10/10 here so I wanted to give explanation to que questions that someone may need. Q1: Byte is the minimum amount of information accessible via pointers, so it's also the minimum measuring unit of memory in C. Q5: C does not define a standard size for any data type other than char, but it specifies a minimum amount of bytes that any compliant implementation must use. In the case of the int, minimum size is 2 bytes (16 bits) and the most usual value nowadays is 4 bytes (32 bits). You can always check the size by using sizeof: printf("%d", sizeof(int)); Q6: A standard (C89) variable name can have any combination of letters, underscores and numbers as long at it DOESN'T START WITH A NUMBER. The variable names "39", " var_39" and "_" are all composed of just valid characters, but "39" starts with a number so it is the only incorrect one. Q7: Well, a little correction to the creator of the video in this one: There is a small lack of information in the question, wich actually leads to TWO VALID ANSWERS in this case. The first valid answer (wich is assumed in the video) comes from the assumption that there are no other statements involving "var" in the program, so it is right to say that "int var;” is both defining (reserving space for the variable) and declaring (making the variable visible in the current context). Meanwhile, the second line " extern int var;" wouldn't need to reserve space anymore so it just declares (redundantly, but not incorrectly) "var" again. However, the question never specifies that the two statements are unique in the program (although it was obviously meant like that), so we can hack the things a little to come to a different solution; What if these statements are present in the global context just after a previous declaration of "var" just like this: int var; // bla int var; extern int var; // bla Or these are (again in global context) before a more specific declaration like in the following: int var; extern int var; // bla int var = 0; In both those cases, the tentative definition "int var;" never turns into a proper definition like it happened when we assume there weren't more statements. It stays as a mere "declaration". That is how we can say that both statements only declare but not define the variable (option a). Strictly speaking, the correct solution to this problem is that there is not enough information to decide if the correct answer is (a) or (d). Q8: By the moment you assing " int var = var", the internal declaration of this variable has already taken effect so it is no longer the "global and initialized var", but a " local and uninitialized var". In C, uninitialized variables are not guaranteed to contain any meaningful value, so when you assign an undefined variable into the same place, it stays undefined. Q9: Notice that "int var = 10;” and the printf appears inside a different internal block. "var" is only available inside the block it was declared in so trying to use it in another block causes a compilation error saying that var was not declared. Q10: There is already a few people that explained this so short story: signed numbers in C are represented in "two's complement", and when you negate the bits of a number in this representation you get that number in negative minus one. But why am I talking about signed numbers when the program clearly uses unsigneds? Because the format " %d" forces a number to be interpreted as signed. Note that the number being unsigned is important, though, because it assures that the most significant bit is treated correctly by the "~" operator, which may otherwise be undefined behavior.
Thanks. I needed q8. Only I got confused at the beginning because you wrote down q8 twice. But then I knew the first q8 was actually q7. But thanks.. you just taught me the difference Between define and declare. I'll keep researching though.
Q5. It only defines sizeof to be 1, but not that it's 8 bits though. Q7. Good points. Expect the definitions separately though. Q8. That got me, in other languages it would have taken the variable that was there before initializing. It's really experience with certain languages that allow shadowing and in the initializer of a shadowing variable you can access the shadowed variable. That's funny. Q10. %d is a signed integer specifier. And I don't think the ~ operator really does funky stuff on signed integers but do the exact same shit. Languages such as Python actually *define* ~x to be (-x-1) as they have bignums.
@@paulstelian97 Q5: Yeah, I'm aware, but I kinda felt lazy to explain. Maybe I'll add corrections to further edits, but I'd like to not only adress your note, but several other comments with concept errors that I've se over here. Q10: Nobody expects bit-operators to do funky stuff with the numbers, but the reality is that the behavior of the sign bit in signed numbers after using such operators is left to the implementation, meaning that the overall result is totally able to change (it is not common but still possible.). I don't remember if it is actually called "undefined behavior" tough, it could be "platform depended behavior", so I'll have to re-read the standard after this.
