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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.



Still have any doubt ? Drop a comment 👇

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;
}