в 14:21, 25 мая 2010

2010-05-25T14:21:19Z

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2010-05-25T10:32:46Z

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==Accessing an array using pointers==

<p>The array name is the pointer constant.</p>

<source lang="cpp">#include <stdio.h>
main(){
int a[5];
int i;
for(i = 0;i<5;i++){
a[i]=i;
}
int *b;

b=a;
for(i = 0;i<5;i++){
printf("value in array %d and address is %16lu\n",*b,b);
b=b+2;
}
}</source>
<pre class=codeResult>value in array 0 and address is 631652
value in array 2 and address is 631660
value in array 4 and address is 631668
value in array 631676 and address is 631676
value in array 42920 and address is 631684</pre>

==Address of second element in array (value of arraypointer+1)==

<source lang="cpp">#include <stdio.h>
int main(void)
{
char multiple[] = "another string";
printf("\nAddress of second element: %p", &multiple[1]);
printf("\nValue of multiple+1 : %p\n", multiple+1);
return 0;
}</source>
<pre class=codeResult>Address of second element: 9a372
Value of multiple+1 : 9a372</pre>

==Arrays and pointers: get address of an array==

<source lang="cpp">#include <stdio.h>
int main(void)
{
char multiple[] = "My string";
char *p = &multiple[0];
printf("\nThe address of the first array element : %p", p);
p = multiple;
printf("\nThe address obtained from the array name: %p\n", p);
return 0;
}</source>
<pre class=codeResult>The address of the first array element : 9a372
The address obtained from the array name: 9a372</pre>

==Arrays and pointers: get array value through array pointer==

<source lang="cpp">#include <stdio.h>
int main(void)
{
char multiple[] = "a string";
char *p = multiple;
int i;
for(i = 0 ; i<strlen(multiple) ; i++)
printf("\nmultiple[%d] = %c *(p+%d) = %c &multiple[%d] = %p p+%d = %p",
i, multiple[i], i, *(p+i), i, &multiple[i], i, p+i);
return 0;
}</source>
<pre class=codeResult>The address of the first array element : 9a372
The address obtained from the array name: 9a372</pre>

==Deal with array pointer of long integer==

<source lang="cpp">#include <stdio.h>
int main(void)
{
long multiple[] = {15L, 25L, 35L, 45L};
long * p = multiple;
int i;
for(i = 0 ; i<sizeof(multiple)/sizeof(multiple[0]) ; i++)
printf("\naddress p+%d (&multiple[%d]): %d *(p+%d) value: %d", i,i, p+i,i,*(p+i));
printf("\n Type long occupies: %d bytes\n", sizeof(long));
return 0;
}</source>
<pre class=codeResult>address p+0 (&multiple[0]): 631656 *(p+0) value: 15
address p+1 (&multiple[1]): 631660 *(p+1) value: 25
address p+2 (&multiple[2]): 631664 *(p+2) value: 35
address p+3 (&multiple[3]): 631668 *(p+3) value: 45
Type long occupies: 4 bytes</pre>

==Move array pointer to the next element==

<source lang="cpp">#include <stdio.h>
int main(void)
{
char multiple[] = "another string";
printf(" first element: %p\n", multiple);
printf("second element: %p\n", multiple + 1);
printf(" third element: %p\n", multiple + 2);
return 0;
}</source>
<pre class=codeResult>first element: 9a371
second element: 9a372
third element: 9a373</pre>

==The name of an array is the same as &array[ 0 ]==

<source lang="cpp">#include <stdio.h>
int main()
{
char array[ 5 ];
printf( " array = %p\n&array[0] = %p\n"
" &array = %p\n",
array, &array[ 0 ], &array );
return 0;
}</source>
<pre class=codeResult>array = 9a37b
&array[0] = 9a37b
&array = 9a37b</pre>

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