10. Given, var=10 In this given question we have to store the 1st complement of var So, binary form of var is 0000 0000 0000 1010 1's 1111 1111 1111 0101 Here, MSB is 1 i. e number is negative And we all know, any negative numbers store in computer as 2's complement So we have to change this binary number in 2's complement 1111 1111 1111 0101 1's 1000 0000 0000 1010 2's 1000 0000 0000 1011 After converting it into decimal number we will get - 11 So , correct answer will be - 11 Thank you! ❤️
Naveen Chowdary MSB stands for *most significant bit*. In a binary number , the leftmost bit is referred to as MSB . And it has the largest place value . In simple words if the binary number is 0001 MSB is 0 . And if it is 1001 then MSB is 1 being the leftmost .
6/10. That was an excellent and challenging quiz! I am 37 and have never taken a timed quiz and am learning C for fun. This is the highest quality education I've ever received!
10th question solution : k=10 In binary form : 0000 0000 0000 1010 now apply ~ to k , result : 1111 1111 1111 0101 but in the printf function they use %d instead of %u so the above value is too large to store in %d format , so compiler will find its equivalent 2's complement value and stores . Now 1111 1111 1111 0101 ( compiler check left most bit if it is 1 it put negative sign behind it ) 2's complement of above value : 0000 0000 0000 1010 ( 1's complement ) +1 0000 0000 0000 1011 which equivalent to 11 .( since left most bit in 1111 1111 1111 0101 is 1 , compiler put negative sign) so answer is -11 .
8/10✌... Its very helpful for every one who is having the knowledge of source code languages and need reasons for wrong answers we have choosed.... mainly helpful for ENGINEERS 👩🎓👨🎓
8/10. C has been my scariest nightmare I don't want to look back at. And now I feel kinda confident about it. Thank you so much for making me good at this 💜
10) The code defines a main function, which is the entry point of the program. Inside the main function, an unsigned integer variable var is declared and initialized with the value 10. The printf function is used to print the bitwise complement of the variable var. The ~ operator is the bitwise NOT operator, which flips all the bits of the operand. In this case, since var is initialized to 10, which in binary is 0000 1010, applying the bitwise NOT operation will result in 1111 0101. The printf function with the format specifier %d is used to print the result of the bitwise complement operation. Therefore, the output of the program will be the decimal value of ~var, which is -11.
10th Question By Kranthi Kumar Format specifier in printf function is "%d", so the value of ~var is asked in signed integer. So as we know that, for a signed integer var + ~var = -1 Now from given var = 10 ~var = -11. If it is asked as printf("%u", ~var); As %u is for unsigned format specifier Then var + ~var = 2^32-1 Here 2^32-1 is maximum value for an integer in 4bytes. And the answer becomes, ~var = 2^32-1-10 ~var = 2^32-11.
Q8. should have output 5 as int var = var will find var in global scope and will get value 5 and assign it to int var in local scope .So shouldn't be output 5??
Sir, plz upload more videos regularly, don't leave 1 video in a week or 2-3, bcoz we need to complete it fast.. N u'r videos are mosttt helping So, we need more lectures on C only byy U.. ThankU..
in binary 10 is writen as 00001010, but the meaning of the sign ~ means a bitwise inversion on a value. wherever you see 0, it will be 1 and wherever you see 1, it will be 0 therefore from the above binary for 10, it will be 00001010 ==== 11110101 which in decimal representation is 11 is that clear?
9/10 I enjoyed a lot I've mistaken in question no 8, I think global variable and local variable contains the global value but I was wrong it's contains the garbage value🙂
since var is declared as an int it will look for the value of "var" no ascii table to it will be not assigned as 5 but as a sum of other values in the ascii table
Pedro Lopes Are you serious? what's up.with ascii here??? as you can see var is declared in both global scope and local scope. the local one is initialized from itself because the fact that when declaring a variable it can be used before the end of declaration settlement. int i = 10, y = i; i haa been allocated in memory then its value in memory is assigned to y before i came to life.
because its declared int var = var for example int var = A; if you printf var you'll get var = 65, however what you want to do is char var = 'var'; then it will print u the string var.
