Computer Architecture HW1

Topics: Machine code, Mnemonic, Assembly language Pages: 3 (375 words) Published: September 28, 2014
Machine-Level Representation of Programs #1

4190.308 Computer Architecture
Name:

Due Date:

Wednesday, September 11, 2014, 23:59

Student-Number:

Submission:

in paper form.
There will be a drop off box in class and in front of the CSAP Lab in building 301, room 419.

Assignment
Update your information on eTL. (your picture, e-mail address and mobile phone number) Question 1
Memory operations
Explain the difference of “reg to mem”, “mem to reg” and “mem to mem” ISAs. Text
ans:
They all make reference to a movement of data from one place to another. REG to REG is the movement of data from one register into another. In contrast MEM to REG is the movement of data from memory to a register. Finally, MEM to MEM represent data transfer from one place in memory into another.

Question 2
Operand Specifiers
Assume the following values are stored at the indicated memory addresses and registers: Address

Value

Register

Value

0x100

0xFF

%eax

0x104

0x104

0xAB

%ecx

0x1

0x108

0x13

%edx

0x3

0x10C

0x11

0x110

0xB5

0x114

0x3B

0x118

0x1A

0x11C

0x03

Fill in the following table showing the values for the indicated operands: Operand

Value

%edx

0x3

0x114

0x3B

$0x11C

0x11C

(%eax)

0xAB

8(%eax)

0x11

0xD(%eax,%edx)

0x3B

264(%ecx,%edx)

0x11

0xFC(,%ecx,8)

0xAB

(%eax,%edx,8)

0x03

Question 3
Data movement instructions
You are given the following information. A function with prototype void decode(int *xp, int *yp, int *zp);
is compiled into assembly code. The body of the code is as follows: xp at %ebp+0x8, yp at %ebp+0xc, zp at %ebp+0x10
mov
mov
mov
mov
mov
mov
mov
mov
mov

0x8(%ebp),%edx
0xc(%ebp),%eax
0x10(%ebp),%ecx
(%eax),%esi
(%ecx),%ebx
(%edx),%edi
%edi,(%ecx)
%esi,(%edx)
%ebx,(%eax)

Parameters xp, yp, and zp are stored at memory locations with offsets 0x8, 0xc, and 0x10, respectively, relative to the...
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