C Programming (Important Questions Set 1)

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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[@ParthKawatra]

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

[@duffyduck4065]

You can do that on your KZbin app settings

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[@user-cl3vv9vz8x]

char = -128 ~ 127 unsigned char = 0 ~ 255 char a = 255 a(2) = 1111 1111 1111 1111 = -1 answer a + 10 = 9

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[@nitinagrawal3184]

Finally You tube recommend something worth to watch.Thanks for the wonderful explanation.

[@impulsiveDecider]

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.

[@nandannswamy4250]

I never saw someone explains like this👏

[@connorhorman]

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).

[@sasishushma]

Yes..u r correct..the question should signed char c or unsigned char c to avoid any complier specific results..

[@royarijit998]

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..

[@gogogalian]

@@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.

[@java_shaileshkushwaha878]

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

[@connorhorman]

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.

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this is from the legendary category course on youtube , explaination on peek

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[@oviya.n1317]

The 2s comp representation of a signed num is -3 = 101 only so when treated as a positive num it is 5 .

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Thank you sir..

[@leoneruri6857]

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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.

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[@himanshubillorePeoplesoftHRMS]

in Q1, no need to write %s in the printf. Can just pass the string directly like printf("Hello\a ");

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Thank your for your lesson!

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awesome advanced questions on basics

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[@rupeshshete6162]

6:50 IMP 265mod256=9 remainder

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Thank you sir

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ربنا يباركلك يغالي

[@shreedharchavan7033]

Awesome questions

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great explanation sir, keep it up

[@bharath_rbp]

Suggest from where i can find more questions like these (covering conceptual and hidden meaning of a syntax)

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[@sid9145]

2:49 no one teach us this one...its very helpfull

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Superb explanation..👌🏻

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what if we changed the specifier %d to other specifier.. will it be the same or it depends on the specifier?

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Just Amazing....

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Super explanation

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Thank you for this 🙂

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Thank u

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thanks

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Sir please upload more videos for 2019 gate sir .nice explanation sir

[@phv7185]

Hi sir, I would like to ask a doubt in qs no. 4. Is it okay to declare i; only instead of int i; ? Will the compiler implicitly assumes the datatype int to the statement?

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Awesome Tutorial.....

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Very informative video

[@riannahenningham341]

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?

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[@Rahul-xm1vn]

What is the meaning of

[@shefatabassum5807]

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

[@funwithcomedy5859]

Can you make a question, answer video of c++ programming exact this type sir

[@accessdenied9393]

"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.

[@impulsiveDecider]

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)

[@learncseasily3385]

Nice Questions

[@aryanpatial650]

Sir in question number 3 maximum value of char should be 127 ?. I am confused sir

[@pratibhaias3629]

No! Char take 1 byte space having 8 bits ... and so the max range will be 255 ...

[@bharath_rbp]

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