Pedro Lopes Totally wrong. it doesn't have any relation with ASCII. as you can see printf("%d", var); which means printing a variable of type integer. where did you find ASCII here??!!!
#include int main() { unsigned int var = 10; printf("%d", ~var); } a) 10 b) -10 c) -11 d) -5 from above question var=10(1010) and ~var=~10, that is (0101) which is equal to 5 but the answer is option C how it is possible?????
Well , this is the first time i am learning programming and the quiz taking method is amazing. Well, in this quiz , i got 6/10 but, in shaa allah , in the future i will definitely get good marks. And thanks by the way your method of teaching is excellent.
Please do update us with the new videos. Don't Leave us Learning in the middle. Please release the new videos in the series after operators or do notify us by when they will be available on your channel. Please Don't hinder the learning process in between. The videos and content are awesome and easy to understand. Please upload next videos in the series please.
8/10 NOW I'M QUITE CONFIDENT THANKS A LOT SIR.EVERY STUDENT MUST REFER TO THIS CHANNEL FOR BETTER UNDERSTANDING OF C LANGUAGE.THE WAY OF TEACHING IS QUITE EASY AND SIMPLE.
9/10 Q7 i wasnt sure if int a; was a declaration and a definition if there was no assigment but i know it has either 0 or garbage value, i have in mind that assigment changes sth, i dont know what those mean: stack, heap, and that is confusing me
~ (tilde) bitwise operator means works on bits (binary value) of a value.. 00001010 When applied ~ 1 becomes 0 and 0 becomes 1 So Step a: 11110101 On this you have to apply 2's compliment 1's compliment 00001010 2's compliment + 1 00001011 Which is binary value of 11 but as we actually got a negative value in step a so final answer is - 11.
A uint can't become negative, no matter what it's binary value is. The uint of decimal value 10 is stored in the memory as : 0000 1010; The tilde(~) operator flips all bits of the "var" variable, which yields this binary value : 1111 0101. This equals to 4294967295-8-2 = 4294967285, which does not show up in the answers.
@@rafca7408 The %d tells printf to treat argument as signed int, so you will get -11 on machines that use 2-complement to represent negative integers. But to really nitpick, the C standard does not mandate 2-complement, so correct answer is actually "output is undefined".
10.Assume that thecompiler uses 4 bytes or 32 bit. The value of 10 in binary is(for 4 bytes):0000 0000 0000 0000 0000 0000 0000 1010 so when you use ~var, the value becomes in binary is(for 4 bytes): 1111 1111 1111 1111 1111 1111 1111 0101. So the decimal value is : 2^31+2^30+2^29+2^28+2^27+2^26+2^25+2^24+2^23+2^22+2^21+2^20+2^19+2^18+2^17+2^16+2^15+2^14+2^13+2^12+2^11+2^10+2^9+2^8+2^7+2^6+2^5+2^4+2^2+2^0=4294967285, this is unsigned int(%u). But we don't have this range in signed int(%d). We need to get the equivalent number in signed range. Since the MSB is 1, we need to add negative to the first bit. So the decimal value becomes: = -(2^31)+2^30+2^29+2^28+2^27+2^26+2^25+2^24+2^23+2^22+2^21+2^20+2^19+2^18+2^17+2^16+2^15+2^14+2^13+2^12+2^11+2^10+2^9+2^8+2^7+2^6+2^5+2^4+2^2+2^0=-11
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10/10 here so I wanted to give explanation to que questions that someone may need.
Q1: Byte is the minimum amount of information accessible via pointers, so it's also the minimum measuring unit of memory in C.
Q5: C does not define a standard size for any data type other than char, but it specifies a minimum amount of bytes that any compliant implementation must use. In the case of the int, minimum size is 2 bytes (16 bits) and the most usual value nowadays is 4 bytes (32 bits).
