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.
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
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☺
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
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 💟💟💟💟💟💟💟💟💟💟💟💟💟
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
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.
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.
@@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. 🙂
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?
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 & ,+ .
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?
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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 ❤️🙏
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.
Nice session Ma'am
My doubts are always cleared by watching your session.
Thanks a lot for this awesome session.
Thank you mamm for the wonderful explanation. a&b = 2, a | b = 14, a ^ b = 12 and and the last is 1
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☺
1 and b = 6... bcoz 0 && b++ then first false then second not checked in and op.
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 👍👍
mam your classes are super I am listening u r class 50 % for subject and 50% for you
I learned alot
Thank You ma'am :)
Now let's got for part 2
Mam thankyou you are better than my teacher it's very helpfully for me
Very good teaching every topic is covered 🤗
I love you ma'am as a student and teacher relationship ☺️💛
Thank you mam very nice gide & very nice best information bitwise operator teaching video.👍
Far better than my college professor..she is just a noob in front of you..😂
Ma'am please upload videos of pointers, structure,union,enum, files handling, command line argument of c.❤️
Yes 🙏
@@princechoudhary1025 mam has uploaded it
Mam ,you use very simple English words.
Thanks 😊
You are such a cute teacher i like all your vedios that understand able 😊😊😊
Mam ur explanation is so nice like you😍😍
Best teacher 🎉
God bless you a million times
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
Thankyou so much mam it’s an amazing experience , learning from you ❤
like seriously ma'am u are really good, u just thought me a shortcut for this...
Tqq so much mam for ur clear cut explanation 🎉🎉❤
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
Great ma'am ❤️.
Excellent ma'am ❤️
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 .
ruclips.net/video/M1GwBX0zhC4/видео.html
Excellent explanation
Mam apko dekh ke hi sb samajh me a gya🤤🤤🤤
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 💟💟💟💟💟💟💟💟💟💟💟💟💟
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
You are very beautiful ❤️
And the video helped me a lot
Im with zero doubts mam thats bcoz of you tq so much mam ❤
Hi, you are the best.. thanks for your help...
Really like your explanation ma'am superb 🔥🔥🔥🔥👌👌👌👌👌👌👌
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
Thanks for help Ma'am 🙏🙏🙂
I am impressed on your 💕
It's very cleary😊
mam your teaching method is very impressive
amazing
thanks mam
BANGLADESH
The guy named Ladesh: ○_○
Good explanation
Thank You for clarified our doubts
Yes mam please take some good, hard, and really tricky examples on Logical and Bitwise, and Relational operators combined
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
It's a very helpful video mam thank you
thank you ma'am😊😊😊
Our sir in our college explained the same topic for 40mins but I understood after watching your video mam
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.
You are my best guru
Thank you❤🌹😊 so much for your lectures madam...... Gratitude from bottom of my heart❤💖
Thanks a lot mam
super explanation madam
thank you for this video mam😃
Video starts at 2:26
a&b= 2, a|b= 14, a^b= 12, lastOne= 1
Thank you so much ma'am
This was helpful
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.
a&b =2
a|b = 14
a^b = 12
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
print1=2
print2=14
print3=12
print4=1
Very beautifully explained 👌
Int a=10 b=6
a&b is 2
a | b is 1
a^b is 1
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
will you make a separate video for bitwise operation examples??
Great mam
very helpful
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 & ,+ .
You can fast forward the video, Actual content started from 2:30,before this all just the unacademy promotion.
Mam you are looking so beautiful
Ans for three Q
2
14
12
Mam do one more video on bitwise operator with more examples
A&B = 2
A|B = 14
A^B = 12
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
Done 🙏🙏😃.
BIGFAN MAM
First comment😍
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.
U ar unique
16:20 output is 1
Beful so interested
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
Thanks
@jenny's lectures.Mam please explain detailed programme like that programme
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
In the very last example question, printf is 1 and b is 7.
Dress++ 😎, *+beauty 🥰
Yahh😂❤️
In last you said that b will remain 6,but wrong madam it will be 7 madam
Mam last me thoda kam samagh aaya h but well done 😊
starts at 2:26
Searching for some one commenting the start time ! Thnx
Chechii, umma... Eniku chechine kadikanam..., finger licking good aanu chechiii...❤😊
mam Give suggestions to MCA students , scope is fit for software
Output of three statement are:-
O
7
7
All the best for your Exam 👍
Ans:2,14,12