The way you're teaching, presenting problems to us, and claiming for answers in the comment box is quite convenient to grasp the concepts and make us feel the classroom environment.
Output will be 1 and b will be 7...Also you are best Ma'am..Today is my exam and I was so tense now my mood is good too after watching your video...Your teaching is best☺
dear mam thank you will be very less for your videos , praying for all happiness n success to u n to ur family , i m benefited a lot by your videos , you are an excellent teacher
No brother the precedence changes and 0 && b++ will be evaluated first and then 1 || (value of 0 && b++) …so output will be 1 but again b value will not be affected
in the last case after adding && b++ and printing b only gives value of 6 so that means it doesnt execute the expression && b++ as per left presidency and OR operator before on left
@@SaraswotiKhadka at last mam changed the question i.e ( a&b && b+1 || 0 || b++) first mam explain with this example later, The question is ( a&b && b+1 || 0 && b++) bitwise & has higher precedence so, now the expression on solving ( 2 & 7 || 0 && b++) so, 2&7 = 1 So expression changes as follows (1 || 0 && b++) here && has higher precedence and also b is updated to 7 here as we used b++ in our expression. so, 0 && b++ is done first i.e 0&&6 output is 0, now ( 1 || (0 && 6)) BECOME (1 || 0) Finally output of this expression is one bcoz of logical or.
@@hariparuchuru3858 bro, in logical and (&&) , if 0 is one of the operand then we don't fetch the second operand and the result is 0 , as in this 0 && 6 output is 0 and we don't consider the second operand So b++ value remains same i.e. 6
@@AnkitKumar-ud4vt @Neeraja Neeraja Because at second last evaluation of expression, first && operator will be evaluated then || operator will be evaluated and there is 0 as first operand in && operator then there is no need to check for the second operand and consequently b value will not be incremented. I hope you will get it, what I tried to explain. 🙂
Technically you r correct and you are technically sound. Don't get nervous or fast while teaching. But teaching is good. But you can make it better because you are technically sound.
mam in the logical AND(&&) operator the final output will be 1 and 7 right. I didn't understood how did we got 2 there. Can you please clear this doubt mam?
Good day Ma'am. 2 years later and your course is still very much relevant! Thank you for what you do. So I have a question, when performing certain operations on while working with binary in jpeg mode, we usually have to check if the first four digits of the 4th byte of a specific amount of data is a certain value. Here, we use the bitwise & operator. My question is, when checking to see if the first four values are similar to 0xe0, why do we use the & operator with 0xf0 on one side and 0xe0 on the other? Why don't we have 0xe0 on both sides? An example code is as shown below: if (buffer[0] == 0xff && buffer[1] == 0xd8 && buffer[2] == 0xff && (buffer[3] & 0xf0) == 0xe0) Look forward to hearing from you soon ma.
Thank you for teaching us in an excellent manner you are extremely better than my college faculty they don't taught me all this so that I'm stuck with lot of doubts So because of you I'm able to i understand everything thank a lot mam 💟💟💟💟💟💟💟💟💟💟💟💟💟
hi mam, nice explanation , in this video , which first prior , & ,+ , pervious video you explain high priorty each by binary.but while combination how we can identity which first for example & ,+ .
I didn't understood that why the output of b would be 6 and not 7, as precedence of increment operator is more than other operators, so logically the increment operator should be executed first and the value of b must be 7 in both && case and || case.
@@vrajalingam3487 I got my answer. See precedence is checked only when a operant(or any number) has more than one operator, and not checked at first for whole statement. And in case of || only first part is checked and if that part returns true then other part is not check., same in case of &&, if first part returns false, other part is not checked.
