Computer Knowledge

Programming Fundamentals

2,611 Questions

Programming fundamentals form the core logic of software development and computer science. This includes variable declarations, pointer assignments, loop iterations, and exception handling. These technical topics are regularly tested in computer knowledge and IT officer competitive examinations.

Variables and arraysPointer assignmentsLoop iterationsException handling blocksCompile time errorsFunction references

Programming Fundamentals Questions

Multiple choice

Which of the following statements is FALSE regarding functions in C?

  1. A function cannot return more than 1 value.

  2. It is a group of statements that performs a task.

  3. Functions cannot be defined by the user.

  4. The return statement is used to return a value from a function.

  5. A function can call any number of functions.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Although most of the programs can be made using inbuilt functions like printf(), scanf(), the user has the option of declaring his own functions. Such functions are known as user defined functions. The main objective of a user defined function is to save time and effort since,  a method can be called as many times as required.

Multiple choice

When we declare a pointer to a structure, which of the following operators is used to access the structure variable through the pointer object?

  1. .(dot)

  2. * (Asterisk)

  3. & (Ampersand)

  4. -> (Arrow)

  5. # (Hash)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The -> (arrow) operator is used to reference individual members of a structures.The arrow operator is used with a pointer to an object. For example, if 'stud' is a pointer object pointing to a structure 'student', then the following statement can be used to access the member of the structure. printf("%d",*s->rollno);

Multiple choice

Which of the following statements is FALSE regarding switch case statement?

  1. We can have any number of case statements within a switch.

  2. When a break statement is reached, the switch terminates.

  3. Using default in switch case is optional.

  4. The value to be compared can be of any type.

  5. Every case need not have a break statement.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The value to be compared should be of integral type. We cannot compare a floating point value using a switch-case.

Multiple choice

Which of the following statements is FALSE regarding structures?

  1. Structures are user defined data types.

  2. Structures cannot be nested.

  3. Structures can contain different types of variables.

  4. Structures are used to represent a record.

  5. The dot operator is used to access the structure members.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Structures can be nested. A structure can contain a reference of another structure. For example : struct person {   char name[10]:   int age; }; struct emp { int empno; personal person; }; In the above code the 'person' variable has a reference of the 'person' structure as its datatype. If an object of the structure 'emp' is declared, we can not only access the member of 'emp' but also of the 'person' structure.

Multiple choice

Which of the following is NOT an operator in C?

  1. sizeof

  2. !

  3. ~

  4. | |

  5. scope resoultion operator (: :)

Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The scope resoultion operator is not a valid C operator. It is used in C++ to specify the scope of a variable or object.

Multiple choice

The default parameter passing mechanism is called __________.

  1. call by value

  2. call by name

  3. call by reference

  4. call by address

  5. call by default
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The call by value method of passing arguments to a function copies the actual value of an argument into the formal parameter of the function. In this case, changes made to the parameter inside the function have no effect on the argument. By default, C uses call by value to pass arguments. In general, this means that code within a function cannot alter the arguments used to call the function. 

Multiple choice

Which of the following statements is TRUE regarding the 'switch' statement?

  1. continue can be used but break cannot be used

  2. continue cannot be used but break can be used

  3. Both continue and break can be used

  4. Neither continue nor break can be used

  5. We cannot use switch statement in nested form
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The break is used to break out of the case statements. Break is a keyword that breaks out of the code block, usually surrounded by braces, which it is in. In this case, break prevents the program from falling through and executing the code in all the other case statements. 

Multiple choice

In a bottom-up evaluation of a syntax directed definition, inherited attributes can

  1. always be evaluated

  2. be evaluated only if the definition is L-attributed

  3. be evaluated only if the definition has synthesized attributes

  4. never be evaluated
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Every S (synthesized) -attributed definitions is L- attributed. So in a bottom-up evaluation of SDD inherited attributes can be evaluated only if the definition has synthesized attributes.

Multiple choice

Which of the following statements is FALSE?