You can always check the size by using sizeof: printf("%d", sizeof(int));
Q6: A standard (C89) variable name can have any combination of letters, underscores and numbers as long at it DOESN'T START WITH A NUMBER. The variable names "39", " var_39" and "_" are all composed of just valid characters, but "39" starts with a number so it is the only incorrect one.
Q7: Well, a little correction to the creator of the video in this one: There is a small lack of information in the question, wich actually leads to TWO VALID ANSWERS in this case.
The first valid answer (wich is assumed in the video) comes from the assumption that there are no other statements involving "var" in the program, so it is right to say that "int var;” is both defining (reserving space for the variable) and declaring (making the variable visible in the current context). Meanwhile, the second line " extern int var;" wouldn't need to reserve space anymore so it just declares (redundantly, but not incorrectly) "var" again.
However, the question never specifies that the two statements are unique in the program (although it was obviously meant like that), so we can hack the things a little to come to a different solution; What if these statements are present in the global context just after a previous declaration of "var" just like this:
int var;
// bla
int var;
extern int var;
// bla
Or these are (again in global context) before a more specific declaration like in the following:
int var;
extern int var;
// bla
int var = 0;
In both those cases, the tentative definition "int var;" never turns into a proper definition like it happened when we assume there weren't more statements. It stays as a mere "declaration".
That is how we can say that both statements only declare but not define the variable (option a).
Strictly speaking, the correct solution to this problem is that there is not enough information to decide if the correct answer is (a) or (d).
Q8: By the moment you assing " int var = var", the internal declaration of this variable has already taken effect so it is no longer the "global and initialized var", but a " local and uninitialized var". In C, uninitialized variables are not guaranteed to contain any meaningful value, so when you assign an undefined variable into the same place, it stays undefined.
Q9: Notice that "int var = 10;” and the printf appears inside a different internal block. "var" is only available inside the block it was declared in so trying to use it in another block causes a compilation error saying that var was not declared.
Q10: There is already a few people that explained this so short story: signed numbers in C are represented in "two's complement", and when you negate the bits of a number in this representation you get that number in negative minus one.
But why am I talking about signed numbers when the program clearly uses unsigneds? Because the format " %d" forces a number to be interpreted as signed.
Note that the number being unsigned is important, though, because it assures that the most significant bit is treated correctly by the "~" operator, which may otherwise be undefined behavior.
Thanks. I needed q8. Only I got confused at the beginning because you wrote down q8 twice. But then I knew the first q8 was actually q7. But thanks.. you just taught me the difference Between define and declare. I'll keep researching though.
@@arturocervantes6366 ah, you're right. It's corrected now.
Q5. It only defines sizeof to be 1, but not that it's 8 bits though.
Q7. Good points. Expect the definitions separately though.
Q8. That got me, in other languages it would have taken the variable that was there before initializing. It's really experience with certain languages that allow shadowing and in the initializer of a shadowing variable you can access the shadowed variable. That's funny.
Q10. %d is a signed integer specifier. And I don't think the ~ operator really does funky stuff on signed integers but do the exact same shit. Languages such as Python actually *define* ~x to be (-x-1) as they have bignums.
@@paulstelian97
Q5: Yeah, I'm aware, but I kinda felt lazy to explain. Maybe I'll add corrections to further edits, but I'd like to not only adress your note, but several other comments with concept errors that I've se over here.
Q10: Nobody expects bit-operators to do funky stuff with the numbers, but the reality is that the behavior of the sign bit in signed numbers after using such operators is left to the implementation, meaning that the overall result is totally able to change (it is not common but still possible.). I don't remember if it is actually called "undefined behavior" tough, it could be "platform depended behavior", so I'll have to re-read the standard after this.
