Comments in the codes are used to explain what is happening in the code. It is helpful as if one coder writes code he also writes comments so that other developers can understand easily about the coding logic.
Please login to submit your explanation
She want to use comments in the program.
You can check your performance of this question after Login/Signup
pointer c, d
a = 30c = &a
c = &a
d = c
a = a + 10
Currently there is no PrepInsta Explanation. Did you know?
You can also submit your own version of explanations under user explanation section and also view other submitted explanations by other users.
option b: is correct
&a = 4165 //given
a = 30, c = &a = 4165 // 3rd line
c = &a = 4165 //(again) 4th line
d = c = 4165 // 5th line
a = a+10 = 40 // 6th line
print *c = *(4165) = 40 // 7th line
so finally answer is (B) 40
there is one mistake in this question, question should be in this format...
pointer c, d
a = 30
c = &a
d = c
a = a + 10
they asked us to print *c which means value not address of pointer.so c points to a value which is first 30 and after increment 40
here initially the value of a is 30 and c store a\'s address now if we update the value of a new value of c is updated.
option b is correct beacuse c refer to a ultimately.
For signed integer one bit is reserved for sign and the range is -32 to 31 .Option 3 is correct ie 32 can not be represented in this data type.
Ok let\'s start calculation. 6 bits means total combinations of bits are 2^6 = 64. Now values up to 63 you can store including 0.
But your data type is signed means -32 to +31 you may store which is exactly 64 combinations.
Now come to the question, you can not store: 32 .
32 option c
32 or 64
This may be the answer you seek:
1 bit -> 2 possible states
2 bits -> 4 possible states
3 bits -> 8 possible states
4 bits -> 16 possible states
5 bits -> 32 possible states
(I hope you see the pattern here)
You need 5 bits to represent all possible letters in the language (28 > 16 & 28 < 32). Since you are told you are storing an array of letters, you will need a total of 5*7 bits = 35 bits to represent all words in the language. (You actually need 29 symbols per array element, not 28, since you need a \'null\' symbol to signify shorter than 7 letter words, but that doesn\'t change the answer.)
Now, in reality no sane person would actually do this in practice on a real computer. They would just store 7 bytes (or possibly 8, for a terminating null character (C-style string) to simplify processing.)
0 to 1000
0 to 1024
0 to 1023
0 to 1025
2^10=1024 so range of unsigned integer will be 0 to 1023
pow(2,n)=pow(2,10)=> 0 to 1023
a = 200 b = 10
a = 10 b = 200
a = 50 b = 100
a = 100 b = 50
0 to 255 integers are accepted when a=10 b=200 c =410 > 255 when a=200 b=10 c=220<255 when a=50 b=100 c=250<255 when a=100 b=50 c=200<255 so, option 1 gives wrong output
answer is B.a=10 b=200
so c became 410 which is greater than 255
You can read more about it here - https://prepinsta.com/c-programming/assembler-compiler-interpreter-linker-loader/
a program used with a compiler or assembler to provide links to the libraries needed for an executable program.
A loader is a major component of an operating system that ensures all necessary programs and libraries are loaded, which is essential during the startup phase of running a program. It places the libraries and programs into the main memory in order to prepare them for execution.
As boolean can store only two values. True or False
Answers A and D are the same
Stack is an Application of Arrays. It is not a data type
Please login to report
Buy TCS NQT Paid Materials
Join TCS NQT Online Classes
Personalized Analytics only Availble for Logged in users
Analytics below shows your performance in various Mocks on PrepInsta
Your average Analytics for this Quiz
Login/Signup to comment