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Write a M68000 assembly language subroutine MULSUM that takes an array named A containing n bytes of positive numbers, and fills two arrays, array B containing n words and array C containing n long words as follows:
Sum = 0;
for(i = 0; i) Sum = A[i] + Sum; if(Sum is an EVEN number) then
B[i] = Sum; else C[i] = Sum * A[i]; end if end for
In evaluating the expression Sum * A[i] you are NOT allowed to use the MULTU, or MULTS instructions, hence, you are to find another way to calculate this expression.
Main program along with data section is shown below. This program passes all the required parameters to subroutine MULSUM by pushing them on stack.
Note: There is no overflow occurring during any operation.
a) Write the main program
b) Write the subroutine MULSUM
c) Show the stack frame for entire program and clearly label SP at different stages of program.
General Bus Operation The 8086 has a joined data and address bus commonly referred to as a time multiplexed address and data bus. The major reason behind multiplexing address
ali is impressed_____ ahmed''s grades.
to separate positive and negative numbers
SCAS : Scan String Byte or String Word:- This instruction scans a string of words or byte for an operand word or byte specified in the register AL or register AX. The string i
Write a program that performs the addition, subtraction, multiplications, division of the given operands. Perform BCD operation for addition and subtraction.
RCR: Rotate Right through Carry:- This instruction rotates the contents bit-wise of the destination operand right by the specified count through carry flag (CF). For each operati
Arithmetic Instruction : These instructions are usually perform the arithmetic operations, like subtraction ,multiplication, addition, and division along with th
Explain the architecture of the file transfer protocol ftp in terms of clients, servers, sockets
There are two parts to this assignment. The first part has you reading 4 integers representing; #QUARTERS, #DIMES, #NICKELS & #PENNIES, respectively. Your program should compute t
Segment Registers The 8086 addresses a segmented memory unlike 8085. The complete 1 megabyte memory, which 8086 is capable to address is divided into 16 logical segments.Thusea
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