What is the output the given C language code snippet? #include<stdio.h> int main() { int x[] = {10,20,30}; int *p = x; p++; printf("%d",*p); return 0; }
20
This snippet demonstrates pointer arithmetic on an array. In C, the name of an array (here x) decays to the address of its first element, and adding to a pointer moves it by whole elements, not by single bytes.
Tracing the execution:
int x[] = {10, 20, 30};— three integers are stored contiguously, at indices 0, 1 and 2.int *p = x;—pis set to the address ofx[0], so*pwould be 10.p++;— this is the key step. Incrementing anintpointer advances it bysizeof(int)bytes, i.e. by exactly one array element. Sopnow points tox[1].printf("%d", *p);— dereferencing givesx[1], which is 20.
Why the other outputs are wrong: 10 would be printed only if p had not been incremented (still pointing at x[0]). 30 would need two increments to reach x[2], but p++ advances just one element. A garbage value would arise only if p were moved past the end of the array (out of bounds); here it still points to a valid element, so the result is the well-defined value 20.
Important Questions from Array
Which of the following expression will delete the entire array pointed to by q?
Consider the following C program segment.
#include <stdio.h>
int main()
{
char s1[7] = "1234";
char *p;p = s1 + 2;
*p = '0';printf("%s", s1);
return 0;
What will be printed by the program?
}#include <stdio.h> void foo(int *p, int x){ *p = x; } int main(){ int *z; int a = 20, b = 25; z = &a; foo(z, b); printf("%d", a); return 0; }The output of the given C program is __________. (Answer in integer)- Consider an array $A$ of integers of size $n$. The indices of $A$ run from 1 to $n$. An algorithm is to be designed to check whether $A$ satisfies the condition given below.
$\forall i,j \in \{1, \dots, n-1\}$ such that $i > j$, $(A[i + 1] - A[i]) > (A[j + 1] - A[j])$
Which one of the following gives the worst case time complexity of the fastest algorithm that can be designed for the problem?