@@lokesh6828bro l also know that even everyone but here the main problem is when we r doin b+1 i.e 6+1=7, then value of b is updated here only & in b++ we have to consider 7 & not 6 (coz the value of b is updated in the memory after doing b+1 ) i.e why when we'll print 'b' we should get 7 but still we r getting 6 . How's that possible, l can't understand
@@Helly04 Because b++ and b+1 is not the same thing, in case of b+1 the value doesn't change in the memory it just changes where it is assigned to. So if b=6, c=b+1 then c will become 7 but b will remain 6 in the question, && has higher precedence so, after initial solving the question comes to this, 1 || 0 && 6, here && will be evaluated first due to higher precedence, in logical and if first operand is 0, second one isn't evaluated final output becomes 0. So now it is 1||0, in logical or if first operand is 1, final output becomes 1, second operand isn't evaluated, so the second operand b++ was never checked in any of the steps, that's why the value of b remains 6 and the final result is 1
NOTE: Precedence will come into picture only if expressions will be evaluated. In case of && or || its possible that some expressions may not be evaluated at all because we obtain 0 or 1 respectively. Hence, no matter how much higher precedence operators are there in those expressions, those expressions would never be evaluated
b=6 Even if && is there, it didn't get incremented while executing #include int main() { int a=10,b=6,c; c=a&b&&b+1||0&&b++; printf("%d %d %d",a,b,c); return 0; } output==> 10 6 1
@@syedalifathimaa9618 because logical && operator precident higher then logic ll operator C = a & b && b+1 || 0 && b++; C = 10 & 6 && 6+1 || 0 && b++; C = 10 & 6 && 7 || 0 && b++; C = 10 & 6 = 2 && 7 || 0 && b++; C = 2 && 7 = 1 || 0 && b++; C = 1 || 0 && b++;(logical && operator incounter 0, it will not move on b++ , it give ans 0 and b++ will not exicuted. C = 1 || 0 = 1 I hope my explanation you will understand 😊
Mam i have a doubt i study python in class 12 but now i study c in 1st year of college its very difficult to jump from high to mid level language so how can i resolve the problem there is lots of change in them
In the expression printf("%d" , a&b && b+1 || 0 || b++) after the logical && the value of b gets replaced by 7 (b+1)...Then while printing the value of b at the end should be 7 right?...Computer is showing 6 pls clear this doubt !
What mam told at the end, if we use logical AND then also the output will be 1 as 1 && 6 is 1 and since the first operand is true which is 1. so will b = 6 or 7? in next printf statement. Can u please explain?
@@krishnapriya3872 bcoz may be it has gone like this a&b && b + 1 || 0 && b++ 2 && 7 || 0 && 6 first both logical AND operations will operate 1 || 0 So in logical OR we got 1 then no need to go further so b++ will not be executed.
@@sumaiyasimi1443 bcoz may be it has gone like this a&b && b + 1 || 0 && b++ 2 && 7 || 0 && 6 first both logical AND operations will operate 1 || 0 So in logical OR we got 1 then no need to go further so b++ will not be executed.
![[Live สด] การออกรางวัลสลากกินแบ่งรัฐบาล งวดวันที่ 16 ธันวาคม 2567](http://i.ytimg.com/vi/dNTiBBthKgQ/mqdefault.jpg)
The way you're teaching, presenting problems to us, and claiming for answers in the comment box is quite convenient to grasp the concepts and make us feel the classroom environment.
I can proudly say that this mam is the reason why i scored good marks while i was doing my engineering. Lots of love from nepal ❤️🙏
Output will be 1 and b will be 7...Also you are best Ma'am..Today is my exam and I was so tense now my mood is good too after watching your video...Your teaching is best☺
dear mam thank you will be very less for your videos , praying for all happiness n success to u n to ur family , i m benefited a lot by your videos , you are an excellent teacher
It very good lecturing about general concept of C++ programming and she continuous this the work teacher.
Thank you mamm for the wonderful explanation. a&b = 2, a | b = 14, a ^ b = 12 and and the last is 1
Nice session Ma'am
My doubts are always cleared by watching your session.
Thanks a lot for this awesome session.
13:07 output of first three statement is
2
14
12
bro can you plz explain y 14 and 12
bro hear we get
2
15
12
In second the output will be 1 in(&&)case
And b++ will print 7
No brother the precedence changes and 0 && b++ will be evaluated first and then 1 || (value of 0 && b++) …so output will be 1 but again b value will not be affected
in the last case after adding && b++ and printing b only gives value of 6 so that means it doesnt execute the expression && b++ as per left presidency and OR operator before on left
Far better than my college professor..she is just a noob in front of you..😂
Ans for 1st 3 Q are 2, 14, 12
🎉🎉🎉
Hello, want your insta id 👋🏻
Which part can u mention it?
