Lesson #21

Passing pointers and arrays into functions

Overview

In this lesson I will cover the following topics

1. Passing pointers to functions and using pointer parameters to lift some restrictions normally placed on functions.

2. Passing arrays to functions.

Body

Passing pointers into functions

Pointers are perfectly ordinary variables. They just happen to hold the address of something else. However, there is nothing unusual about the way you pass pointers into functions. Here is how it might look.

void some_function(int *p)
{
// I'm giving the address of some integer. I'll call that address p.

*p = 8;
// Here I'm modifying the integer using its address to find it.
}

int main(void)
{
int age;  // Your typical integer.

some_function(&age);
return 0;
}

This example is rather silly since it does nothing useful. However, it shows you how things work. The function some_function has been defined to take a pointer to integer as its one and only parameter. The name of the parameter is just p. In main I created an integer named age and then called some_function using age's address as the argument. When some_function executes it will store an 8 into the memory location pointed at by that address. In other words, it will store an 8 into age.

So what is the "point" (ha, ha) of this? There are several things to notice:

1. Normally functions can't modify their arguments. C uses only "call by value" to pass arguments to functions. That is still true. However, by sending the address of a variable to a function, the function is able to modify that variable. Thus there is, in effect, a way for functions to modify their arguments. The scanf function in the standard library uses this technique. It is very common. When you say

int age;

printf("What is your age? ");
scanf("%d", &age);

You are providing scanf with the address of the place where it is to deposit the value it scans. You give scanf the address of age and it will use that address to modify age before it returns. If you just did scanf("%d", age) there would be no way for scanf to modify the age variable since it would only be getting a copy of that variable.

2. Normally functions can only return a single value. This is still true. However, by using pointers you can effectively return several values. Here's how it might look

void some_function(int *p1, int *p2, int *p3)
{
// blah

*p1 = return_1;
*p2 = return_2;
*p3 = return_3;
}

int main(void)
{
int result1, result2, result3;

some_function(&result1, &result2, &result3);
return 0;
}

In this example some_function wants to return three different integers. Instead of just returning only one in the usual way, it requires that its caller provide three integer "place holders" to hold the results. The caller then must give some_function the addresses of those place holders so that some_function can fill them in with appropriate values. This technique is very common.

Other languages do these things differently. Some languages have several different ways of specifing parameters. Pascal has a concept of "VAR" parameters and Ada uses "IN OUT" parameters. Such parameters provide a value to the function and can be modified by the function. C does not have these features. It doesn't need them. C just uses pointers.

The classic swapping example

I really can't emphasize too much how function parameters work. I would be remiss in my duties as an instructor if I didn't show you the example of a swapping function.

Suppose you wanted to write a function that could swap two integers. At first you might try this

void swap(int A, int B)
{
int temp;

temp = A;
A = B;
B = temp;
}

The temporary integer is necessary. Without it you end up overwriting one of the values before you have used it.

Now you try to use this handy swap function like this

int main(void)
{
int top, bottom;

// Put something interesting in top and bottom.

swap(top, bottom);

// etc...
}

This does not work. Do you know why? If not think it over a bit...

Did you think about it?

Okay... remember that in C parameters are passed by value. That means the arguments are copied. In this case the values stored in top and bottom are copied and the copies are used to initialize A and B in the function. The function happily swaps A and B but the original values in top and bottom are not changed.

To do this correctly in C you have to use pointers.

void swap(int *A, int *B)
{
int temp;

temp = *A;
*A = *B;
*B = temp;
}

int main(void)
{
int top, bottom;

// etc...

swap(&top, &bottom);
// etc...
}

This is fine. First swap has been modified so that it takes two pointers to int instead of two integers. The temporary variable needs to still be an integer, however. I'm not trying to swap addresses! Inside swap I have to use the indirection operator a lot to get at the integers A and B are pointing at. That's annoying, but not a problem.

The calling function is very much like it was except that instead of passing top and bottom into swap it passes the address of top and the address of bottom. It is true that those addresses get copied, but so what? The swap function uses those addresses to reach back into the calling function and modify the variables that are stored there. Pretty cool, huh?

Passing arrays to functions

C does not allow you to pass an array to a function. This might seem surprising, but consider the implications of doing otherwise. Since C only passes copies of arguments to a function, passing an array would entail coping the entire array. Since arrays tend to be large copying the whole array would be very time consuming. Many functions don't really need the copy so the time spent creating it would be a waste of time.

