Question
#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)
#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 given C program demonstrates the use of pointers and function calls to modify values. Here's a step-by-step explanation:
1. The function
2. In
3. A pointer
4. The function
5. The
6. Since
Final Output: 25
1. The function
foo(int *p, int x) takes a pointer to an integer p and an integer x. It sets the value pointed to by p to x. This operation means that the integer at the memory location p is updated to x.2. In
main(), two int variables a and b are initialized with values 20 and 25 respectively.3. A pointer
z is assigned the address of a (z = &a;), making z point to a.4. The function
foo is called with arguments z and b. Since z points to a, calling foo(z, b) effectively sets a to the value of b, which is 25.5. The
printf statement outputs the value of a.6. Since
a is modified to 25, the output of the program is 25, which falls within the specified range (25, 25).Final Output: 25
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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?
}- 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? 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;
}
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