Question

Se al A c and code frogments MUST use 32 ぁ ar movennes (anor be "nedfra-ne ibrary function or procedural calls ino- Pe Peue c for ll codr ond code frogments only othu ey document your code with Comments unless the code size is Drow Fow Chort for any flow of control involving branches and loops Use PoCedures as modules to organize most of the code and code fragments. In doing be sure to follow stack discipline, and use of activation frames 1 Assume the registers are initialized to the indicated values: EAX: 12345D, EBX: 9528D, ECX:-1275D, EDX:-3001D. Determine the value in the destination register, (and the values of the ZF, PF, SF, OF, CF flags), aftes execution of each of the following instructions.Choose and indicate whether the instructions are executed in sequence; or considered executed independently of each other. a) And EAX, EBX b) Not EDX c) SHL EAX, CL d) SHR EAX, CL e) Or EAX, ECX f XOR EAX, ECX 8) SAL EBX, 9 h) SAR EAx, 5 i) ROR EAX, CL j) ROL EAX, CL k) RCL EAX, CL l) RCR EAX, CL Do the following problems. a) What are the disadvantages of passing parameters via the stack? (Hint: compare advantages of the alternative methods. What these other options?) b) Write a code fragment that illustrates how value/s) can be returned from a PRO stack, and then manipulated in the calling procedure. Write a code fragment to illustrate how storage for local variables in a PROC

1) Assume the registers are initialized to the indicated values:

Answer 1

Solution: Note: More than one question is asked, please post as separate questions 2. The values of registers in binary form are shown below: EAX: 11000000111001 EBX: 10010100111000 EDX: 1111010001000111 a) And EAX, EBX The result of And The result is 8248D The values of status registers are shown below: operation is 10000000111000 . . . o ZF -0, as result is non zero o PF0, as result has odd parity o SF 0, the result is positive o OF 0, as there is no overflow o CF 0, as there is no carry present b) Not EDX . The result of NOT . The result is 3000D . The values of status registers are shown below: operation is 101110111000 o ZF = 0, as result is non zero o PF 1, as result has even parity o SF 1, as result is negative o OF 0, as there is no overflow o CF = 0, as there is no carry present