However, you do often want to write functions that work on an array. If you can't pass an array to a function, how can you write such a function? The answer is to use pointers. C's handling of arrays is such that the pointer solution is very easy. Here is how it might look

void sort(int *array)
{
// Sort the elements of array.
}

int main(void)
{
int some_numbers;

// Load up some_numbers with values.

sort(some_numbers);
return 0;
}

Here I've defined a sorting function that accepts the address of a single integer as a parameter. However, the function will assume that address is actually the beginning of an array of integers. In function main I created an array of 100 integers. After loading the array with some (presummably unsorted) data I call sort on that array. When I call sort I just use the name of the array without an index. The compiler takes this to mean the address of the zeroth element of the array---just what the sort function is expecting.

You might be wondering how sort knows the size of the array. After all, sort is just getting a single address. There is nothing in that address to indicate the array's size. In fact, the sort function that I'm showing above does not know the size of the array! It would just have to assume some size and would thus only be able to sort arrays of exactly that size. That is a problem. A more realistic sort function might look like this

void sort(int *array, int size)
{
// Do the sorting.
}

int main(void)
{
int some_numbers;

// Load up some_numbers.

sort(some_numbers, 100);
return 0;
}

This sort function takes a second parameter that defines the size of the array. Now when sort is called, the size must be provided. This approach is actually quite flexible. What if I only wanted to sort the second half of the array? No problem:

sort(&some_numbers, 50);

I could sort the last quarter of the array like this

sort(&some_numbers, 25);

In these calls, I have to use the "address of" operator and the indexing operator (the square brackets). The cute trick with the array name only works if you want the address of the beginning of the array. It is very important that I provide the right count. Whatever happens, I don't want to tell sort to manipulate array elements that don't exist. For example, the following would be bad

sort(some_numbers, 101);

In this case I'm telling sort that there are 101 elements. In fact the array has only 100. Eventually as sort works it will get around to manipulating that non-existant array element and then undefined things will happen (core dump!). Something like this has the same problem

sort(&some_numbers, 26);

Do you see why?

Another thing to watch out for are errors like the following

int main(void)
{
int *array;

sort(array, 100);
return 0;
}

The compiler will not object to this. The sort function takes a pointer to integer and an integer as parameters and that is exactly what you are giving it. However, array is an uninitialized pointer. It contains an undefined address. Thus sort will attempt to sort 100 random integers at a random place in memory. Core dump! People write code like this fairly often. They see that sort expects a pointer to integer and they assume that they have to declare a pointer to integer in order to satisfy it. Yet what sort really wants is a pointer to the first integer in an array of integers. You need to declare an array, not a pointer to make sort happy.

Read the above paragraph again. The error I'm illustrating here is extremely common and even seasoned C programmers have been known to make it. When you see a function that expects a pointer parameter don't assume that you should declare a pointer for it. Many such functions are really designed to work on arrays.

Using an integer is one obvious way to specify the size of an array. Another way is to give sort a pointer to both the beginning and the end of the array. It might look like this

void sort(int *first, int *last)
{
// sort the array between first and last.
}

int main(void)
{
int array;

sort(array, array + 100);
return 0;
}

It turns out to be a lot easier for everyone if you define the array by using a pointer to the beginning and a pointer just past the end. The name array represents the address of the zeroth element and array + 100 is the address of the first non-existant element (array). The sort function needs to be written with the understanding that last is just past the end of the array, but that turns out to be easier anyway.

In any case, no matter how you do it, sort will need to be told how large the array is. The address of the beginning of the array is not enough information for sort to figure that out on its own. This is different than the way some other languages work. In many languages you can pass an array to a function and the size of the array is information that is "attached" to the array in some way. C is not like that.

Summary

1. You can pass a pointer to a function the same way you pass any other type of variable. For example,

void some_function(int *p)
{
...
}

This defines some_function to take a pointer to integer as a parameter. When you call some_function you either have to give it a pointer to integer or you can take the address of some suitable integer variable (depending on your needs). some_function will get a copy of the pointer but it can still access whatever original variable that pointer was pointing at. In this way some_function is able to, in effect, modify the argument you give it.

You can pass a pointer to a function as a way of getting around the restriction that functions only get copies of their arguments. You can also pass (several) pointers to a function to allow the function to effectively return several different values. The function would use the pointers to access "place holder" variables defined in the calling program.

2. You can't pass arrays to functions directly, but you can give a function a pointer to the first element of an array. When the function is called, you just pass it the array name without an index. Inside the function you can treat the pointer as a pointer, or you can apply an index to it again and use it as an array.

Since a pointer to the first element of an array does not tell you how large the array is, you have to arrange some other way of getting that information to your function. You can either pass the size as a separate parameter, or pass two pointers that define the start and end of the array.

© Copyright 2003 by Peter C. Chapin.
Last Revised: July 2, 2003