C Programming (Important Questions Set 1)
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- Опубликовано: 8 сен 2024
- C Programming Lectures: goo.gl/7Eh2SS
C Programming & Data Structures: Important questions in C programming.
Topics discussed:
1. Important questions in C language.
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Hello Neso Academy, Could you please enable all English closed captioning for all of your C programming lectures? I would be grateful if you do that. Because I am deaf so I cannot hear your explanations. Thank you
You can do that on your RUclips app settings
@@duffyduck4065 Buddy, some videos in the playlist are not haviing option to enable captions. Nevermind, I already learned from mycodeschool.
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Finally You tube recommend something worth to watch.Thanks for the wonderful explanation.
char = -128 ~ 127
unsigned char = 0 ~ 255
char a = 255
a(2) = 1111 1111
1111 1111 = -1
answer
a + 10 = 9
In C, Q3 is not guaranteed to work if char is signed, as the result of the narrowing value conversion is either implementation-defined (as the signed-type representation is implementation-define). On compilers which define char as unsigned, it is guaranteed to work and print 9. And this all of course assumes that CHAR_BIT is 8. Just to make you aware that the results are by no means guaranteed by the standard. (if CHAR_BIT is say 16, then it prints 265), and if char is signed using sign-and-magnetude, it prints -117 (as 255 converts in that narrowing value conversion to -127, then -127+10=-117).
Yes..u r correct..the question should signed char c or unsigned char c to avoid any complier specific results..
255 in binary comes out to be 0b11111111, which comes out to be the same for the binary representation of -1 (0b11111111).. So, for signed representation, 'c' will have a value of -1.. Upon adding 10 to that, it assumes the value 9.. The other case of unsigned representation of the character 'c', has been explained in the video.. However, by default, upon declaration, the 'signed' modifier is chosen for the character..
@@royarijit998 C does not specify a default sign or integer encoding. The binary representation of -1 could be 0b10101010 for all you know. The default sign could be both signed or unsigned. It's even possible that char has 50 bits.
In Q3.
In declaration statement
Char c=255;
U can not e specify modifier
THEN compiler implicit assume by default it is signed
Range would be
-128 to 127
But u assign c=255; na
This value is already out of range then compiler do cyclic process
255 is equivalent to - 1
Then finally c contain - 1
C=c+10;
C= - 1 + 10
C=9;
Out put is 9
( in hole explanation we assume character take 1 byts= 8 bits
Thank you neso academy
shailesh kushwaha This isn't guaranteed. This assumes a) The compiler defines signed integers as 2s compliment, b) the compiler has bitwise conversion for out-of-range singed integers Both of which are unspecified.
Also, char on its own is a distinct type from signed char, and it is implementation defined if the type is signed or unsigned. (See en.cppreference.com/w/c/language/arithmetic_types for details).
Further, this also depends on having 8-bit bytes (CHAR_BIT=8), which is not always true.
this is from the legendary category course on youtube , explaination
on peek
6:50 IMP 265mod256=9 remainder
This is so great. Pronunciation is better than every other Hindis
Sir i am preparing for gate 2020... ur videos whr in recommendation .. u hv no idea how good u are at teaching.... totally worth watching...
Hey wht you doing now
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This questions were just too helpful.
Glad that I find your channel sir 😀
I would strongly advise you to include to which standard you are adhering. After C99 using implicit integer will at least throw a compiler warning and depending on parameters will not even compile.
The 2s comp representation of a signed num is -3 = 101 only so when treated as a positive num it is 5 .
I am so lucky to have found your channel 😊
nice lectures by you sir
that too for free!
appreciated your efforts.
Nice
Sir in question number 3 maximum value of char should be 127 ?. I am confused sir
No!
Char take 1 byte space having 8 bits ... and so the max range will be 255 ...
Char occupies 1 byte of memory and there is no point of signed and unsigned, since the binary representation will be same for both. Look into video 9
Q3 : char c is signed character , so c= 255 is equivalent to c = -1 and -1 + 10 = 9.
