo; foo_ptr is declared as a pointer to int. We have initialized it to point to foo . As I said, foo occupies some memory. Its location in memory is called its address. &foo is the address of foo (which is why & is called the “address-of operator”). Think of every variable as a box. foo is a box that is sizeof(int) bytes in size. The location of this box is its address. When you access the address, you actually access the contents of the box it points to. This is true of all variables, regardless of type. In fact, grammatically speaking, there is no such thing as a “pointer variable”: all variables are the same. There are, however, variables with different types. foo 's type is int. foo_ptr 's type is int *. (Thus, “pointer variable” really means “variable of a pointer type”.) The point of that is that the pointer is not the variable! The pointer to foo is the contents of foo_ptr . You could put a different pointer in the foo_ptr box, and the box would still be foo_ptr . But it would no longer point to foo . The pointer has a type, too, by the way. Its type is int. Thus it is an “int pointer” (a pointer to int). An int **'s type is int * (it points to a pointer to int). The use of pointers to pointers is called multiple indirection . More on that in a bit.

Interlude: Declaration syntax The obvious way to declare two pointer variables in a single declaration is: int* ptr_a , ptr_b ; If the type of a variable containing a pointer to int is int * ,

is , and a single declaration can declare multiple variables of the same type by simply providing a comma-separated list ( ptr_a, ptr_b ),

), then you can declare multiple int -pointer variables by simply giving the int -pointer type ( int * ) followed by a comma-separated list of names to use for the variables ( ptr_a, ptr_b ). Given this, what is the type of ptr_b ? int *, right? *bzzt* Wrong! The type of ptr_b is int. It is not a pointer. C's declaration syntax ignores the pointer asterisks when carrying a type over to multiple declarations. If you split the declaration of ptr_a and ptr_b into multiple declarations, you get this: int *ptr_a; int ptr_b; Think of it as assigning each variable a base type (int), plus a level of indirection, indicated by the number of asterisks ( ptr_b 's is zero; ptr_a 's is one). It's possible to do the single-line declaration in a clear way. This is the immediate improvement: int *ptr_a, ptr_b; Notice that the asterisk has moved. It is now right next to the word ptr_a . A subtle implication of association. It's even clearer to put the non-pointer variables first: int ptr_b, *ptr_a; The absolute clearest is to keep every declaration on its own line, but that can take up a lot of vertical space. Just use your own judgment. Finally, I should point out that you can do this just fine: int *ptr_a, *ptr_b; There's nothing wrong with it. Incidentally, C allows zero or more levels of parentheses around the variable name and asterisk: int ((not_a_pointer)), (*ptr_a), (((*ptr_b))); This is not useful for anything, except to declare function pointers (described later). Further reading: The right-left rule for reading C declarations.

Assignment and pointers Now, how do you assign an int to this pointer? This solution might be obvious: foo_ptr = 42; It is also wrong. Any direct assignment to a pointer variable will change the address in the variable, not the value at that address. In this example, the new value of foo_ptr (that is, the new “pointer” in that variable) is 42. But we don't know that this points to anything, so it probably doesn't. Trying to access this address will probably result in a segmentation violation (read: crash). (Incidentally, compilers usually warn when you try to assign an int to a pointer variable. gcc will say “ warning: initialization makes pointer from integer without a cast ”.) So how do you access the value at a pointer? You must dereference it.

Dereferencing int bar = *foo_ptr; In this declaration, the dereference operator (prefix *, not to be confused with the multiplication operator) looks up the value that exists at an address. (This is called a “load” operation.) It's also possible to write to a dereference expression (the C way of saying this: a dereference expression is an lvalue , meaning that it can appear on the left side of an assignment): *foo_ptr = 42; (This is called a “store” operation.)

Interlude: Arrays Here's a declaration of a three-int array: int array[] = { 45, 67, 89 }; Note that we use the [] notation because we are declaring an array. int *array would be illegal here; the compiler would not accept us assigning the { 45, 67, 89 } initializer to it. This variable, array , is an extra-big box: three ints' worth of storage. One neat feature of C is that, in most places, when you use the name array again, you will actually be using a pointer to its first element (in C terms, &array[0] ). This is called “ decaying ”: the array “decays” to a pointer. Most usages of array are equivalent to if array had been declared as a pointer. There are, of course, cases that aren't equivalent. One is assigning to the name array by itself ( array = … )—that's illegal. Another is passing it to the sizeof operator. The result will be the total size of the array, not the size of a pointer (for example, sizeof(array) using the array above would evaluate to ( sizeof(int) = 4) × 3 = 12 on a current Mac OS X system). This illustrates that you are really handling an array and not merely a pointer. In most uses, however, array expressions work just the same as pointer expressions. So, for example, let's say you want to pass an array to printf . You can't: When you pass an array as an argument to a function, you really pass a pointer to the array's first element, because the array decays to a pointer. You can only give printf the pointer, not the whole array. (This is why printf has no way to print an array: It would need you to tell it the type of what's in the array and how many elements there are, and both the format string and the list of arguments would quickly get confusing.) Decaying is an implicit &; array == &array == &array[0] . In English, these expressions read “ array ”, “pointer to array ”, and “pointer to the first element of array ” (the subscript operator, [], has higher precedence than the address-of operator). But in C, all three expressions mean the same thing. (They would not all mean the same thing if “ array ” were actually a pointer variable, since the address of a pointer variable is different from the address inside it—thus, the middle expression, &array , would not be equal to the other two expressions. The three expressions are all equal only when array really is an array.)

