C Programming - Pointers - Discussion
Discussion Forum : Pointers - Find Output of Program (Q.No. 13)
13.
What will be the output of the program?
#include<stdio.h>
int main()
{
int arr[3] = {2, 3, 4};
char *p;
p = arr;
p = (char*)((int*)(p));
printf("%d, ", *p);
p = (int*)(p+1);
printf("%d", *p);
return 0;
}
Discussion:
87 comments Page 3 of 9.
Ankita said:
1 decade ago
What is this endian method?
Raghav said:
1 decade ago
(char*)((int*)(p))
Could someone explain this?
Could someone explain this?
Garvita said:
1 decade ago
Why we type casted p again to char?
p=(char*)((int*)(p));
p=(char*)((int*)(p));
Sanjoy said:
1 decade ago
What is type casted?
Vaibhav said:
1 decade ago
p = (int*)(p+1) ; what it does actually ?
Shivam varshney said:
8 years ago
The answer is wrong here.
The compiler will produce an error.
The compiler will produce an error.
Abdul said:
4 years ago
Well done @Mayur.
Akash said:
5 years ago
All droughts are cleared, Thanks for explaining @Puru.
Pushparaj said:
5 years ago
Just, (p+1) is NULL. (p+2) is NULL. (p+2) is NULL. (p+4) is NULL. etc
Otherwise, p=(int*)(p+1) is NULL or ZERO.
Otherwise, p=(int*)(p+1) is NULL or ZERO.
Shivani Mundra said:
5 years ago
It will give a warning instead of error hence p=arr is illegal but this code will compile and give output.
But my question is how to know that these integers will be stored in little-endian firm or big-endian?
Please, anyone, help me.
But my question is how to know that these integers will be stored in little-endian firm or big-endian?
Please, anyone, help me.
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