@@sumityadavazamgarh8023Bruh where it is?
@@ananthu414111:11
after changing the logical or operator to logical and the final output of the statement will be 1
and B value will be updated as 7.
how 7? can you explain??I'm confused
@@SaraswotiKhadka at last mam changed the question
i.e ( a&b && b+1 || 0 || b++) first mam explain with this example later,
The question is
( a&b && b+1 || 0 && b++) bitwise & has higher precedence so, now the expression on solving
( 2 & 7 || 0 && b++) so, 2&7 = 1
So expression changes as follows
(1 || 0 && b++) here && has higher precedence and also b is updated to 7 here as we used b++ in our expression. so, 0 && b++ is done first i.e
0&&6 output is 0, now ( 1 || (0 && 6)) BECOME
(1 || 0) Finally output of this expression is one bcoz of logical or.
@@hariparuchuru3858 may be i got it. Thank you
@@hariparuchuru3858 bro, in logical and (&&) , if 0 is one of the operand then we don't fetch the second operand and the result is 0 , as in this 0 && 6 output is 0 and we don't consider the second operand
So b++ value remains same i.e. 6
@@udaysingh6621 we don't get 7 but result is 1 only
mam your classes are super I am listening u r class 50 % for subject and 50% for you
Mam thankyou you are better than my teacher it's very helpfully for me
Again jenny ma'am you nailed it up to mark very clear and smart teaching
Mam the way you teach to us is the best way and we understand and resolve our problems easily 👍👍
I learned alot
Thank You ma'am :)
Now let's got for part 2
Ma'am please upload videos of pointers, structure,union,enum, files handling, command line argument of c.❤️
Yes 🙏
@@princechoudhary1025 mam has uploaded it
I love you ma'am as a student and teacher relationship ☺️💛
Very good teaching every topic is covered 🤗
Video starts at 2:26
You can fast forward the video, Actual content started from 2:30,before this all just the unacademy promotion.
Mam in this above video u said ,printf("%d
%d",a&b&&b+1|0&&b++), it will print b value as7 but it is giving only b++ old value as 6
Yes ,I also got the same answer
@@neerajaneeraja6971 me too
@@AnkitKumar-ud4vt @Neeraja Neeraja Because at second last evaluation of expression, first && operator will be evaluated then || operator will be evaluated and there is 0 as first operand in && operator then there is no need to check for the second operand and consequently b value will not be incremented.
I hope you will get it, what I tried to explain.
🙂
@@farzanaashraf8136 bhai insta id chahiye aapka
@@madhubantu5180 Why
Technically you r correct and you are technically sound. Don't get nervous or fast while teaching. But teaching is good. But you can make it better because you are technically sound.
like seriously ma'am u are really good, u just thought me a shortcut for this...
Mam ,you use very simple English words.
Thanks 😊
Thank you mam very nice gide & very nice best information bitwise operator teaching video.👍
Yes mam please take some good, hard, and really tricky examples on Logical and Bitwise, and Relational operators combined
mam you best from my all c teachers . i saw your all videos and inspiration from your video love you mam and make new best videos on c language .
th-cam.com/video/M1GwBX0zhC4/w-d-xo.html
Mam ur explanation is so nice like you😍😍
You are such a cute teacher i like all your vedios that understand able 😊😊😊
Our sir in our college explained the same topic for 40mins but I understood after watching your video mam
Thankyou so much mam it’s an amazing experience , learning from you ❤
mam in the logical AND(&&) operator the final output will be 1 and 7 right. I didn't understood how did we got 2 there. Can you please clear this doubt mam?
The one is below 2 so
Mam apko dekh ke hi sb samajh me a gya🤤🤤🤤
Im with zero doubts mam thats bcoz of you tq so much mam ❤
mam your teaching method is very impressive
Good day Ma'am. 2 years later and your course is still very much relevant! Thank you for what you do.
So I have a question, when performing certain operations on while working with binary in jpeg mode, we usually have to check if the first four digits of the 4th byte of a specific amount of data is a certain value. Here, we use the bitwise & operator. My question is, when checking to see if the first four values are similar to 0xe0, why do we use the & operator with 0xf0 on one side and 0xe0 on the other? Why don't we have 0xe0 on both sides?