Pointer arithmetic (or: why 1 == 4) Say we want to print out all three elements of array . int * array_ptr = array; printf(" first element: %i

", *(array_ptr++)); printf("second element: %i

", *(array_ptr++)); printf(" third element: %i

", *array_ptr); first element: 45 second element: 67 third element: 89 In case you're not familiar with the ++ operator: it adds 1 to a variable, the same as variable += 1 (remember that because we used the postfix expression array_ptr++ , rather than the prefix expression ++array_ptr , the expression evaluated to the value of array_ptr from before it was incremented rather than after). But what did we do with it here? Well, the type of a pointer matters. The type of the pointer here is int. When you add to or subtract from a pointer, the amount by which you do that is multiplied by the size of the type of the pointer. In the case of our three increments, each 1 that you added was multiplied by sizeof(int) . By the way, though sizeof(void) is illegal, void pointers are incremented or decremented by 1 byte. In case you're wondering about 1 == 4 : Remember that earlier, I mentioned that ints are four bytes on current Intel processors. So, on a machine with such a processor, adding 1 to or subtracting 1 from an int pointer changes it by four bytes. Hence, 1 == 4 . (Programmer humor.)

Indexing printf("%i

", array[0]); OK… what just happened? This happened: 45 Well, you probably figured that. But what does this have to do with pointers? This is another one of those secrets of C. The subscript operator (the [] in array[0] ) has nothing to do with arrays. Oh, sure, that's its most common usage. But remember that, in most contexts, arrays decay to pointers. This is one of them: That's a pointer you passed to that operator, not an array. As evidence, I submit: int array[] = { 45, 67, 89 }; int *array_ptr = &array[1]; printf("%i

", array_ptr[1]); 89 That one might bend the brain a little. Here's a diagram: array points to the first element of the array; array_ptr is set to &array[1] , so it points to the second element of the array. So array_ptr[1] is equivalent to array[2] ( array_ptr starts at the second element of the array, so the second element of array_ptr is the third element of the array). Also, you might notice that because the first element is sizeof(int) bytes wide (being an int), the second element is sizeof(int) bytes forward of the start of the array. You are correct: array[1] is equivalent to *(array + 1) . (Remember that the number added to or subtracted from a pointer is multiplied by the size of the pointer's type, so that “ 1 ” adds sizeof(int) bytes to the pointer value.)

Interlude: Structures and unions Two of the more interesting kinds of types in C are structures and unions. You create a structure type with the struct keyword, and a union type with the union keyword. The exact definitions of these types are beyond the scope of this article. Suffice to say that a declaration of a struct or union looks like this: struct foo { size_t size; char name[64]; int answer_to_ultimate_question; unsigned shoe_size; }; Each of those declarations inside the block is called a member. Unions have members too, but they're used differently. Accessing a member looks like this: struct foo my_foo; my_foo.size = sizeof(struct foo); The expression my_foo.size accesses the member size of my_foo . So what do you do if you have a pointer to a structure? (*foo_ptr).size = new_size; But there is a better way, specifically for this purpose: the pointer-to-member operator . foo_ptr->size = new_size; Unfortunately, it doesn't look as good with multiple indirection. (*foo_ptr_ptr)->size = new_size; (**foo_ptr_ptr).size = new_size; Rant: Pascal does this much better. Its dereference operator is a postfix ^: foo_ptr_ptr^^.size := new_size; (But putting aside this complaint, C is a much better language.)

Multiple indirection I want to explain multiple indirection a bit further. Consider the following code: int a = 3; int *b = &a; int **c = &b; int ***d = &c; Here are how the values of these pointers equate to each other: *d == c;

**d == *c == b;

***d == **c == *b == a == 3; Thus, the & operator can be thought of as adding asterisks (increasing pointer level, as I call it), and the *, ->, and [] operators as removing asterisks (decreasing pointer level).

Pointers and const The const keyword is used a bit differently when pointers are involved. These two declarations are equivalent: const int *ptr_a; int const *ptr_a; These two, however, are not equivalent: int const *ptr_a; int *const ptr_b; In the first example, the int (i.e. *ptr_a ) is const; you cannot do *ptr_a = 42 . In the second example, the pointer itself is const; you can change *ptr_b just fine, but you cannot change (using pointer arithmetic, e.g. ptr_b++ ) the pointer itself.

