Assume For Each Part Of This Problem That The EAX Register
A Assume For Each Part Of This Problem That The Eax Register Contains
Assume for each part of this problem that the EAX register contains F and the doubleword referenced by value contains FF FF FF 38. Determine whether each of the conditional jump statements causes a jump to dest.
#1 cmp eax, 04fh je dest
#2 add eax, 200 js dest
Paper For Above instruction
The assignment involves understanding conditional jumps and their behavior in the context of a specific register and value, as well as translating high-level control structures into assembly language. The task is divided into three parts: evaluating conditional jumps, implementing an if-structure in assembly, and translating a while loop into assembly.
Part A: Evaluating Conditional Jumps in Assembly
Given that the EAX register contains the value F (hexadecimal) and the doubleword referenced by the label "value" contains FF FF FF 38, we analyze whether each jump instruction will cause a jump to "dest".
First, interpret the values involved: EAX contains F, which in hexadecimal is 0x0F, and "value" holds FF FF FF 38, which is a 32-bit number. Converting "value" to decimal gives 4294967240, considering it as an unsigned 32-bit number.
For the first comparison: cmp eax, 04fh, the number 0x04F equals decimal 79. Since EAX is 0x0F (decimal 15), the comparison indicates whether 15 equals 79. Specifically, the instruction checks if EAX == 0x4F (79).
Since EAX (15) != 79, the je dest (jump if equal) will not cause a jump because the comparison results in inequality.
For the second instruction: add eax, 200, the value 200 in decimal (0xC8 hexadecimal) is added to EAX. After the addition, EAX equals 0x0F + 0xC8 = 0xD7, or decimal 215.
The js dest stands for jump if sign (negative). The sign flag is set if the result of the addition is negative in two's complement. Since 0xD7 in binary begins with a 1 in the most significant bit (indicating a negative number in signed representation), the sign flag will be set, and the jump to "dest" will occur.
Part B: Assembly Code for an If-Structure
The design specifies a simple conditional: if count > value, then count = 0. To implement this in assembly, we assume:
- Variables are stored appropriately in memory, with
countin a register or memory location. - The
valueis stored at a known address or as a constant.
Below is a fragment of x86 assembly code implementing this logic, using eax for count and assuming value is stored at label value.
; Assume count is in EAX, value in memory at [value_address]
mov ebx, [value_address] ; load value into ebx
cmp eax, ebx ; compare count with value
jg set_zero ; if count > value, jump to set_zero
; continue with other code
jmp continue_execution
set_zero:
xor eax, eax ; set count (EAX) to 0
continue_execution:
; rest of the code continues
Part C: Assembly Code for a While Loop
The high-level design:
sum := 0;
count := 1;
while (sum
sum := sum + count;
count := count + 1;
}
To implement this in 80x86 assembly, assuming sum is stored in memory at sum, and count is in register ecx, the following code fragment can be used:
; Initialize sum and count
xor edx, edx ; sum = 0
mov ecx, 1 ; count = 1
while_loop:
cmp edx, 1000
jge end_while ; exit loop if sum >= 1000
add edx, ecx ; sum = sum + count
inc ecx ; count++
jmp while_loop
end_while:
; continue with the rest of the program
This implementation initializes the sum to zero and count to 1. It then repeatedly checks whether the sum is less than 1000, adds the count to the sum, increments count, and repeats until the condition no longer holds.
Conclusion
The provided assembly snippets effectively translate high-level conditional structures and loops into low-level instructions, demonstrating how control flow is managed explicitly in assembly language for the x86 architecture. Such translations are fundamental to understanding how high-level programming languages are executed at the hardware level, emphasizing the importance of meticulous handling of registers, memory, and control flow instructions.
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