2:49 no one teach us this one...its very helpfull
I never saw someone explains like this👏
I must say that this is best channel on youtube to learn programming
Thank you very much. You are a genius 👍👍🔝🔝
In question 3 there sholud be unsigned, otherwise it can not hold 255
App agar c programming sikhane lag gaye to case hi nehi hain... Many engineering graduate will be beneficial from thiz... With the help ur videos we passed the exam of digital logic, basic eletronics subject... Who ever you r... U r in the in the heart sir....
in Q1, no need to write %s in the printf. Can just pass the string directly like printf("Hello\a
");
Thank u so much sir actually i like tge way u were teaching it helped me a lot🙇
Thank you do much sir it was very helpful for me
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Great variety of questions! Loving the content
Character is capable of holding the information upto 1byte only which is equal to 8bits (that means the max value it can hold is 255)
Data types obey the law of arithmetic modulo 2 raised to the power n where n is equal to 8 in this case
Here ans is 265 mod 2 raised to the power 8
Very nice and very useful thank you sir
"signed i = value" equivalent to "int i = value", why? because when you write the signed modifier without any data-type after it, your compiler assume that i is either a variable of type int or it's a variable of type char, and that assumption depends on the value stored in i.
Depending on compiler and standard, this will probably either give you a warning or not compile at all (implicit type conversion is non standard since C 99)
in this Q why 256divide by 255
char c=255;
c=c+10;
printf("%d",c);
return0;
out put is 9
let me be frank, I like your videos and i've learnt a lot from you. Many at times i've been excited and some depressed, like in this video, i thought i understood the concepts until you asked the tough question. In short, thank you for sharpening our experience
Really great videos one can refer to these videos for certain competitive exam like gate too. Helped me a lot throughout. Thanks neso academy for such wonderful videos.
Please slove this question..
write a program to compute the value of u for different values of x at distinct instances of time and plots the series of graphs on u vs plot.
When I made the c fragment as like :
#include
Int main()
{
Printf("%d",printf("%10s","jee"));
}
Xpected output as:
[. jee3]
But the Output as shown as like this:
[ jee10 ]
Plse anyone let me know how's it possible?
& If I'm wrong then tell me wh're my concept was wrong ...
As it also count spaces, output will have 10 not 3.
Its correct what's ur doubt
A beautiful explanation
Happy to listen in a simple manner
These videos are really helpful, I watch them with my friends on Voicely whenever we practice coding. We all agree that we learned more from RUclips videos than from school lol
You got a new.. keen curious learner (beginner)of C as subscriber today✌
Suggest from where i can find more questions like these (covering conceptual and hidden meaning of a syntax)
Too good for the c programming language.well done
Such an amazing lecture. Keep it up sir ☺
Thank you! I had a pending program since long time.
Today i was just watching your video when a concept clicked in my mind because of a small concept which learned from your video. My program is completed,i am thankful.
Thanks I learned something new!
Unfortunately, the answer to Question 3 is a _lot_ more complicated than presented here.
Depending on the implementation, plain "char" can be either signed _or_ unsigned. The encodings for characters in the _basic character set_ (upper- and lowercase Latin characters, digits, most punctuation) are guaranteed to be non-negative, but encodings for additional characters may be negative or positive. If plain "char" is unsigned, then the value will "wrap" as described. If it's signed, then the behavior is *undefined* . You _may_ get a result of 9. You may get something different.
Signed or unsigned char will give same answer.
actually i get 19
@@duggirambabu7792 It _may_ give the same answer, but it isn't guaranteed. Unsigned overflow is well-defined; signed overflow is not. If you're on a system that uses two's complement to represent signed values and doesn't do anything funky on signed overflow, then yes, you are likely to get the same result. On a system that uses ones' complement or sign-magnitude to represent signed values, or raises some exceptional condition on signed overflow, you'll likely get something completely different.
@@johnbode5528 if the range of signed char is from -128 to 127, aren't we exceeding the range in this case which means we should divide 265 by 127?
I got a garbage value
woah so nice questions. i appreciate it
Thank you so much for this video. It is very helpful video 🙂
This is a really important video lecture, thank you very much Neso Academy.
qus no (3) we use clock analogy method we also get an answer 9
Happy to find your lecture which is very understandable and clear...
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10:55 Q5 answer 2^n - 3.
In the first question,Have u given space after "!" ,Then the output should be 13 :)
He hasnt given space after '!' char.
