What is the output of this program?

#include <stdio.h>
int main() {
printf("%d", printf("ABCD"));
return 0;
}

In this question we use formal a×2^k and
-(2^n -k) here k is the number and n is the no. of bit

Therefore
We have n = 8 , (char size)
K = 21

- ( 2⁸ -168) = - (256-168)
= -88

#include <stdio.h>
int main() {
int i = -1;
for (++i; i++; i++)
printf("Untoldcoding");
return 0;
}

Think and give answer ... In quiz

Explanation

1. Initialization : The loop starts with ++i, which increments i from -1 to 0.

2. Condition : The condition i++ is checked next. Since i++ returns the value of i before incrementing, the loop checks if 0 is true, which it is not (0 is considered false in C).

3. Increment: If the condition were true, the loop would proceed to increment i with i++ again, but this part doesn't get executed in this case because the condition i++ evaluates to false on the first check.

Due to the condition i++ being false when i is 0, the loop doesn't execute even once. Therefore, "Untoldcoding" is never printed.



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Consider the following C program:

#include <stdio.h>
int main( )
{
int i, j, k = 0;

j=2*3/4+2.0/5+8/5;

k -= --j;

for (i=0; i<5; i++)
{
switch(i+k)
{
case 1:

case 2: printf("\n%d", i+k);

case 3: printf("\n%d", i+k)

default: printf("\n%d", i+k);

}

}
return 0;
}

Explanation

Total printf Calls

Summing up all the outputs from each iteration:

Iteration 1: 1 time
Iteration 2: 1 time
Iteration 3: 3 times
Iteration 4: 3 times
Iteration 5: 2 times

Total number of printf calls: 1 + 1 + 3 + 3 + 2 = 10

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#include <stdio.h>

int main()

{
int i;
for(i=1; i<=10; i++){
if(i>5){
break;
}
}
printf("%d",i);

return 0;
}

Consider the C Programming

#include<stdio.h>

void print (int n) {
if (n <= 0) return;
print(n--);
printf ("%d", n);

}

int main() {
print(5);
return 0;

}

Consider the following sample of numbers: 9, 18, 11, 14, 15, 17, 10, 69, 11, 13 The median of the sample is

(a) 14

(b) 13.5

(c) 12

(d) 11