Programming with 8085 MCQ Quiz in తెలుగు - Objective Question with Answer for Programming with 8085 - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 10, 2025
Latest Programming with 8085 MCQ Objective Questions
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Programming with 8085 Question 1:
Internal details of entity are specified by architecture body using _______ modeling style
Answer (Detailed Solution Below)
Programming with 8085 Question 1 Detailed Solution
Explanation:
Architecture body
The internal details of an entity are specified by an architecture body using any of the following modeling styles:
- As a set of interconnected components (to represent structure)
- As a set of concurrent assignment statements (to represent dataflow)
- As a set of sequential assignment statements (to represent behavior)
- Any combination of the above three.
Structural style
In the structural style of modeling, an entity is described as a set of interconnected components. Such a model for the HALF_ADDER entity, shown in Fig., is described in an architecture body as shown below.
architecture HA_STRUCTURE of HALF_ADDER is
component XOR2
port (X, Y: in BIT; Z: out BIT);
end component;
component AND2
port (L, M: in BIT; N: out BIT);
end component;
begin
X1: XOR2 port map (A, B, SUM);
A1: AND2 port map (A, B, CARRY);
end HA_STRUCTURE;
The name of the architecture body is HA_STRUCTURE.
Data flow style
In this modeling style, the flow of data through the entity is expressed primarily using concurrent signal assignment statements.
The structure of the entity is not explicitly specified in this modeling style, but it can be implicitly deduced.
Consider the following alternate architecture body for the HALF.ADDER entity that uses this style.
architecture HA_CONCURRENTof HALF_ADDER is
begin
SUM <= A xor B after 8 ns;
CARRY <= A and B after 4 ns;
end HA_CONCURRENT;
- The dataflow model for the HALF_ADDER is described using two concurrent signal assignment statements.
- In a signal assignment statement, the symbol <= implies an assignment of a value to a signal. The value of the expression on the right-hand side of the statement is computed and is assigned to the signal on the left-hand-side, called the target signal.
- Delay information is included in the signal assignment statements using after clauses
Behavioral style
The behavioral style of modeling specifies the behavior of an entity as a set of statements that are executed sequentially in the specified order.
- This set of sequential statements, which are specified inside a process statement, do not explicitly specify the structure of the entity but merely specifies its functionality.
- A process statement is a concurrent statement that can appear within an architecture body.
Conclusion:
Any one of the above modeling styles can be used in VHDL to describe a digital model.
Programming with 8085 Question 2:
Which one of following is not synthesizable VHDL statement?
Answer (Detailed Solution Below)
Programming with 8085 Question 2 Detailed Solution
wait statement can not generate any hardware directly so it is non-synthesizable statements in VHDL.OPTION (3) is correct.
VHDL- It stands for very-high-speed integrated circuit hardware description language. It is a programming language used to model a digital system by dataflow, behavioral and structural style of modeling. In VHDL there are two types of statements which are following-
Synthesizable statements- The statement which directly can be used to generate the Hardware called Synthesizable statements.Ex- A 3-input AND gate, A 32-bit output bus.
Non-Synthesizable statements- The statements which can't make any hardware is known as Non-synthesizable. For generating delay we have a counter. So 'wait' as such will not do anything. As far as the simulation is concerned it will show the same output but after synthesis, it will not. Ex: wait, after statements.
IMPORTANT POINTS-
1. wait until- It suspends a process until a change occurs on one or more of the signals in the statement and the condition is evaluated to be true.
2. Wait for– It is useful in behavioral models and test benches.
3. Generate- VHDL provides the GENERATE statement to create well-patterned structures easily
4. Case- It works exactly the way that a switch statement in C works. Given an input, the statement looks at each possible condition to find one that the input signal satisfies. They are useful to check one input signal against many combinations.
Programming with 8085 Question 3:
__________ is a primitive that can execute code. It contains an instruction pointer (=program counter) and sometimes has its own stack
Answer (Detailed Solution Below)
Programming with 8085 Question 3 Detailed Solution
THREAD:
A thread is a basic unit of CPU utilization, consisting of a program counter, a stack, and a set of registers.
A thread of execution results from a fork of a computer program into two or more concurrently running tasks, a thread is contained inside a process.
Multiple threads can exist within the same process and share resources such as memory, while different processes do not share these resources.
So option (D) is correct.
Important Points
Process: A process is an instance of a computer program that is being executed. It contains the program code and its current activity.
Depending on the operating system (OS), a process may be made up of multiple threads.
Task: A task is a set of program instructions that are loaded in memory.
Kernel: A kernel is the central part of an operating system.
It manages the operations of the computer and the hardware, most notably memory and CPU time.
Programming with 8085 Question 4:
In VHDL all the statements written inside a process statement are _________
Answer (Detailed Solution Below)
Programming with 8085 Question 4 Detailed Solution
CONCURRENT STATEMENT:
Those statements which start execution or simulation at the same point in time.
The primary concurrent statement in VHDL is a process statement. A number of processes may run at the same simulated time.
Eg: process, component instance, concurrent signal assignment.
SEQUENTIAL STATEMENT:
Those statements which start execution in order or one after other.
Within a process, sequential statements specify the step-by-step behavior of the process, or, essentially, the behavior of an architecture.
They execute in the order in which they appear in the process.
Eg: if, for, switch-case, signal assignment.
So inside the process statement, all statements are concurrent and sequential both but sequential statements written only inside the process statement.
Programming with 8085 Question 5:
For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is
3000 MVI A, 45H
3002 MOV B, A
3003 STC
3004 CMC
3005 RAR
3006 XRA B
Answer (Detailed Solution Below)
Programming with 8085 Question 5 Detailed Solution
-
MVI A, 45H: This line of code loads the Accumulator (A) with the hexadecimal value 45.
