Hi Friends, I am new to C. I am trying to learn it, I got stuck some where in arrays. Please check following program
#include <stdio.h>
#include <stdlib.h>
int arr1[] = {10,20,30,40,50};
int arr2[] = {5,15,25,35,45};
int *main_arr[] = {arr1,arr2};
int main()
{
printf("in first array 0th locatin value is: %d\n",*main_arr[0]);
system("PAUSE");
return 0;
}
By using printf i can print the value at 0th location, but not getting how to access rest of the element ...please help me!
You want
...: %d\n",(main_arr[0])[0]);
------------- ->arr1
--- ->arr1[0]
main_arr is pointing to both arrays arr1, arr2. So main_arr[0] points to the first element of the first array. To access other elements modify the 2nd [0].
The other alternative, uglier but closer to your current code, is to use pointer arithmetic.
...s: %d\n",*(main_arr[0]+1));
Remember that arr[1] is the same as *(arr+1).
#include <stdio.h>
#include <stdlib.h>
int arr1[] = {10,20,30,40,50};
int arr2[] = {5,15,25,35,45};
int *main_arr[] = {arr1,arr2};
int main()
{
int iter1, iter2;
for(iter1 = 0; iter1 < 2; iter1++){
for(iter2 = 0; iter2 < 5; iter2++){
printf("in first array nth locatin value is: %d\n",(main_arr[iter1][iter2]));
}
}
system("PAUSE");
return 0;
}
I guess the code is simple enough to be understood?
There are only two pointers in the main_arr, pointing to address of arr1 and arr2.
main_arr| ptr to an array | -> arr1
| ptr to an array | -> arr2
So you can use main_arr[0][1] to access the second element of arr1, because main_arr[0] points to the address of the very first element of arr1.
You should have know that in C, if p is a pointer, then both p[3] and 3[p] will evaluate to *(p + 3 * sizeof(type)), so let's assume p = main_arr[0], then p[1], which is main_arr[0][1], will evaluate to *(main_arr[0] + 1 * sizeof(int)), which is the same value with arr1[1].
