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C Interview Questions And Answers - Part 3.

What is the meaning of NULL?
Answer: NULL is a macro constant which has been defined in several header files such as stdio.h, alloc.h, mem.h, stddef.h and stdlib.h as
#define NULL 0

Example 1: What is the output of following c program?
#include "stdio.h"
int main()`{
if(!NULL)
printf("I know preprocessor");
else
printf("I don't know preprocessor");
}
Output: I know preprocessor
Explanation: !NULL = !0 = 1, i.e. any non-zero number means true.

Example 2:  What is the output of following c program?
#include "stdio.h"
int main() {
int i;
static int count;
for(i=NULL;i<=5;)
{
count++;
i+=2;
}
printf("%d",count);
}
Output: 3 

What is the output if you execute following c code?
#include<stdio.h>
int main() {
int i;
for(i=0;i<5;i++) {
int i=10;
printf(" %d",i);
i++;
}
return 0;
}
Output: 10 10 10 10 10

Explanation: The default storage class of local variable is auto. Scope of auto variables are block in which it has been declared, i.e. they are dead out of the scope. So variable i which has been declared inside “for” loop has scope only within it and after each iteration, variable i is dead and re-initialized.

Note: If we have declared two variables of same name but different scope then local variable will have higher priority.

What is the output if you execute following c code?
#include<stdio.h>
int main() {
int arr[3] = {10,20,30};
int x=0;
x = ++arr[++x] + ++x + arr[--x];
printf("%d ",x);
return 0;
}
Output: 44

Explanation:
= ++arr[++x] + ++x + arr[--x] //x = 0 + 1
= ++arr[1] + ++x + arr[--x] //x = 1 + 1
= ++arr[++x] + 2 + arr[--x] //x = 2 - 1
= ++arr[1] + 2 + arr[1] //arr[1] = 20+1
= arr[1] + 1 + arr[1] //arr[1] = 21
= 21 + 2 + 21
= 44

What is the output if you execute following c program?
#include<stdio.h>
int main() {
int a[]={10,20,30,40};
int i=3,x;
x=1*a[--i]+2*a[--i]+3*a[--i];
printf("%d",x);
return 0;
}
Output: 90 or 100 (depends on compiler)

Explanation:
= 1 * a[--i] + 2 * a[--i] + 3 * a[--i] //i = 3 - 2
= 1 * a[--i] + 2 * a[--i] + 3 * a[--i] //i = 2 - 1
= 1 * a[1] + 2 * a[1] + 3 * a[--i] //i = 1 - 1
= 1 * a[1] + 2 * a[1] + 3 * a[0]
= 1 * 20 + 2 * 20 + 3 * 10
= 20 + 40 + 30
= 90

What is the output if you execute following c code?
#include<stdio.h>
int f(int);
int main() {
int i=3, val;
val=sizeof (f(i)+ +f(i=1)+ +f(i-1));
printf("%d %d",val,i);
return 0;
}
int f(int num) {
return num*5;
}
Output: 4 3

Explanation: Any expression inside “sizeof” operator never changes the value of the any variable. Thus the value of variable i will remain 3. After the evaluation of expression inside “sizeof” operator, we will get an integer value. So the value of variable val will be size of “int” data type.

Note: Size of “int” in Linux gcc complier is four byte.

What is the output if you execute following c code?
#include<stdio.h>
int main() {
int x, a=3;
x=+ +a+ + +a+ + +5;
printf("%d  %d",x,a);
return 0;
}
Output: 11 3

Explanation: Consider this expression + +a
Here both + are Unary plus operation. So
= + +a+ + +a+ + +5;
= + +3+ + +3+ + 5
= 3+ 3+ 5
= 11

Note: Increment operator ++ cannot have space between two plus symbol.