-
MOV B, A: This line moves the contents of the Accumulator (A) into register B. So now, B also contains the value 45H.
-
STC: This command sets the carry flag.
-
CMC: This command complements the carry flag. Since the carry flag was set by the STC command, the CMC command will now unset it.
-
RAR: This command rotates the Accumulator (A) and the carry flag right through carry. Since the carry was unset, the 45H (binary 0100 0101) will be rotated right to become 22H (binary 0010 0010).
-
XRA B: This command performs an exclusive OR operation between the contents of the Accumulator (A) and register B. Since A now contains 22H and B contains 45H, the result of the exclusive OR operation is 67H.
So, after the execution of the program, the content of the accumulator will be 67H.
Programming with 8085 Question 6:
Consider the following 8085 instructions
XRA A
MVI B, 4AH
SUI 4FH
ANA B
HLT
The content of register A and B are respectively ________
Answer (Detailed Solution Below)
Programming with 8085 Question 6 Detailed Solution
Let's analyze each instruction step by step to determine the final contents of registers A and B.
XRA A:
This instruction performs an XOR operation between register A and itself (A = A XOR A).
XORing a number with itself results in 0.
So, after this instruction, register A = 00.
MVI B, 4AH:
This instruction moves the immediate value 4A into register B.
So, after this instruction, register B = 4A.
SUI 4FH:
This instruction subtracts the immediate value 4F from register A with borrow.
Since register A is 00 after the XRA instruction, performing A = A - 4F results in underflow and the value will be calculated as follows using 2's complement.
00 - 4F in 8-bit arithmetic:
A = 00
Immediate value is 4F
2's complement of 4F = (FF - 4F + 1) = B1
So, after this instruction, register A = B1.
ANA B:
This instruction performs an AND operation between register A and register B (A = A AND B).
Register A is B1 and register B is 4A.
Performing A AND B:
B1: 1011 0001
4A: 0100 1010
Result: 0000 0000
So, after this instruction, register A = 00.
HLT:
This instruction halts the processor, and thus no further operations are performed.
Summarizing the final contents:
Register A = 00 (after ANA B instruction)
Register B = 4A (after MVI B, 4AH)
So the final contents of register A and B are 00 and 4A respectively.
Since none of the provided answer options match with the result we've calculated:
05, 4A
4F, 00
B1, 4A
None of these
The correct option is: 4) None of these
Programming with 8085 Question 7:
An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is
MVI A, 07H
RLC
MOV B, A
RLC
RLC
ADD B
RRC
Answer (Detailed Solution Below)
Programming with 8085 Question 7 Detailed Solution
MVI A, 07 H ⇒ 0000 0111 ← The content of 'A'
RLC ⇒ 0000 1110 ← The content of 'A'
MOVT B, A ⇒ 0000 1110 ← The content of 'B'
RLC ⇒ 0001 1100 ← The content of 'B'
RLC ⇒ 0011 1000 ← The content of 'B'
ADD B
\(\frac{A \space 0000 \space 1110\\\ + \\\ B \space 0011 \space 1000}{0100 \space 0110}\)
\(RCC \rightarrow \frac{0010}{2}\frac{0011}{3} 23 H\)
Programming with 8085 Question 8:
Consider the following VHDL code. What does it represent?
process
begin
CLK <= '0' ;
wait for 20 ns;
CLK <= '1' ;
wait for 12 ns;
end process;Answer (Detailed Solution Below)
Programming with 8085 Question 8 Detailed Solution
Explanation:
It is possible to use case or loop statements within a process. The semantics and structure of these statements are very similar to those in other high-level programming languages like C or Pascal.
- An explicit wait statement can also be used to suspend a process.
- It can be used to wait for a certain amount of time or to wait until a certain condition becomes true, or wait until an event occurs on one or more signals.
Here is an example of a process statement that generates a clock with a different on-off period. Figure 2.6 shows the generated waveform.
This process does not have a sensitivity list since explicit wait statements are present inside the process. It is important to remember that a process never terminates.
It is always either being executed or in a suspended state.
NOTE:
All processes are executed once during the initialization phase of simulation until they get suspended. Therefore, a process with no sensitivity list and with no explicit wait statements will never suspend itself.
Conclusion:
Option 2 is correct.Programming with 8085 Question 9:
Insertion of a record in a circularly linked list involves the modification of:
Answer (Detailed Solution Below)
Programming with 8085 Question 9 Detailed Solution
It will require modification of 2 pointers, next of node N1 after which insertion is to be done and next of new node.
We can understand by given example-
Consider ACLL having 4 Nodes A, B, C, D
A New Node E is inserted between B & C:
So 2 pointer modification required.
Programming with 8085 Question 10:
The data type defined by user is:
Answer (Detailed Solution Below)
Programming with 8085 Question 10 Detailed Solution
Abstract data type-
An abstract data type is defined by its behavior (semantics) from the point of view of a user. For example, integers are an ADT, defined as the values ..., −2, −1, 0, 1, 2.
Important Points
1. Built-in data types-
These are the most basic data types. The term built-in means that they are pre-defined and can be used directly in a program. char, int, float, and double are the most common built-in data types.
2. Homogeneous data type-
It contains only a similar type of data. E.g- An array containing only integer values.
3. Real data type-
It is a data type used in a computer program to represent an approximation of a real number because the real numbers are not countable, computers cannot represent them exactly using a finite amount of information.
E.g-Rational numbers, Fixed-point numbers, Floating-point numbers, Decimal numbers.