In a 3 byte instruction of 8085, the first, second and third byte respectively indicate ?->(Show Answer!)

Question Set

1. In a 3 byte instruction of 8085, the first, second and third byte respectively indicate

Show Similar Question And Answers

QA->There are 3 numbers .The product of the first and the second is The product of second and the third is The product of the first and the third is Find the numbers....

QA->P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?....

QA->A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?....

QA->A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?....

QA->Present ages of Kiran and Syam are in the ratio of 5:4 respectively. Three years hence, the ratio of their ages will become 11:9 respectively. What is Syam"s present age in years?....

MCQ->In a 3 byte instruction of 8085, the first, second and third byte respectively indicate....

MCQ->Assertion (A): A two byte instruction of 8085 has an operation code in first byte and operand / address in the second byte.Reason (R): If an instruction is too long it is possible to reduce its length by a number of methods.

MCQ->What will be the output of the C#.NET code snippet given below? byte b1 = 0xAB; byte b2 = 0x99; byte temp; temp = (byte)~b2; Console.Write(temp + " "); temp = (byte)(b1 << b2); Console.Write (temp + " "); temp = (byte) (b2 >> 2); Console.WriteLine(temp);....

MCQ->Consider the following statements about 8085 The instruction LDA uses direct addressing mode.The instruction MVI uses direct addressing mode.The instruction MVI uses immediate addressing mode.The instruction LDA uses immediate addressing mode. Which of the above are correct?....

MCQ->What will be the output of the C#.NET code snippet given below? byte b1 = 0xF7; byte b2 = 0xAB; byte temp; temp = (byte)(b1 & b2); Console.Write (temp + " "); temp = (byte)(b1^b2); Console.WriteLine(temp);....

Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use | Powered By:Omega Web Solutions
© 2002-2017 Omega Education PVT LTD...Privacy | Terms And Conditions Question ANSWER With Solution