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

What are the types of variable f2 and p2 in the following declaration?

#define floatptr float * #define int * intptr

floatptr f1,f2; intptr p1,p2;

  1. float, int pointer

  2. float pointer, int pointer

  3. float, int

  4. float pointer, int
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The macro floatptr expands to 'float *', so 'floatptr f1,f2' becomes 'float *f1,f2'. This declares f1 as a pointer to float, but f2 as a regular float (the * only binds to f1). Similarly, 'intptr p1,p2' becomes 'int *p1,p2', making p2 a regular int. This is a classic macro pitfall.

Multiple choice

struct node nptr,*sptr; /*pointers for linked list/ for(nptr=sptr;nptr;nptr=nptr->next) { free(nptr); }

Releases memory from a linked list. Which of the following accurately describes how it will work?

  1. It will work correctly since the for loop covers the entire list.

  2. It may fail since each node “nptr” is freed before its next address can be accessed.

  3. This is invalid syntax for freeing memory.

  4. The loop will never end.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The loop frees nptr BEFORE reading nptr->next in the next iteration. After free(nptr), accessing nptr->next is undefined behavior - the memory has been deallocated. The correct approach is to save nptr->next BEFORE freeing nptr: next = nptr->next; free(nptr); nptr = next;

Multiple choice

#define big(a,b) a > b ? a : b #define swap(a,b) temp=a; a=b; b=temp; void main() { int a=3,b=5,temp=0; if ((3+big(a,b)) > b) swap(a,b); printf("%d %d",a,b); }

  1. 3 0

  2. 5 3

  3. 3 5

  4. 5 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The macro big(a,b) returns the larger value, so big(3,5) returns 5. The condition (3+5)>5 evaluates to true (8>5). When swap(a,b) expands, it executes temp=a; a=b; b=temp; making a=5 and b=3. The printf outputs "5 3". The key point is that the swap macro lacks braces, but it still executes sequentially.

Multiple choice

What does the line containing "break;" do in the following code?

void afunction() { if(1) { break; afunction(); printf("Error"); } }

  1. Breaks out of the if statement

  2. Exits the function

  3. Compiler error

  4. Nothing
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The break statement can only be used inside loops (for, while, do-while) or switch statements. Using break inside an if statement that is not inside a loop or switch causes a compilation error. The compiler will reject this code because break has no valid target to break out of.