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Pointers And Memory - Contents

Simple Reference Parameter Example


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    Swap(): The standard example of reference parameters is a Swap() function which exchanges the values of two ints. It's a simple function, but it does need to change the caller's memory which is the key feature of pass by reference.

    Swap() Function The values of interest for Swap() are two ints. Therefore, Swap() does not take ints as its parameters. It takes a pointers to int (int*)'s. In the body of Swap() the parameters, a and b, are dereferenced with * to get at the actual (int) values of interest.

void Swap(int* a, int* b) {
int temp;
temp = *a;
*a = *b;
*b = temp;

    Swap() Caller To call Swap(), the caller must pass pointers to the values of interest...

void SwapCaller() {
int x = 1;
int y = 2;
Swap(&x, &y); // Use & to pass pointers to the int values of interest
// (x and y).

    The parameters to Swap() are pointers to values of interest which are back in the caller's locals. The Swap() code can dereference the pointers to get back to the caller's memory to exchange the values. In this case, Swap() follows the pointers to exchange the values in the variables x and y back in SwapCaller(). Swap() will exchange any two ints given pointers to those two ints.

    Swap() With Arrays Just to demonstrate that the value of interest does not need to be a simple variable, here's a call to Swap() to exchange the first and last ints in an array. Swap() takes int*'s, but the ints can be anywhere. An int inside an array is still an int.

void SwapCaller2() {
int scores[10];
scores[0] = 1;
scores[9[ = 2;
Swap(&(scores[0]), &(scores[9]));// the ints of interest do not need to be
// simple variables -- they can be any int. The caller is responsible
// for computing a pointer to the int.

The above call to Swap() can be written equivalently as Swap(scores, scores+9)
due to the array syntax in C. You can ignore this case if it is not familiar to you it's
not an important area of the language and both forms compile to the exact same thing

    Is The & Always Necessary? When passing by reference, the caller does not always need to use & to compute a new pointer to the value of interest. Sometimes the caller already has a pointer to the value of interest, and so no new pointer computation is required. The pointer to the value of interest can be passed through unchanged.

For example, suppose B() is changed so it calls a C() function which adds 2 to the value
of interest...
// Takes the value of interest by reference and adds 2.
void C(int* worthRef) {
*worthRef = *worthRef + 2;
// Adds 1 to the value of interest, and calls C().
void B(int* worthRef) {
*worthRef = *worthRef + 1; // add 1 to value of interest as before
C(worthRef); // NOTE no & required. We already have
// a pointer to the value of interest, so
// it can be passed through directly.

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