Function pointers Note: The syntax for all of this seems a bit exotic. It is. It confuses a lot of people, even C wizards. Bear with me. It's possible to take the address of a function, too. And, similarly to arrays, functions decay to pointers when their names are used. So if you wanted the address of, say, strcpy, you could say either strcpy or &strcpy . ( &strcpy[0] won't work for obvious reasons.) When you call a function, you use an operator called the function call operator . The function call operator takes a function pointer on its left side. In this example, we pass dst and src as the arguments on the interior, and strcpy as the function (that is, the function pointer) to be called: enum { str_length = 18U }; char src [str_length] = "This is a string.", dst [str_length]; strcpy(dst, src); There's a special syntax for declaring variables whose type is a function pointer. char *strcpy(char *dst, const char *src); char *(*strcpy_ptr)(char *dst, const char *src); strcpy_ptr = strcpy; strcpy_ptr = &strcpy; Note the parentheses around *strcpy_ptr in the above declaration. These separate the asterisk indicating return type (char *) from the asterisk indicating the pointer level of the variable ( *strcpy_ptr — one level, pointer to function). Also, just like in a regular function declaration, the parameter names are optional: char *(*strcpy_ptr_noparams)(char *, const char *) = strcpy_ptr; The type of the pointer to strcpy is char *(*)(char *, const char *); you may notice that this is the declaration from above, minus the variable name. You would use this in a cast. For example: strcpy_ptr = (char *(*)(char *dst, const char *src))my_strcpy; As you might expect, a pointer to a pointer to a function has two asterisks inside of the parentheses: char *(**strcpy_ptr_ptr)(char *, const char *) = &strcpy_ptr; We can have an array of function-pointers: char *(*strcpies[3])(char *, const char *) = { strcpy, strcpy, strcpy }; char *(*strcpies[])(char *, const char *) = { strcpy, strcpy, strcpy }; strcpies[0](dst, src); Here's a pathological declaration, taken from the C99 standard. “[This declaration] declares a function f with no parameters returning an int, a function fip with no parameter specification returning a pointer to an int, and a pointer pfi to a function with no parameter specification returning an int.” (6.7.5.3[16]) int f(void), *fip(), (*pfi)(); In other words, the above is equivalent to the following three declarations: int f(void); int *fip(); int (*pfi)(); But if you thought that was mind-bending, brace yourself… A function pointer can even be the return value of a function. This part is really mind-bending, so stretch your brain a bit so as not to risk injury. In order to explain this, I'm going to summarize all the declaration syntax you've learned so far. First, declaring a pointer variable: char *ptr; This declaration tells us the pointer type (char), pointer level ( * ), and variable name ( ptr ). And the latter two can go into parentheses: char (*ptr); What happens if we replace the variable name in the first declaration with a name followed by a set of parameters? char *strcpy(char *dst, const char *src); Huh. A function declaration. But we also removed the * indicating pointer level — remember that the * in this function declaration is part of the return type of the function. So if we add the pointer-level asterisk back (using the parentheses): char *(*strcpy_ptr)(char *dst, const char *src); A function pointer variable! But wait a minute. If this is a variable, and the first declaration was also a variable, can we not replace the variable name in THIS declaration with a name and a set of parameters? YES WE CAN! And the result is the declaration of a function that returns a function pointer: char *(*get_strcpy_ptr(void))(char *dst, const char *src); Remember that the type of a pointer to a function taking no arguments and returning int is int (*)(void). So the type returned by this function is char *(*)(char *, const char *) (with, again, the inner * indicating a pointer, and the outer * being part of the return type of the pointed-to function). You may remember that that is also the type of strcpy_ptr . So this function, which is called with no parameters, returns a pointer to a strcpy-like function: strcpy_ptr = get_strcpy_ptr(); Because function pointer syntax is so mind-bending, most developers use typedefs to abstract them: typedef char *(*strcpy_funcptr)(char *, const char *); strcpy_funcptr strcpy_ptr = strcpy; strcpy_funcptr get_strcpy_ptr(void);

Strings (and why there is no such thing) There is no string type in C. Now you have two questions: Why do I keep seeing references to “C strings” everywhere if there is no string type? What does this have to do with pointers? The truth is, the concept of a “C string” is imaginary (except for string literals). There is no string type. C strings are really just arrays of characters: char str[] = "I am the Walrus"; This array is 16 bytes in length: 15 characters for "I am the Walrus", plus a NUL (byte value 0) terminator. In other words, str[15] (the last element) is 0. This is how the end of the “string” is signaled. This idiom is the extent to which C has a string type. But that's all it is: an idiom. Except that it is supported by: the aforementioned string literal syntax

the string library The functions in string.h are for string manipulation. But how can that be, if there is no string type? Why, they work on pointers. Here's one possible implementation of the simple function strlen, which returns the length of a string (not including the NUL terminator): size_t strlen(const char *str) { size_t len = 0U; while(*(str++)) ++len; return len; } Note the use of pointer arithmetic and dereferencing. That's because, despite the function's name, there is no “string” here; there is merely a pointer to at least one character, the last one being 0. Here's another possible implementation: size_t strlen(const char *str) { size_t i; for(i = 0U; str[i]; ++i); return i; } That one uses indexing. Which, as we found out earlier, uses a pointer (not an array, and definitely not a string).