Really amazing.... Plz make more videos on this topic
In Q3: The range of signed char is -127 to 128 and for unsigned it is 0 to 255. But in the question it is signed as default. But you said the range is 255 here also? Enlighten me
Dear Sir, when you were teaching us signed integer and in that you were telling us how to represent -ve no. (Like -127), you were using place value method i.e. -2^7 and 2^0 to set -127. So my question is to you, can we apply same method to represent -3 rather than using 1's comp. Of 3, then adding 1 and finally gets 2's comp. Of -3??
Regards
at 6:37, how did you explain n =8, how didi you know 2 raise to the power of 8?
Superb
awesome advanced questions on basics
Nice this is very good and really helpfull video sir , Thank you very much.
have a great sir this topic is very helpful video sir.
Thanks you so much sir
In question 5
After converting 3 to two's compliment we had only one zero .
If we had 2 zeros then how to subtract
so DS/sometimes water controlled and damshello sirwell done your speech and thank you very much
Thank you for this complete and well ordganized Course about C. Very good explained with funny indian accent ;). Long story short : AMAZING. Thanks for sharing.
But sir!!
Consider the code:
Printf("%10s","hello") it prints 5spaces and hello.
Printf("%1s","hello") it prints 4 characters but not 1 mentioned,i didn't understand that.
Sir in question 3 there would be default modifier signed and therefore the range would be -128 to +127 so it can't hold 255 I thought so, after watching your answer I get confused...clear this please.
it's showing me no such file or directory in the commnt box in code blocks can u tell me what's is the error
Thanking you sir
I learn something new...
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your teaching is awesome
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oe, my god, again a co-incidence or what.
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U r always the best man
What is the meaning of
sir, with question 3, when i tried working out the mod on my calculator i keep getting 10, does that mean that 10 are the number of characters and 9 is the answer due to starting with 0?
But in the question there is asked according to definition of integer so Option a is correct .Compiler or Computer will assume as int
sir your teaching is excellent😀😀;
and sir please clarify my doubt,it would be very helpful😁.
5:48 dude, char is usually signed
ربنا يباركلك يغالي
Sir for question no.5, I tried the following code with long int and therefore, the size of long int is 8 bytes. Theoretically, I should get an answer different from the one you got but still I got the same answer as you described. Why isn't long int considering 8 bytes rather just 4 bytes?
I have an doubt in same question... am just changing the second argument expression as i-j and i get 5,why? Instead of getting 4294967291
For ur doubt... May be ur computer holds 4 bytes for long int else long long int holds 8 bytes.
While it is true that most of the loosey goosey C compilers will accept the integer declaration shown in Q4 it is quite bad form to assume that every compiler will behave the same and much better for code readability and portability to explicitly and fully define your types. There is a big difference between what will compile w/o error and good coding practice.
Yeah, exactly. C dropped implicit type conversion a long time ago and it's just plain better than supporting legacy code like that. Especially in uni you usually have strict compilation parameters which are specific to the standard you are using and -pedantic will definitely not let you compile this.
Great effort sir....no words...for what u had made!!!!
Can you make a question, answer video of c++ programming exact this type sir
I executed [printf("%d","Hello World");] The Output is 4210688 . Why the output is like this ? What is the link between the output and the Input ?
Hi sir, i have a doubt regarding q no:1 printf("%s","HELLO WORLD"); will print only HELLO right? ie till space is encountered? Answer should be HELLO 5 ??
Cant understand rhe Q3 where did you get the 2raise to 5*8?
Super explanation
Awesome questions
NO WORD'S TO SAY JUST I LOVE U❤
Sir if printf also returns the no. Of characters it successfully prints on the screen then why don't we get 12 twice(once from the internal printf and other from external)??
Because they are two separate calls. It’s equivalent to writing
int tmp = printf( "%s", "Hello, World!" ); // returns 12
printf( "%d", tmp ); // returns 2
Good work, I enjoy it.
in first que, why the answer is not 12Hello World! ? since we put the %d at the beginning of the printf function shouldn't the ans be 12Hello World! ?
Thank your for your lesson!
Q5: why is there 32 bits in -3?
thanks you so much for the video
Thanku Neso for this amazing explanation.
Mind blowing