Untold Coding
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👍5
We should start imp gate questions quiz?
Anonymous Poll
85%
Yes
15%
No
👍21
Char ch = 21;
ch = ch<<3;
printf("%d",ch); What is the output ( ) //Ignore typing error
Anonymous Quiz
27%
34
24%
-88
14%
-68
35%
18
👍11👎21🥰1👏1
What is the output of this program?

#include <stdio.h>
int main() {
printf("%d", printf("ABCD"));
return 0;
}
👍13🤔10
Output of above program is ?
Anonymous Quiz
24%
ABCD4
14%
ABCD 4
21%
4
40%
ABCD
👍81
Untold Coding
Char ch = 21;
ch = ch<<3;
printf("%d",ch); What is the output ( ) //Ignore typing error
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
👍11
#include <stdio.h>
int main() {
int i = -1;
for (++i; i++; i++)
printf("Untoldcoding");
return 0;
}
👍12🔥1
How many time loop will executed ?
Anonymous Quiz
26%
0
15%
1
53%
Infinity
6%
2
👍14🥰21
Untold Coding
#include <stdio.h> int main() { int i = -1; for (++i; i++; i++) printf("Untoldcoding"); return 0; }
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 👇
👍13
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;
}
👍12
How many number of times printf statement executed?
Anonymous Quiz
35%
5
22%
9
33%
10
10%
11
👍101👏1
Untold Coding
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); …
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

Still have any confusion, drop a comment 👇
👍81
How many quiz needed in a day ?
Anonymous Poll
22%
1
78%
2
👍4
#include <stdio.h>

int main()

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

return 0;
}
👍141
How many times printf will executed?
Anonymous Quiz
34%
4
34%
5
20%
6
12%
11
👍15
👍7