Programming in C MCQ Quiz in मल्याळम - Objective Question with Answer for Programming in C - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Option 2 is correct.

C language stands between low-level and high-level language because it can create system-level programs for the operating system as like low-level languages, but it provides more abstraction than a low-level language.

C can also create application programs like high-level languages. C also have great readability and machine independence features like high-level languages. Consequently, C is a middle-level language.

Additional Information

• Low-level languages are those which are designed according to the machine and work according to the processor. It is similar to machine instruction.
• An assembly language is a low-level programming language in which there is a very strong correspondence between the program's statements and the architecture's machine code instructions
• A machine language is a binary code and binary code is a sequence of instructions and operands in binary that list the exact representation of instructions as they appear in computer memory

extern is not a valid identifier in C++. Because it is a keyword.

• Combination of letter (A-Z) and digit (0-9) are used as identifier but identifier should not start with digit

           Example: ext321ern and extern21 are valid identifier

• extern is a reserved word (keyword) in C++ programming language, hence it cannot be used as an identifier
• Underscore “_” is counted as a letter so it can be used as a variable name
• extern is a reserved word (keyword) in C++ but _extern is not reserved hence could be used as a variable name

           e.g. int extern; / is not a valid variable name
Additional Information 

Extern:-
The variables declared using keyword extern are called global variables.

These variables can be accessed throughout the program.

Keyword extern is used to declare a variable as a global variable inside any function.

Syntax is :

Extern datatype variable_name;

• Zero(0) appended in the beginning of an integer data type, says that variable ‘a’ is explicitly assign a octal constant
• It is used as a conversion specifier for decimal, 42 in octal number will get converted into decimal number; (42)₍₈₎= (4× 8)+2=34

int val[2] [4] = {1, 2, 3, 4, 5, 6, 7, 8} is a 2- dimensional array

treating the array declaration in the row-major ‘4’ comes under the first row and fourth column which has the index value val [0] [3] hence (d) is correct option.

The correct answer is 17.

Solution:-

At the last 1 return to the fun(5).

fun(5)+4=5, 5 is return to fun(4).

fun(4)+4=9, 9 is return to fun(3).

fun(3)+4=13, 13 is return to fun(2).

fun(2)+4=17, 17 is return to fun(1)

Now understand the working of the recursion function:-​

1. n=1 v=0
2. n=2 v=1
3. n=3 v=2
4. n=4 v=3
5. n=5 v=4, now n>=5 conditions true hence the function returns 1.

Finally, the fun() returns 17 to the printf().

Note:-

• The v=4 is used everywhere because it is a static variable.
• Recursion questions can be solved in many ways like with stack, tree, etc.
• But I prefer to do such questions with the tree.
• One thing here you should notice is that v is a static variable means memory is allocated only once at compile time.

Array:

Modified Array:

sum = 1 + 2 + 3 + 3 + 5 + 5 + 7 + 7 + 9 + 10 = 52.

Output: 52. 

Character ‘a’ has ASCII value 97, adding 2 will result in 99, which is the ASCII value for ‘c’.

Hence dereferencing q + 1 and q + 2 will give 0 and c respectively.

char x = 'a' + 2, that is, x == 'c'

So, p={'1','0','c'};

*((char*)q+1) = *(address of data '1' + 1) = *(address of data '0')  = 0; 

*((char*)q + 2) = *(address of data '1' + 2) = *(address of data 'c')  = c;

printf("%c, %c",*((char*)q+1),*((char*)q+2)) will print 0, c.

Concept:-

Meaning of i++: [post increment]

First, we read i then after reading the value of i we are going to increment i this is nothing but post-increment.

Meaning of ++i: [pre increment]

First we increment i then we read the value of I, this is nothing but pre-increment:

Analysis:-

• int m = ++p; /before storing the value increments the value of p by 1 therefore m = 11 and p = 11

• int n = p--; /value of p is stored in n then it is decremented by 1 therefore n = 11 and p = 10

• m + n = 11 + 11 = 22

Note:-

p = i++ , Here i value first stored in p and then i value incremented. 

p= ++i , here i value first incremented and then stored in p.

Answer: Option 3

Explanation: 

- arr denote the Base address of the array.

- &arr will not work the same way as some of you think.

printf("%ld\n",arr+1);

- In the First printf ; arr +1 , addition of 1 resulted in an address increment of 1 int data size hence 2000 + 4 = 2004 will be printed

printf("%ld",(&arr + 1));
- &arr will return a pointer to the whole array of 10 int, addition of 1 resulted in an address with an increment of 4 x 10 = 40 and hence 2040 will be printed in next line.

Hence Option 3 is the correct answer.

int  *ip = arr + 4 /ip is a integer pointer pointing to 5 element of the array