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The value of a is 7

The value of *aPtr is 7

Notice that the address of a and the value of aPtr are identical in the output, confirming that the address of a is assigned to the pointer variable aPtr.

Calling functions by reference with pointer arguments

In C++, programmers can use pointers and the dereference operator to simulate call-by-reference. When calling a function with arguments should be modified, the addresses of the arguments are passed. This is normally achieved by applying the address-of operator (&) to the name of the variable whose value will be used. A function receiving an address as an argument must define a pointer parameter to receive the address.

Example

// Cube a variable using call-by-reference

// with a pointer argument

#include<iostream.h>

void cubeByReference( int * ); // prototype

int main()

{

int number = 5;

cout<<"The original value of number is "<<number;

cubeByReference(&number );

cout<<"\nThe new value of number is "<<number<<endl;

return 0;

}

void cubeByReference( int *nPtr )

{

*nPtr = (*nPtr) * (*nPtr) * (*nPtr); // cube number in main

}

The output of the above propgram:

The original value of number is 5

The new value of number is 125

Pointers and arrays

Notice that the name of an array by itself is equivalent to the base address of that array. That is, the name z in isolation is equivalent to the expression&z[0].

Example

#include<iostream.h>

int main()

{

int z[] = { 1, 2, 3, 4, 5};

cout<<“The value return by ‘z’ itself is

the addr “<<z<<endl;

cout<<“The address of the 0th element of

z is “<<&z[0]<<endl;

return 0;

}

The output of the above program:

The value return by ‘z’ itself is the addr 0x0065FDF4

The address of the 0th element of z is 0x0065FDF4

Accessing Array Element Using Pointer and Offset

Now, let us store the address of array element 0 in a pointer. Then using the indirection operator, *, we can use the address in the pointer to access each array element.

For example, if we store the address of grade[0] into a pointer named gPtr, then the expression *gPtr refers to grade[0].

One unique feature of pointers is that offset may be included in pointer expression.

For example, the expression *(gPtr + 3) refers to the variable that is three (elements) beyond the variable pointed to by gPtr.

The number 3 in the pointer expression is an offset. So gPtr + 3 points to the element grade[3] of the grade array.

Example

#include<iostream.h>

int main()

{

int b[] = { 10, 20, 30, 40 }, i, offset;

int *bPtr = b; // set bPtr to point to array b

cout<<"Array b printed with:\n"

<<"Array subscript notation\n";

for ( i = 0; i<4; i++ )

cout<<"b["<<i<<"] = "<<b[ i ]<<'\n';

cout<<"\nPointer/offset notation\n";

for ( offset = 0; offset<4; offset++ )

cout<<"*(bPtr + "<<offset<<") = "

<<*( bPtr + offset )<<'\n';

return 0;

}

The output of the above program is:

Array b printed with:

Array subscript notation

b[0] = 10

b[1] = 20

b[2] = 30

b[3] = 40

Pointer/offset notation

*(bPtr + 0) = 10

*(bPtr + 1) = 20

*(bPtr + 2) = 30

*(bPtr + 3) = 40

Pointers and strings

In C++ we often use character arrays to represent strings. A string is an array of characters ending in a null character (‘\0’). Therefore, we can scan through a string by using a pointer. Thus, in C++, it is appropriate to say that a string is a constant pointer – a pointer to the string’s first character.

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Source:  OpenStax, Programming fundamentals in c++. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10788/1.1
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