practice concept:
An array can be considered a constant pointer
Assume we have a code fragment as below:
#include <stdio.h>
#include <stdlib.h>
/** This is the second practice of C/C++ pointer,
* practice concept:
* An array can be considered a constant pointer
*/
int main () {
// declare an int array that contains 5 elements
int num_arr[5];
// get the length of the array num
// we know it is '5', just show how we can get it
// if we don't know it
int length = sizeof(num_arr)/sizeof(num_arr[0]);
// the num_arr can be considered a constant pointer points the num[0]
int* num_arr_ptr = num_arr;
int i;
// assign values to num_arr by the pointer
// where *(num_arr_ptr+i) denotes
// 'the ith int element after the first element in num_arr'
for (i = 0; i < length; i++)
*(num_arr_ptr+i) = i*10 + i;
// print the elements of num_arr
for (i = 0; i < length; i++)
printf("%d\n", num_arr[i]);
printf("\n");
// print the elements of num_arr through pointer
for (i = 0; i < length; i++)
printf("%d\n", *(num_arr_ptr+i));
printf("\n");
// change the num_arr_ptr that points the
// 3rd element of num_arr and assign value
num_arr_ptr = &num_arr[2];
*num_arr_ptr = 55;
// change the num_arr_ptr that points the
// 4th element of num_arr and assign value
num_arr_ptr = num_arr+3;
*num_arr_ptr = 66;
// increase the num_arr_ptr so it points
// the 5th element and assign value
num_arr_ptr++;
*num_arr_ptr = 77;
// print the elements of num_arr again
for (i = 0; i < length; i++)
printf("%d\n", num_arr[i]);
printf("\n");
system("pause");
}
The result will be:
Download:
The file pointer_practice_002.c of this practice is available at github
https://github.com/benbai123/C_Cplusplus_Practice/tree/master/C_Pointer
Reference:
http://www.cplusplus.com/doc/tutorial/pointers/
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