C Programming - Arrays - Discussion
Discussion Forum : Arrays - Find Output of Program (Q.No. 5)
5.
What will be the output of the program ?
#include<stdio.h>
int main()
{
static int arr[] = {0, 1, 2, 3, 4};
int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};
int **ptr=p;
ptr++;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
*ptr++;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
*++ptr;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
++*ptr;
printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
return 0;
}
Discussion:
38 comments Page 2 of 4.
HARI said:
9 years ago
How ptr points to 0? It should point to the address of first element in p.
Prashant said:
9 years ago
Hi, I am not getting this. Please, anyone explain me.
Jyoti said:
8 years ago
I don't understand.
Why takes p as 0, and then 0-array as 1?
Why takes p as 0, and then 0-array as 1?
Raji said:
8 years ago
Thank you @Leenaja.
Vishwa said:
8 years ago
#include<stdio.h>
int main()
{
int a[][2]={3,6,8,9};
printf("%d\n",a[1][-1]);
return 0;
}
I got output 6.
Can anyone explain this?
int main()
{
int a[][2]={3,6,8,9};
printf("%d\n",a[1][-1]);
return 0;
}
I got output 6.
Can anyone explain this?
Neeraj said:
8 years ago
What if the array arr is not declared as static?
Rohit said:
8 years ago
I have considered 100 as the base address.
We can write a[1][-1] as *(*(100+4)-2).it will evaluate to *102 which is nothing but the value 6.
We can write a[1][-1] as *(*(100+4)-2).it will evaluate to *102 which is nothing but the value 6.
Sweety said:
7 years ago
Hi, I didn't get it. Could you guys make it as simple as possible and explain?
Archana said:
7 years ago
What about that 'static' ?
Why are we using that static here? What can it do?
Without static in that line what will happen?
Can anyone answer me?
Why are we using that static here? What can it do?
Without static in that line what will happen?
Can anyone answer me?
Karthick said:
1 decade ago
Step 4 confused please help.
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