Thanks for providing solutions
10. Given, var=10
In this given question we have to store the 1st complement of var
So, binary form of var is
0000 0000 0000 1010
1's 1111 1111 1111 0101
Here, MSB is 1
i. e number is negative
And we all know, any negative numbers store in computer as 2's complement
So we have to change this binary number in 2's complement
1111 1111 1111 0101
1's 1000 0000 0000 1010
2's 1000 0000 0000 1011
After converting it into decimal number we will get - 11
So , correct answer will be - 11
Thank you! ❤️
Wat's MSB
Naveen Chowdary MSB stands for *most significant bit*. In a binary number , the leftmost bit is referred to as MSB . And it has the largest place value . In simple words if the binary number is 0001 MSB is 0 . And if it is 1001 then MSB is 1 being the leftmost .
Thank u soo much for ur effort...i understood so clearly..🤗🤗
I understood tq for explanation
@Ashutosh Naytiyal Y becoz ,
The MSB gives the sign of the number (sign bit), 0 for positive and 1 for negative.
6/10. That was an excellent and challenging quiz! I am 37 and have never taken a timed quiz and am learning C for fun. This is the highest quality education I've ever received!
That's a responsible citizen
Huh
Me too
10th question solution :
k=10
In binary form : 0000 0000 0000 1010
now apply ~ to k , result : 1111 1111 1111 0101
but in the printf function they use %d instead of %u so the above value is too large to store in %d format ,
so compiler will find its equivalent 2's complement value and stores .
Now 1111 1111 1111 0101 ( compiler check left most bit if it is 1 it put negative sign behind it )
2's complement of above value : 0000 0000 0000 1010 ( 1's complement )
+1
0000 0000 0000 1011
which equivalent to 11 .( since left most bit in 1111 1111 1111 0101 is 1 , compiler put negative sign)
so answer is -11 .
😢😢😭😭 I am trying my best to understand this question but I cannot understand it....
Pls Any one explain this
Why 2'complement is 0000 instead of 1111 last 4 bits ?
@@diyadey1654 same yrr samjha bhi hai aur nahi bhi🥲
I was never going to get that tf
thank you!
7/10
Thank you.... I am so afraid of C.. But now it's just fun to understand new concepts... Thank you guys.... Means a lot..
How do you clear all concept of C .becoz I am afraid of c.
Same 7/10
7/10
It could be useful for everyone, if you explain the reasons too.....
7
Ya it's true
10/10😎
Exactly
Yes. It would be more helpful with reasons.... 8/10
Good questions but include the explanation for that questions
8/10✌... Its very helpful for every one who is having the knowledge of source code languages and need reasons for wrong answers we have choosed.... mainly helpful for ENGINEERS 👩🎓👨🎓
5/10. It will be great if you please provide with a little explanation of each of these questions, sir
5/10 All The Questions Were Interesting 😊
Same😭
Can't thank you enough sir. Till now I've learnt a lot from you. Your explanations are so vivid and crisp.
No it's good..
8/10
Now I realize that I learned something.. Thank You so much😊
Last (10.) question : ~x = -x - 1 (since we have x = 10 in the video, the answer is -11)
thnx
7/10
Raised my confidence😊
10/10. Very easy because i am learning C in deep.
3/10... DONT MAKE FUN, I'M JUST A BEGINNER!
Keep it up bro
@@ronitbagade1384 thanks buddy!
Me 3/10
nisha same level 😅. Keep it up, one day we both will make history 😁
@@MJamesBond007 Ahaaa kkk
I am learning c++ and never learnt c but I secured 7/10
You should learn C before C++, you might get a better understanding of things.
That doesn't make sense bro..
if u were asian then u would be called worthless in these scores...
@@shujatalikhan5582 C and C++ are both the same compilers, except that C++ has something on top of it
If you know C++, then you basically know C too. There's a few minor differences in C, but C++ builds on C.
8/10. C has been my scariest nightmare I don't want to look back at. And now I feel kinda confident about it. Thank you so much for making me good at this 💜
7/10 is my score. Badhiya 👍
10) The code defines a main function, which is the entry point of the program.