What is the output if you execute following c code?
#include<stdio.h>
int main() {
int num, a=15;
num=- - - -a--;
printf("%d  %d",num,a);
return 0;
}
Output: 15 14

What is the output if you execute following c code?
#include<stdio.h>
int sq(int);
int main() {
int a=1, x;
x=sq(++a)+sq(a++)+sq(a++);
printf("%d",x);
return 0;
}
int sq(int num) {
return num*num;
}
Output: 17

Explanation:
= sq(++a) + sq(a++) + sq(a++) //a = 1 + 1
= sq(2) + sq(2) + sq(a++) //a = 2 + 1
= sq(2) + sq(2) + sq(3) //a = 3 + 1
= 4 + 4 + 9
= 17

Note: Pre-increment first increment, then assign while post increment operator first assign then increment.

What will be output if you will execute following c code?
#include<stdio.h>
int main() {
printf("%c",*"abcde");
return 0;
}
Output: a

Explanation: String constant "abcde" returns memory address of first character of the string constant. *"abcde" returns the first character of string constant.

How does increment and decrement operators work in C programming?
Answer: Increment and decrement operators are used to increase and decrease the value of the variable by one in C programs.

Increment operator:  ++i;    i++;
Decrement operator:  --i;    i--;

Example 1: In this program, value of “i” is incremented one by one from 1 up to 9 using “i++” operator.
#include <stdio.h>
int main() {
int i=1;
while(i<10) {
printf("%d ",i);
i++;
}}
Output: 1 2 3 4 5 6 7 8 9

Example 2: In this program, value of “i” is decremented one by one from 20 up to 11 using “i--” operator.
#include <stdio.h>
int main() {
int i=20;
while(i>10) {
printf("%d ",i);
i--;
}}
Output: 20 19 18 17 16 15 14 13 12 11

Difference between pre and post increment and decrement operators:
Pre-increment
++i

value of i is incremented before assigning it to variable i.
Post–increment
i++
value of i is incremented after assigning it to variable i.
Pre-decrement
--i
value of i is decremented before assigning it to variable i.
Post-decrement
i--
value of i is decremented after assigning it to variable i.

Example 3:
#include <stdio.h>
int main() {
int i=0;
while(++i < 5 ) {
printf("%d ",i);
}
return 0;
}
Output: 1 2 3 4

Explanation: In above program, value of “i” is incremented from 0 to 1 using pre-increment operator. This incremented value “1” is compared with 5 in while expression. Then, this incremented value “1” is assigned to the variable “i”. These 3 steps are continued until while expression becomes false and output is displayed as “1 2 3 4”.

Example 4:
#include <stdio.h>
int main() {
int i=0;
while(i++ < 5 ) {
printf("%d ",i);
}
return 0;
}
Output: 1 2 3 4 5

Explanation: In this program, value of i “0” is compared with 5 in while expression. Then, value of “i” is incremented from 0 to 1 using post-increment operator. Then, this incremented value “1” is assigned to the variable “i”. These 3 steps are continued until while expression becomes false and output is displayed as “1 2 3 4 5”.

Example 5:
#include <stdio.h>
int main() {
int i=10;
while(--i > 5 ) {
printf("%d ",i);
}
return 0;
}
Output: 9 8 7 6

Explanation: In above program, value of “i” is decremented from 10 to 9 using pre-decrement operator. This decremented value “9” is compared with 5 in while expression. Then, this decremented value “9” is assigned to the variable “i”. These 3 steps are continued until while expression becomes false and output is displayed as “9 8 7 6”.

Example 6:
#include <stdio.h>
int main() {
int i=10;
while(i-- > 5 ) {
printf("%d ",i);
}
return 0;
}
Output: 9 8 7 6 5
Explanation: In this program, value of i “10” is compared with 5 in while expression. Then, value of “i” is decremented from 10 to 9 using post-decrement operator. Then, this decremented value “9” is assigned to the variable “i”. These 3 steps are continued until while expression becomes false and output is displayed as “9 8 7 6 5”.

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