The first things to do with pointers are to declare a pointer variable, set it to point somewhere, and finally manipulate the value that it points to. A simple pointer declaration looks like this:
int *ip;This declaration looks like our earlier declarations, with one obvious difference: that asterisk. The asterisk means that ip, the variable we're declaring, is not of typeint, but rather of type pointer-to-int. (Another way of looking at it is that *ip, which as we'll see is the value pointed to by ip, will be an int.)
We may think of setting a pointer variable to point to another variable as a two-step process: first we generate a pointer to that other variable, then we assign this new pointer to the pointer variable. We can say (but we have to be careful when we're saying it) that a pointer variable has a value, and that its value is ``pointer to that other variable''. This will make more sense when we see how to generate pointer values.
Pointers (that is, pointer values) are generated with the ``address-of'' operator &, which we can also think of as the ``pointer-to'' operator. We demonstrate this by declaring (and initializing) an int variable i, and then setting ip to point to it:
int i = 5; ip = &i;The assignment expression ip = &i; contains both parts of the ``two-step process'': &i generates a pointer to i, and the assignment operator assigns the new pointer to (that is, places it ``in'') the variable ip. Now ip ``points to'' i, which we can illustrate with this picture:
i is a variable of type int, so the value in its box is a number, 5. ip is a variable of type pointer-to-int, so the ``value'' in its box is an arrow pointing at another box. Referring once again back to the ``two-step process'' for setting a pointer variable: the & operator draws us the arrowhead pointing at i's box, and the assignment operator =, with the pointer variable ip on its left, anchors the other end of the arrow in ip's box.
We discover the value pointed to by a pointer using the ``contents-of'' operator, *. Placed in front of a pointer, the * operator accesses the value pointed to by that pointer. In other words, if ip is a pointer, then the expression *ip gives us whatever it is that's in the variable or location pointed to by ip. For example, we could write something like
printf("%d\n", *ip);
which would print 5, since ip points to i, and i is (at the moment) 5.(You may wonder how the asterisk * can be the pointer contents-of operator when it is also the multiplication operator. There is no ambiguity here: it is the multiplication operator when it sits between two variables, and it is the contents-of operator when it sits in front of a single variable. The situation is analogous to the minus sign: between two variables or expressions it's the subtraction operator, but in front of a single operator or expression it's the negation operator. Technical terms you may hear for these distinct roles are unary and binary: a binary operator applies to two operands, usually on either side of it, while a unary operator applies to a single operand.)
The contents-of operator * does not merely fetch values through pointers; it can also set values through pointers. We can write something like
*ip = 7;which means ``set whatever ip points to to 7.'' Again, the * tells us to go to the location pointed to by ip, but this time, the location isn't the one to fetch from--we're on the left-hand sign of an assignment operator, so *ip tells us the location to store to. (The situation is no different from array subscripting expressions such as a[3]which we've already seen appearing on both sides of assignments.)
The result of the assignment *ip = 7 is that i's value is changed to 7, and the picture changes to:
If we called printf("%d\n", *ip) again, it would now print 7.
At this point, you may be wondering why we're going through this rigamarole--if we wanted to set i to 7, why didn't we do it directly? We'll begin to explore that next, but first let's notice the difference between changing a pointer (that is, changing what variable it points to) and changing the value at the location it points to. When we wrote *ip = 7, we changed the value pointed to by ip, but if we declare another variable j:
int j = 3;and write
ip = &j;we've changed ip itself. The picture now looks like this:
We have to be careful when we say that a pointer assignment changes ``what the pointer points to.'' Our earlier assignment
*ip = 7;changed the value pointed to by ip, but this more recent assignment
ip = &j;has changed what variable ip points to. It's true that ``what ip points to'' has changed, but this time, it has changed for a different reason. Neither i (which is still 7) nor j (which is still 3) has changed. (What has changed is ip's value.) If we again call
printf("%d\n", *ip);
this time it will print 3.We can also assign pointer values to other pointer variables. If we declare a second pointer variable:
int *ip2;then we can say
ip2 = ip;Now ip2 points where ip does; we've essentially made a ``copy'' of the arrow:
Now, if we set ip to point back to i again:
ip = &i;the two arrows point to different places:
We can now see that the two assignments
ip2 = ip;and
*ip2 = *ip;do two very different things. The first would make ip2 again point to where ip points (in other words, back to i again). The second would store, at the location pointed to by ip2, a copy of the value pointed to by ip; in other words (if ip and ip2 still point to i and j respectively) it would set j to i's value, or 7.
It's important to keep very clear in your mind the distinction between a pointer and what it points to. The two are like apples and oranges (or perhaps oil and water); you can't mix them. You can't ``set ip to 5'' by writing something like
ip = 5; /* WRONG */5 is an integer, but ip is a pointer. You probably wanted to ``set the value pointed to by ip to 5,'' which you express by writing
*ip = 5;Similarly, you can't ``see what ip is'' by writing
printf("%d\n", ip); /* WRONG */
Again, ip is a pointer-to-int, but %d expects an int. To print what ip points to, useprintf("%d\n", *ip);
Finally, a few more notes about pointer declarations. The * in a pointer declaration is related to, but different from, the contents-of operator *. After we declare a pointer variable
int *ip;the expression
ip = &isets what ip points to (that is, which location it points to), while the expression
*ip = 5sets the value of the location pointed to by ip. On the other hand, if we declare a pointer variable and include an initializer:
int *ip3 = &i;we're setting the initial value for ip3, which is where ip3 will point, so that initial value is a pointer. (In other words, the * in the declaration int *ip3 = &i; is not the contents-of operator, it's the indicator that ip3 is a pointer.)
If you have a pointer declaration containing an initialization, and you ever have occasion to break it up into a simple declaration and a conventional assignment, do it like this:
int *ip3; ip3 = &i;Don't write
int *ip3; *ip3 = &i;or you'll be trying to mix oil and water again.
Also, when we write
int *ip;although the asterisk affects ip's type, it goes with the identifier name ip, not with the type int on the left. To declare two pointers at once, the declaration looks like
int *ip1, *ip2;Some people write pointer declarations like this:
int* ip;This works for one pointer, because C essentially ignores whitespace. But if you ever write
int* ip1, ip2; /* PROBABLY WRONG */it will declare one pointer-to-int ip1 and one plain int ip2, which is probably not what you meant.
What is all of this good for? If it was just for changing variables like i from 5 to 7, it would not be good for much. What it's good for, among other things, is when for various reasons we don't know exactly which variable we want to change, just like the bank didn't know exactly which club member it wanted to send the statement to.