Inside the main function, an unsigned integer variable var is declared and initialized with the value 10.
The printf function is used to print the bitwise complement of the variable var.
The ~ operator is the bitwise NOT operator, which flips all the bits of the operand.
In this case, since var is initialized to 10, which in binary is 0000 1010, applying the bitwise NOT operation will result in 1111 0101.
The printf function with the format specifier %d is used to print the result of the bitwise complement operation.
Therefore, the output of the program will be the decimal value of ~var, which is -11.
It will be much usefull if u give a hlf min xplanation on each qstn.
This is only for those who watched all the previous lecture ! On this channel
4 out of 10 😉 i just improved my self......
Thank you for this rapid fire quiz...
Please upload more videos like this... Thank you Neso Academy😊
My score is 7/10.
Mayank Jadhav
Okk sry and thanks for the reply
My score is 5/10
I want more video and after completing quiz we want description of each and every question
My score is:
#include
#include
Int main ()
{
Int var=5;
Printf("%d",var);
reurn 0;
}
10th Question By Kranthi Kumar
Format specifier in printf function is "%d", so the value of ~var is asked in signed integer.
So as we know that, for a signed integer
var + ~var = -1
Now from given var = 10
~var = -11.
If it is asked as
printf("%u", ~var);
As %u is for unsigned format specifier
Then var + ~var = 2^32-1
Here 2^32-1 is maximum value for an integer in 4bytes.
And the answer becomes,
~var = 2^32-1-10
~var = 2^32-11.
I had a 5/10 but without the time limit I would've gotten a 9/10, thanks Neso academy ❤️
5/10
first attempt was great for learning,
thank you sir.
Not bad i proved my self wat iam 😍my score 6/10
#include
int main()
{
printf("9/10 JUST A BEGINNER");
return 0;
}
Q8. should have output 5 as int var = var will find var in global scope and will get value 5 and assign it to int var in local scope .So shouldn't be output 5??
Woahh 1/10!!
All questions timed up! except first. ;)
5/10
this quiz is so helpful to understand how i learnt c program!!.
9/10 .
last one i totally forgot,
need to see one moe time
Simple instead of doing 1's and 2's compliment,just do -(a+1)=-(10+1)=-11
Q8,9&10-- int main() should return a value (exit status) because return type of main() is int.
Sir, plz upload more videos regularly, don't leave 1 video in a week or 2-3, bcoz we need to complete it fast.. N u'r videos are mosttt helping So, we need more lectures on C only byy U.. ThankU..
Can u explain last questions solution
ya please
in binary 10 is writen as 00001010,
but the meaning of the sign ~ means a bitwise inversion on a value.
wherever you see 0, it will be 1
and wherever you see 1, it will be 0
therefore from the above binary for 10, it will be
00001010 ==== 11110101 which in decimal representation is 11
is that clear?
9/10
I enjoyed a lot
I've mistaken in question no 8, I think global variable and local variable contains the global value but I was wrong it's contains the garbage value🙂
I still don't get it. Please explain.
9/10. I missed question 8. int var = var; I thought var is globally scoped thus its value will be assigned to the locally just created var.
since var is declared as an int it will look for the value of "var" no ascii table to it will be not assigned as 5 but as a sum of other values in the ascii table
Pedro Lopes I think the answer was not for me.
Pedro Lopes Are you serious? what's up.with ascii here??? as you can see var is declared in both global scope and local scope. the local one is initialized from itself because the fact that when declaring a variable it can be used before the end of declaration settlement. int i = 10, y = i; i haa been allocated in memory then its value in memory is assigned to y before i came to life.
because its declared int var = var for example int var = A; if you printf var you'll get var = 65, however what you want to do is char var = 'var'; then it will print u the string var.
Pedro Lopes Totally wrong. it doesn't have any relation with ASCII. as you can see printf("%d", var); which means printing a variable of type integer. where did you find ASCII here??!!!