An example code is as shown below:
if (buffer[0] == 0xff &&
buffer[1] == 0xd8 &&
buffer[2] == 0xff &&
(buffer[3] & 0xf0) == 0xe0)
Look forward to hearing from you soon ma.
Tqq so much mam for ur clear cut explanation 🎉🎉❤
God bless you a million times
Thank You for clarified our doubts
Thank you for teaching us in an excellent manner you are extremely better than my college faculty they don't taught me all this so that I'm stuck with lot of doubts
So because of you I'm able to i understand everything thank a lot mam 💟💟💟💟💟💟💟💟💟💟💟💟💟
Great ma'am ❤️.
Excellent ma'am ❤️
Best teacher 🎉
hi mam, nice explanation , in this video , which first prior , & ,+ , pervious video you explain high priorty each by binary.but while combination how we can identity which first for example & ,+ .
Hi, you are the best.. thanks for your help...
will you make a separate video for bitwise operation examples??
Thanks for help Ma'am 🙏🙏🙂
You are very beautiful ❤️
And the video helped me a lot
amazing
thanks mam
BANGLADESH
The guy named Ladesh: ○_○
can you please tell why when a=10,b=5,a&b is 0. while when a=10 and b=6, a&b is 2.
Excellent explanation
mam Give suggestions to MCA students , scope is fit for software
as it will get true and true ,ouput will be one and value of b is 6 and if it finds another b then b will be 7
a&b= 2, a|b= 14, a^b= 12, lastOne= 1
Really like your explanation ma'am superb 🔥🔥🔥🔥👌👌👌👌👌👌👌
I am impressed on your 💕
Mam last printf statement you said that b value will be 6 it will not be 7 but if i run that program and printf("%d",b); the output is 7 not 6
Mam do one more video on bitwise operator with more examples
It's a very helpful video mam thank you
Thank you❤🌹😊 so much for your lectures madam...... Gratitude from bottom of my heart❤💖
I didn't understood that why the output of b would be 6 and not 7, as precedence of increment operator is more than other operators, so logically the increment operator should be executed first and the value of b must be 7 in both && case and || case.
i also same doubt!
@@vrajalingam3487 I got my answer. See precedence is checked only when a operant(or any number) has more than one operator, and not checked at first for whole statement. And in case of || only first part is checked and if that part returns true then other part is not check., same in case of &&, if first part returns false, other part is not checked.
@@lokesh6828bro l also know that even everyone but here the main problem is when we r doin b+1 i.e 6+1=7, then value of b is updated here only & in b++ we have to consider 7 & not 6 (coz the value of b is updated in the memory after doing b+1 ) i.e why when we'll print 'b' we should get 7 but still we r getting 6 .
How's that possible, l can't understand
@@Helly04
Because b++ and b+1 is not the same thing, in case of b+1 the value doesn't change in the memory it just changes where it is assigned to.
So if b=6, c=b+1 then c will become 7 but b will remain 6
in the question, && has higher precedence so, after initial solving the question comes to this,
1 || 0 && 6, here && will be evaluated first due to higher precedence, in logical and if first operand is 0, second one isn't evaluated final output becomes 0.
So now it is 1||0, in logical or if first operand is 1, final output becomes 1, second operand isn't evaluated, so the second operand b++ was never checked in any of the steps, that's why the value of b remains 6 and the final result is 1
NOTE: Precedence will come into picture only if expressions will be evaluated. In case of && or || its possible that some expressions may not be evaluated at all because we obtain 0 or 1 respectively. Hence, no matter how much higher precedence operators are there in those expressions, those expressions would never be evaluated
Mam, is b+1 is a relational operator?(13:25)
It is arithmetic operator
@venkat sai ,then y mam told b+1 is relational operator (13:25)
arithmetic operator
I think b+1 is a arithmetic operator
@@kumaravel1301 By mistakely mam said that thing ...... there is + between two operands so it is an arithmetic operator only don't get confuse
Good explanation
b=6 Even if && is there, it didn't get incremented while executing
#include
int main()
{
int a=10,b=6,c;
c=a&b&&b+1||0&&b++;
printf("%d %d %d",a,b,c);
return 0;
}
output==> 10 6 1
Can you tell me why?