  1. In statically typed languages, each variable in a program has a fixed type

  2. In un-typed languages, values do not have any types

  3. In dynamically typed languages, variables have no types

  4. In all statically typed languages, each variable in a program is associated with values of only a single type during the execution of the program
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

(1) True for statically typed languages where each variable has fixed type. Similarly (4) is also correct. (2) True, in un-typed languages types of values are not defined. But option (3) is false, since in dynamically typed language variables have dynamically changing types but not that they have no type.

Multiple choice

Consider the syntax directed definition shown below, S $\rightarrow$ id : = E {gen (id.place = E.place;);} E $\rightarrow$ E1 + E2 {t = newtemp( ); gen(t = E1.place + E2.place;); E.place = t;} E $\rightarrow$ id {E.place = id.place;}

Here, gen is a function that generates the output code, and newtemp is a function that returns the name of a new temporary variable on every call. Assume that ti's are the temporary variable names generated by newtemp. For the statement 'X : = Y + Z', the 3-address code sequence generated by this definition is

  1. X = Y + Z

  2. t1 = Y + Z; X = t1

  3. t1 = Y; t2 = t1 + Z; X = t2

  4. t1 = Y; t2 = Z; t3 = t1 + t2; X = t3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In 3-address code we use temporary variables to reduce complex instructions so here $$ t_1 = Y \\ t_2 = Z \\ t_3 = t_1 + t_2 \\ x = t_3 \\ $$

Multiple choice

In a bottom-up evaluation of a syntax directed definition, inherited attributes can

  1. always be evaluated

  2. be evaluated only if the definition is L-attributed

  3. be evaluated only if the definition has synthesized attributes

  4. never be evaluated
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Every S (synthesized) -attributed definitions is L- attributed. So in a bottom-up evaluation of SDD inherited attributes can be evaluated only if the definition has synthesized attributes.

Multiple choice

Which of the following statements is FALSE?

  1. In statically typed languages, each variable in a program has a fixed type

  2. In un-typed languages, values do not have any types

  3. In dynamically typed languages, variables have no types

  4. In all statically typed languages, each variable in a program is associated with values of only a single type during the execution of the program
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

(1) True for statically typed languages where each variable has fixed type. Similarly (4) is also correct. (2) True, in un-typed languages types of values are not defined. But option (3) is false, since in dynamically typed language variables have dynamically changing types but not that they have no type.

Multiple choice

Consider the syntax directed definition shown below, S $\rightarrow$ id : = E {gen (id.place = E.place;);} E $\rightarrow$ E1 + E2 {t = newtemp( ); gen(t = E1.place + E2.place;); E.place = t;} E $\rightarrow$ id {E.place = id.place;}

Here, gen is a function that generates the output code, and newtemp is a function that returns the name of a new temporary variable on every call. Assume that ti's are the temporary variable names generated by newtemp. For the statement 'X : = Y + Z', the 3-address code sequence generated by this definition is

  1. X = Y + Z

  2. t1 = Y + Z; X = t1

  3. t1 = Y; t2 = t1 + Z; X = t2

  4. t1 = Y; t2 = Z; t3 = t1 + t2; X = t3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In 3-address code we use temporary variables to reduce complex instructions so here $$ t_1 = Y \\ t_2 = Z \\ t_3 = t_1 + t_2 \\ x = t_3 \\ $$

Multiple choice

The following program fragment is written in a programming language that allows global variables and does not allow nested declarations of functions. global int i = 100, j = 5; void P(x) { int i = 10; print(x + 10); i = 200; j = 20; print (x); } main() {P(i + j);}

If the programming language uses static scoping and call by need parameter passing mechanism, the values printed by the above program are

  1. 115, 220

  2. 25, 220

  3. 25, 15

  4. 115, 105
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In static scoping the variables are initialized at compile time only So i =100 & j =5 P(i + j) = P(100 + 5) = P(105)

So x =105 x + 10 = 105 + 10 = 115 So 115 and 105 will be printed