#include
int main() {
unsigned int var = 10;
printf("%d", ~var);
}
a) 10
b) -10
c) -11
d) -5
from above question var=10(1010)
and ~var=~10, that is (0101) which is equal to 5
but the answer is option C
how it is possible?????
5:38 8/10... learnt a lot! thankss NA!💌
Sir please upload all the lectures. I'm watching from bangladesh.
Great video but please also give explanations for right answers.
9 bro, But it's awesome you provide such a amazing series of c language.
thankyou brother :)
Well , this is the first time i am learning programming and the quiz taking method is amazing. Well, in this quiz , i got 6/10 but, in shaa allah , in the future i will definitely get good marks.
And thanks by the way your method of teaching is excellent.
Please do update us with the new videos. Don't Leave us Learning in the middle. Please release the new videos in the series after operators or do notify us by when they will be available on your channel. Please Don't hinder the learning process in between. The videos and content are awesome and easy to understand. Please upload next videos in the series please.
6/10 really help full sir
I got 6/10🙄
Same
Same. 😛
same😅
Cheer's 💃for all who get 6 out of 10😜
The last question ❓ is extraordinary 😂😂😂😅😅👍👍😓😔
u are learning it thus yr?
8/10 NOW I'M QUITE CONFIDENT THANKS A LOT SIR.EVERY STUDENT MUST REFER TO THIS CHANNEL FOR BETTER UNDERSTANDING OF C LANGUAGE.THE WAY OF TEACHING IS QUITE EASY AND SIMPLE.
8/10 thanks for the questions:-) those are very interesting:-)
6/10, n i did my best..
8/10,also I've to attempt these questions in half the time as I'm watching this video in 2X, nice problems
Me at 1.75 😂
8|10
6/10 There's scope of improvement for me and I'll try my best to get a better score, thanks to the NESO Academy for making this series so interactive!
7/10
Pls provide explanation also it will help to get more knowledge about the question
There's no point of these questions if you don't explain why. Anyone can google the answers but they won't understand how this works
Its true
Anmol Jain exactly
the questions are from previously discussed topics
If they don't no the answer means then they used to do google search
If you know what you got wrong, you know what topics to focus on for revision.
5/10 and 3 are 50-50choices 2 wrong
There was no concept related to ques 8 was taught... In scoping topic this concept was not mentioned...
8 it was extreme you know? I don!t think i can see this kind of questions anywhere else. Its amazing
Bro 10 th Question answer is -10 right?
7/10 only, I cant understand question 10 even after going through the comment section...someone pls help .
STUDY ABOUT BITWISE ~ OPERATOR YOU WILL UNDERSTAND .
AND YOU WILL ALSO STUDY 2'S COMPLIMENT OF BITWISE OPERATOR
My score is 8/10
Same
so baddddddd..
My score is just 6/10
7/10. I missed Q7,8,10. anyone who can explain please assist a fellow brother. Nevertheless NESO Academy for global president. 😍😍
5/10 I am in between journey of learning C it was fun activity solving unedr time limit.
9/10 ,sir can u please explain the extern keyboard
Its too easy , take a look on google
Plzz give the explanation for last question...~this symbol has a aski velue -1 ?
~ it was for NOT gate or negative value if 101 is then out will be 010 in binary
6/10🥴🥴
I'm a beginner 😂
Same
7/10 This was an excellent test of knowledge🎉. I now have a better understanding of the concepts I failed😊
9/10. Its very intersting and quite challenging task, i've gained some knowledge in c programming
someone pls explain Q.10...
actually go through 1's compliment then again 2's complement but still I don't get the value as -11 .
did u get the ans?
Simple instead of doing 1's and 2's compliment ,negation of a value is simply -(a+1) =-(10+1)=11
3/10 biginer,🤗🙏
😊
Good yaar👏
9/10
Lost to last question
5/10 so funnny though......i m having my semester on tuesday and i am learning the whole syllabus now....thank you so much sir,,,,,
Madhu Reyn same 😂😂
9/10
Q7 i wasnt sure if
int a;
was a declaration and a definition if there was no assigment but i know it has either 0 or garbage value, i have in mind that assigment changes sth, i dont know what those mean: stack, heap, and that is confusing me
My score was 7/10
Sir I'm score 8/10
Sir you don't explain reasons of every questions. So plz give every questions ans with it's proper solutions
I'm waiting for the DS lectures ...I have to cover my syllabus..
