The ADD and SUB operators affect all the status flags according to the result of the operation. Give the outputs of the following sequential operations in assembly language and determine whether overflow flag (OF) or carry flag (CF) will be signaled. al, al, al, al, al, al, mox OFFh add 1 mox 7Fh add 1 80h mox add 80h

The ADD and SUB operators affect all the status flags according to the result of the operation. Give the outputs of the following sequential operations in assembly language and determine whether overflow flag (OF) or carry flag (CF) will be signaled. al, al, al, al, al, al, mox OFFh add 1 mox 7Fh add 1 80h mox add 80h

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter3: Data Representation

Section: Chapter Questions

Problem 8RQ: Why doesnt a CPU evaluate the expression 'A' = 'a' as true?

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Why doesnt a CPU evaluate the expression 'A' = 'a' as true?
1) Write a subprogram that accepts 4 values from the argument registers $a0 through $a3 and returns to the caller programme the largest value and the average of $v0 and $v1. The programme must be set up in the following manner: Subprogram largestAndAverage($a1, $a2, $a3, $a4) { int var0 = $a0, var1 = $a1, var2 = $a2, var3 = $a3; $s0 = getLarger($a1, $a2); $s0 = getLarger($s0, $a3); $v0 = getLarager(s0, $a4); // Largest is in $v0 $v1 = (var0 + var1 + var2 + var3)/ 4; // Aversge is in $v1 return;}Subprogram getLarger($a0, $a1) { $v0 = $a0 if ($a1 > $a0) $v0 = $a1 return;}Take note of how var0...var3 are used. The values of $a0 and $a1 (at least) must be placed on the stack since they are not immediately available when needed to compute the average because they are modified during the call to getLarger. You must use the getLarger subprogram displayed above to compute the greatest value for this issue, and it must be called before the average calculation. This indicates that $a0 and…
1) Write a subprogram that accepts 4 values from the argument registers $a0 through $a3 and returns to the caller programme the largest value and the average of $v0 and $v1. The programme must be set up in the following manner: Subprogram largestAndAverage($a1, $a2, $a3, $a4) { int var0 = $a0, var1 = $a1, var2 = $a2, var3 = $a3; $s0 = getLarger($a1, $a2); $s0 = getLarger($s0, $a3); $v0 = getLarager(s0, $a4); // Largest is in $v0 $v1 = (var0 + var1 + var2 + var3)/ 4; // Aversge is in $v1 return;}Subprogram getLarger($a0, $a1) { $v0 = $a0 if ($a1 > $a0) $v0 = $a1 return;}Take note of how var0...var3 are used. The values of $a0 and $a1 (at least) must be placed on the stack since they are not immediately available when needed to compute the average because they are modified during the call to getLarger. You must use the getLarger subprogram displayed above to compute the greatest value for this issue, and it must be called before the average calculation. This indicates that $a0 and…
Can someone help me draw a flow-diagram of this code in LC3 assembly language. Would appreciate it. .ORIG x3000 isPrime ST R1, SAVE_R1 ; Save the original value of R1 AND R1, R1, #0 ; Clear R1 (to use as a counter) BRz ZERO_CASE ; Handle the case when the input is 0 LOOP ADD R1, R1, #1 ; Increment the counter ADD R1, R1, #1 ; Check for divisibility by counter + counter DIV R0, R1 ; Divide the input by (counter + counter) BRnp IS_PRIME ; If the remainder is not 0, the number is prime BRz NOT_PRIME ; If the remainder is 0, the number is not prime IS_PRIME AND R0, R0, #0 ; Clear R0 ADD R0, R0, #1 ; Set R0 to 1 (prime) BRnzp DONE ; Branch to DONE NOT_PRIME AND R0, R0, #0 ; Clear R0 (not prime) BRnzp DONE ; Branch to DONE ZERO_CASE AND R0, R0, #0 ; Clear R0 (0 is not prime) DONE LD R1, SAVE_R1 ; Restore the original value of R1 RET ; Return from the function SAVE_R1 .BLKW 1 ; Storage for saving…
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Convert the following C++ programs into Pep/9 assembly #include <iostream> using namespace std; void times(int& prod, int mpr, int mcand) { prod = 0; while (mpr != 0) { if (mpr % 2 == 1) prod = prod + mcand; mpr /= 2; mcand *= 2; } } int main(){ int product, n, m; cout << "Enter two numbers: "; cin >> n >> m; times(product, n, m); cout << "Product: " << product << endl; return 0; } Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
Translate the following C++ program to Pep/9 assembly language. Code should also be commented. #include <iostream>using namespace std;int Lesser(int x, int y) { int least; if (x < y) least = x; else least = y; return least;}int main(){ int num1; int num2; cout << "Enter two integers: "; cin >> num1 >> num2 cout << "Lesser: " << Lesser(num1, num2); << endl; return 0;}
choose the correct answers (Multiple Choice Questions) 1- Which of the following is considered a keyword in C++ LANGUAGE?A) ForB) LongC) CinD) All of theseE) None of the above 2- In which part of the for loop termination condition is checked? for (I;II;III)A)IB) II C) IIID)IVE) None of the above 3)IF a= -8, b=3, c1 = --a + b, c2 = a-- + b then the value of c1 and c2 respectively are:A) -7 , -4B) -7 , -3 C) -4 , -4D) -6 , -6E) None of the above 5)The advantage of a SWITCH over an ELSE-IF statement is:A) A default condition can be used in switch B) The switch is easier to understandC) Several different condition can cause one set of statements to be executed in aswitch D) Several different statements can be executed in a switchE) All of the above 6)- In C++ LANGUAGE, cin and cout are the predefined stream …….A) operatorsB) functionsC) data typesD) objectsE) None of the above
Convert the following C++ program into Pep/9 Assembly. #include <iostream>using namespace std;void SetEqualGreater (int &num1, int &num2){ if (num1 < num2) num1 = num2; else num2 = num1;}int main() { int x, y; // Using Local Variables cin >> x >> y; SetEqualGreater(x, y); cout << "x = y = " << x << endl; return 0;} Submit: Source file along with screen captures showing the program running in the Pep/8 simulator.