@@syedalifathimaa9618 because logical && operator precident higher then logic ll operator
C = a & b && b+1 || 0 && b++;
C = 10 & 6 && 6+1 || 0 && b++;
C = 10 & 6 && 7 || 0 && b++;
C = 10 & 6 = 2 && 7 || 0 && b++;
C = 2 && 7 = 1 || 0 && b++;
C = 1 || 0 && b++;(logical && operator incounter 0, it will not move on b++ , it give ans 0 and b++ will not exicuted.
C = 1 || 0 = 1
I hope my explanation you will understand 😊
@@ja3r thnx bro keep it up
It's very cleary😊
In the very last example question, printf is 1 and b is 7.
@jenny's lectures.Mam please explain detailed programme like that programme
Thank you so much ma'am
16:20 output is 1
Thanks a lot mam
First comment😍
thank you for this video mam😃
Int a=10 b=6
a&b is 2
a | b is 1
a^b is 1
你做得非常好！ 您能帮我处理一些不相关的事情吗？ 我有一个SafePal钱包，里面有USDT，并且我有恢复短语.[pride]-[pole]-[obtain]-[together]-[second]-[when]-[future]-[mask]-[review]-[nature]-[potato]-[bulb]. 我该如何将它们转移到Binance？
You are my best guru
a&b =2
a|b = 14
a^b = 12
print1=2
print2=14
print3=12
print4=1
Chechii, umma... Eniku chechine kadikanam..., finger licking good aanu chechiii...❤😊
Mam can I use one operater twice in program 🙏🙏 I mean in the same line
There's a b+1 in expression. By the precedence b will 7 already.
So, will there be change in value of b when we add b++. ??
b+1 doesn't change the value of b, it just returns b+1. So b will be same
How to do right shift and left shift and also more on ternary operators topics ...
thank you ma'am😊😊😊
Mam i have a doubt i study python in class 12 but now i study c in 1st year of college its very difficult to jump from high to mid level language so how can i resolve the problem there is lots of change in them
C should be the first language you should learn and python should be the last
In the expression printf("%d" , a&b && b+1 || 0 || b++) after the logical && the value of b gets replaced by 7 (b+1)...Then while printing the value of b at the end should be 7 right?...Computer is showing 6 pls clear this doubt !
At 16:23... U said to use apparends in the place of logical OR... I applied and I got ans =1. Am I right mam
Yes you are right but the value of b will not be updated, it will remain 6.
super explanation madam
starts at 2:26
Searching for some one commenting the start time ! Thnx
Output:1
b=6
What mam told at the end, if we use logical AND then also the output will be 1 as 1 && 6 is 1 and since the first operand is true which is 1. so will b = 6 or 7? in next printf statement. Can u please explain?
b is 7 bcz in and operator first is 1 then second expression will be evaluated so in this question b is evaluated therefore b is 7
This was helpful
Very beautifully explained 👌
Ma'am u told to replace logical OR by logical AND and then b++
After that if i print b then i am getting b=6
No increment in value of b
B=7 I think so, can u give justification how it will be B=6
Same... I am also getting 6
ya b= 6 there should be no increment of b even though after using logical and instead logical or and the output is 1
@@krishnapriya3872 bcoz may be it has gone like this
a&b && b + 1 || 0 && b++
2 && 7 || 0 && 6
first both logical AND operations will operate
1 || 0
So in logical OR we got 1 then no need to go further so b++ will not be executed.
@@sumaiyasimi1443 bcoz may be it has gone like this
a&b && b + 1 || 0 && b++
2 && 7 || 0 && 6
first both logical AND operations will operate
1 || 0
So in logical OR we got 1 then no need to go further so b++ will not be executed.
Mam can you please do some vedios on python also
keep it up
Mam for this operation
a=10, b=3
a&b&&++b ->1
But for prefix increment we need to use b as 4 then
a&b gives 0 this results 0&&++b as 0 no mam
Tutorial Starts at 2:22.
Mam in these problem ++ has more precedence then + so why first b value not increase can you please explain