Computer Organization and Design--计组作业习题（2）
Computer Organization and Design
----------------------个人作业，如果有后辈的作业习题一致，可以参考学习，一起交流，请勿直接copy
Problem 1：
Consider the following MIPS assembly program:
0: lbu $5,100($0)
1: sh $5,-42($5)
2: lh $4,2($8)
3: sw $4,101($0)
Recall that register $0 always contains the value zero. The memory is organized in big-endian format. The initial value for register 8 is: $8 = 0x00000064
Show the state of the memory and of the two 32-bit registers ($4 and $5) after executing each instruction of the snippet above. The initial values for register file and memory are provided. Fill in the following tables to show your answers (Leave a box empty if its value is the same as the previous one):
|
Reg |
Initial |
After inst 0 |
After inst 1 |
After inst 2 |
After inst 3 |
|
$4 |
0x00000020 |
|
|
0xFFFFC700 |
|
|
$5 |
0x00000074 |
0x00000091 |
|
|
|
|
Memory address |
Initial Value |
After inst 0 |
After inst 1 |
After inst 2 |
After inst 3 |
|
10010 |
0x91 |
|
|
|
|
|
10110 |
0x34 |
|
|
|
0xFF |
|
10210 |
0xC7 |
|
|
|
0xFF |
|
10310 |
0x84 |
|
0x00 |
|
0xC7 |
|
10410 |
0x57 |
|
0x00 |
|
0x00 |
|
10510 |
0xB3 |
|
|
|
|
Problem 2 (7 points)
Given the following C variable declarations:
char ee;
short cs;
struct {
short x3;
struct {
char *x7;
double x0[3];
int r;
} u;
char l[2];
float *e;
} s;
Assuming the data memory starts at location 100 (decimal), answer the following questions:
a. Fill in the size, start and end address for each variable in the table below for the MIPS architecture using the native data type sizes and alignment rules discussed in class. (5 pts)
|
Variable |
size(bytes) |
starting address |
ending address |
|
ee |
1 |
100 |
100 |
|
cs |
2 |
102 |
103 |
|
s |
56 |
104 |
159 |
|
x3 |
2 |
104 |
105 |
|
u |
40 |
112 |
151 |
|
x7 |
1 |
112 |
112 |
|
x0[0]to x0[2] |
24 |
120 |
143 |
|
r |
4 |
144 |
147 |
|
l[0] to l[1] |
2 |
152 |
153 |
|
e |
4 |
156 |
159 |
b. Calculate the total memory used for the data declarations. (0.5 pts)
total memory：（159-100）+1=60bytes;
c. Suggest a reorganization of the above data structures to save maximum possible amount of memory noting that you cannot add or remove fields from the structs. How much memory can be saved? (Hint, the structs can be reorganized such that there is no wasted memory). (1.5 pts)
short cs; ------ 100-101
char ee; ------ 102-102
struct {
short x3; ------ 104-105
char l[2]; ------ 106-107
float *e; ------ 108-111
struct {
double x0[3]; ------ 112-135
int r; ------ 136-139
char *x7; ------ 140-140
} u; ------ 112-143
} s; ------ 100-143
total memory：（143-100）+1=44bytes;
so,16bytes are saved;
Problem 3 (5 points)
Convert the following C code into MIPS assembly code by filling in the blanks below. Assume that the variable data is an array of integers starting at address 400.
int index = 100, counter = 0;
while ( index > 0 ) {
if ( data[index] == data[index+3] ) {
counter+=5;
}
else {
counter-=5;
}
index--;
}
MIPS Assembly Code:
addi $1, __$0_, 100
addi $2, $0, 0
loop ble $1, $0, __exit__
addi $3, $1, 0
sll $3, $3, __2__
lw $4, 400(__$3__)
addi $3, $3, __12__
lw $3, ___400__($3)
bne $3, $4, __diff___
addi $2, $2, __5___
beq $0, $0, next
diff addi $2, $2, __-5___
next addi _$1__, _$1__, -1
beq $0, $0, loop
exit halt
Problem 4 (8 points)
You are using the “Great Old Compiler”(GOC) to compile the three functions below. function1 is called from main once, while function2 and function3 are both called from function1 only. Relevant snippets of the functions‟ code, including all the calls to other functions is shown below. The architecture that GOC is targeting has 2 caller-saved registers and 2 callee-saved registers
int function1(int tintin){
int a,b,c,d;
a = 15;
b = 11;
function2(a);
c = a + b;
while(a > 0) {
//iterates 15 times
function3(a);
a--;
}
d = 9;
d = 8 + tintin;
d++;
return 0;
}
int function2(int haddock){
int k,l,m;
k = 17;
m = 18;
printf(“captain\n”);
l = k + m;
l = l / 2;
return 0;
}
int function3(int snowy){
int w,x,y;
w = 1;
x = 2;
y = w + x;
printf(“dog\n”) ;
y = w;
return 0;
}
GOC has mapped the local variables of the functions as follows:
|
Function |
In caller saved register |
In callee saved register |
|
function 1 |
a,b |
c,d |
|
function 2 |
k,l |
m |
|
function 3 |
w,y |
x |
(a) How many executed save/restore instruction pairs exist in the assembly code of function1 ONLY for the purpose of preserving register values across function calls? Show your calculation per variable. (2 pts)
(b) Can you do better than GOC? Write down the number of executed S/R operation pairs per function basis in the table, if a variable is mapped to caller saved register or in a callee saved register. Provide the best assignment to caller or callee-saved registers on a per-register basis. If caller and callee-save has the same performance for a variable, then put “either” for your answer. State the total number of save/restore instruction pairs inserted by the compiler under the best assignment. (6 pts)
a:16 pairs;
b:1 pairs;
c:1 pairs;
d:1 pairs;
Tatal:2+2+1=5 pairs;
|
|
variable |
Caller-save S/R op pairs |
Callee-save S/R op pairs |
Answer |
|
function1 |
a |
16 |
1 |
callee |
|
b |
1 |
1 |
either |
|
|
c |
0 |
1 |
caller |
|
|
d |
0 |
1 |
caller |
|
|
function2 |
k |
1 |
1 |
either |
|
l |
0 |
1 |
caller |
|
|
m |
1 |
1 |
either |
|
|
function3 |
w |
1 |
1 |
either |
|
x |
0 |
1 |
caller |
|
|
y |
0 |
1 |
caller |
Problem 5 (7 points)
Given the following C code in a file called awesome.c. Note that sizeof is not a function call, the compiler evaluates sizeof to a constant before it actually compiles the code.
0 int g;
1 int func1(char inp) {
2 int *data = malloc( 50*sizeof(int) );
3 static char c;
4 c = ‟b‟;
5 data[3] = g + 5;
6 bar(c);
7 return g;
8 }
9
10 void func2(char arg) {
11 printf(“%c”, arg);
12 }
(a) Indicate the memory region where the following variables/strings are stored (static, stack, heap, text) (3.5 pts)
|
*data |
heap |
|
g |
static |
|
arg |
stack |
|
inp |
stack |
|
c |
static |
|
%c |
Stack |
(b) Circle all symbols that are placed in the symbol table of awesome.o (2.5 pts)
func1 inp c arg data func2
g printf malloc
(c) Please list any two line numbers that will have entries in the relocation table: (1 pt)
Line : 5 ,6
