Computer Knowledge

Programming Fundamentals

2,381 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 is/are false about Just-in-time compilation?

  1. It specifies and converts the byte code into executable format.

  2. It is responsible for the runtime type safety and assembly verification.

  3. It provides the assembly metadata for any illegal accesses and handles violations appropriately.

  4. It handles reading the metadata by reflection.

  5. Options 3 and 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

JIT compilation does not have such reading of metadata information.

Multiple choice

Consider the following program segment for a hypothetical CPU having three user registers Rl R2 and R3. Instruction Operation Instruction Size in word. MOV Rl,5000 ; Rl ¬ Memory[5000] 2 MOV R2,(R1) ; R2 ¬ Memory[(Rl)] ADD R2,R3 ; R2 ¬ R2 + R3 1 MOV 6000, R2 ; Memory[6000] ¬ R2 2 HALT; Machine halts 1 Let the clock cycles required for various operations be as follows. 1 Register to/from memory transfer : 3 clock cycles ADD with both operands in register: 1 clock cycle Instruction fetch and decode : 2 clock cycles per word The total number of clock cycles required to execute the program is

  1. 29

  2. 24

  3. 23

  4. 20
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Instruction fetch/decode: 2 cycles per word. Memory transfer: 3 cycles. ADD: 1 cycle. MOV R1,5000 (2 words): 4 cycles + 3 (mem) = 7. MOV R2,(R1) (1 word): 2 cycles + 3 (mem) = 5. ADD R2,R3 (1 word): 2 cycles + 1 = 3. MOV 6000,R2 (2 words): 4 cycles + 3 (mem) = 7. HALT (1 word): 2 cycles. Total: 7+5+3+7+2 = 24.

Multiple choice

Union definition does not allocate memory.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A union definition itself is just a template or a type declaration; it does not allocate memory until a variable of that union type is instantiated.