7/10 😅😅
5/10 .it is my score .
10/10
Can you explain 10th question please!
can somebody explain question 10?
Yes ! In last question printf is present outside of the main function so it will give the compiler error.
~ (tilde) bitwise operator means works on bits (binary value) of a value..
00001010
When applied ~
1 becomes 0 and 0 becomes 1
So
Step a: 11110101
On this you have to apply 2's compliment
1's compliment 00001010
2's compliment + 1
00001011
Which is binary value of 11 but as we actually got a negative value in step a so final answer is - 11.
Always remember that, the negation of a number is adding a 1 and then writing minus..
Ex ~8=-[8+1]
=-9
~ is a bitwise complement operator
10 = 1010
1010+1=1111=(-11)
B
9 ... Last wala wrong ho gya
Explain 10 question
A uint can't become negative, no matter what it's binary value is.
The uint of decimal value 10 is stored in the memory as :
0000 1010;
The tilde(~) operator flips all bits of the "var" variable, which yields this binary value :
1111 0101.
This equals to 4294967295-8-2 = 4294967285, which does not show up in the answers.
@@rafca7408
The %d tells printf to treat argument as signed int, so you will get -11 on machines that use 2-complement to represent negative integers. But to really nitpick, the C standard does not mandate 2-complement, so correct answer is actually "output is undefined".
@@sverkeren oh cool! Thanks!
Please explain last question
8/10 it's really amazing 🔥
sir ,your teaching method is awesome and i love it.
0/10 😭 iam now working now in microsoft.
not a problem...
6 -> right answers
1) ,6) ,8) wrong
7) unattempted
I got 6/10
Although I know, that extern question but I couldn't finish reading before the time elapse.
8/10 explain the last one please
10.Assume that thecompiler uses 4 bytes or 32 bit. The value of 10 in binary is(for 4 bytes):0000 0000 0000 0000 0000 0000 0000 1010
so when you use ~var, the value becomes in binary is(for 4 bytes): 1111 1111 1111 1111 1111 1111 1111 0101. So the decimal value is : 2^31+2^30+2^29+2^28+2^27+2^26+2^25+2^24+2^23+2^22+2^21+2^20+2^19+2^18+2^17+2^16+2^15+2^14+2^13+2^12+2^11+2^10+2^9+2^8+2^7+2^6+2^5+2^4+2^2+2^0=4294967285, this is unsigned int(%u). But we don't have this range in signed int(%d). We need to get the equivalent number in signed range.
Since the MSB is 1, we need to add negative to the first bit.
So the decimal value becomes: =
-(2^31)+2^30+2^29+2^28+2^27+2^26+2^25+2^24+2^23+2^22+2^21+2^20+2^19+2^18+2^17+2^16+2^15+2^14+2^13+2^12+2^11+2^10+2^9+2^8+2^7+2^6+2^5+2^4+2^2+2^0=-11
7/10 I'm student of IIT bhu and start learning coding from neso academy . Sir thanku from bottom of my heart❤
6/10
Happy with the result. I've learned a lot from this channel
7/10, just started, relearning the concepts
8/10
special thanks to u sir, u r way of expalination is worthfull to many other professors as well as students too
5/10 as a beginner feels k and hope I'll improve my c skills through ur videos
6/10. It's very help me understand more about C language. Thank you So Much :D
10 question is very complicated 😮
My score is 8/10..feeling good because this score was not expected from my side as I am a beginner.
7 sir
6/10 It was interesting quiz I really enjoy to learn this c programming, thanks a lot neso academy!
I love you @Neso Academy ❤️