HW11 1600012736
11.8
void sigchld_handler(int sig)
{
while (waitpid(-1,NULL,WNOHANG|WUNTRACED) > 0); return;
}
/server_dynamic中添加下面一行/
signal(SIGCHLD,sigchld_handler);
11.9
/修改server_dynamic 19-22行/
srcp = malloc(filesize)
Rio_readn(srcfd,srcp,filesize);
Close(srcfd);
Rio_writen(fd,srcp,filesize);
free(srcp);
第一题：有误
第二题：正确
HW 9 1600012736
/*
-
VM作业：编写虚页页号遍历程序，该程序打印出本进程所能够访问的所有虚页页号
-
/proc/pid/maps
*/
#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
int main()
{
unsigned long start, end;
char c;
char s[500];
char filename[50];
FILE *pf = NULL;
pid_t pid = getpid();
sprintf(filename, "/proc/%ld/maps", pid);
pf = fopen(filename, "r");
while(fscanf(pf, "%[^\n]", s) > 0)
{
sscanf(s, "%lx-%lx %c", &start, &end, &c); // 读区间起止和权限
start = start >> 12; // 页号
end = end >> 12;
if(c == 'r')
{
for(;start < end; start++)
printf("%013lx\n", start);
}
fscanf(pf, "%c", &c); // 结尾\n
}
fclose(pf);
return 0;
}
HW7,8 1600012736
Ji Weiminghui
8.9
n y y y y y
8.18
ACE
8.24
#include "csapp.h"
#define N 2
#define MAXLEN 100
int main() {
int status, i;
pid_t pid;
for (i = 0; i < N; i++)
if ((pid = Fork()) == 0) {
char* cptr = NULL;
*cptr = 'a';
exit(100 + i);
}
while ((pid = waitpid(-1, &status, 0)) > 0) {
if (WIFEXITED(status))
printf("child %d terminated normally with exit status=%d\n",
pid, WEXITSTATUS(status));
else if (WIFSIGNALED(status)) {
char buf[MAXLEN];
sprintf(buf, "child %d terminated by signal %d", pid, WTERMSIG(status));
psignal(WTERMSIG(status), buf);
}
else
printf("child %d terminated abnormally\n", pid);
}
if (errno != ECHILD)
unix_error("waitpid error");
exit(0);
}
10.6 fd2 = 4
10.9
if (Fork() == 0) { /* Child */
int fd = Open("foo.txt", O_RDONLY, 0);
Dup2(fd, STDIN_FILENO);
Close(fd);
Execve("fstatcheck", argv, envp);
}
第一题：正确
第二题：正确
第三题：正确
第四题：正确
第五题：正确
HW6 1600012736
7.6
buf yes extern m.o .data
bufp0 yes global swap.o .data
bufp1 yes local swap.o .bss
swap yes global swap.o .text
temp no - - -
incr yes local swap.o .text
count yes local swap.o .bss
7.7
static double x;
void f()
{
x=-0.0;
}
7.12
A. 0xa
B. 0x22
7.13
A. libc.a 1579; libm.c 471
B. 不同 –g选项有debug
C. linux-vdso.so.1 => (0x00007ffd0e58c000)
libc.so.6 =>/lib/x86_64-linux-gnu/libc.so.6 (0x00007f2ab687a000)
/lib64/ld-linux-x86.so.2 (0x0000560061231000)
第一题：有一处错误
第二题：正确
第三题：正确
第四题：正确
HW5
4.47A
void bubble_a(int * data, int count)
{
int i, last;
int * p;
for (last = count - 1; last > 0; last--)
{
for (i =0; p = data; i < last; i++, p++)
{
if (*(p + 1) < *p)
{
int t = *(p + 1);
*(p + 1) = *p;
*p = t;
}
}
}
}
4.47B
#Ji Weiminghui 1600012736
.pos 0
init:
irmovl stack, %esp
irmovl array, %edx #data
pushl %edx
irmovl $5, %edx #count
pushl %edx
call bubble_a
halt
.align 4
array:
.long 0x5
.long 0x4
.long 0x3
.long 0x2
.long 0x1
bubble_a:
#last ecx
#i edi
#p esi
#t
#array 8(%ebp)
#count 4(%ebp)
rrmovl %esp, %ebp
#for (last = count - 1; last > 0; last--)
mrmovl 4(%ebp),%eax
irmovl $1, %edx
subl %edx, %eax
rrmovl %eax, %ecx
andl %ecx, %ecx #ecx=last=count-1
jle End
Loop1:
irmovl $0, %edi #i=0
mrmovl 8(%ebp), %esi#p=data
rrmovl %ecx, %eax
subl %edi, %eax #last-i
jle Test1
Loop2:
mrmovl 4(%esi), %eax # *(p+1)
mrmovl (%esi), %edx # *(p)
subl %eax, %edx
jle Test2
mrmovl (%esi), %edx
rmmovl %eax, (%esi)
rmmovl %edx, 4(%esi)
Test2:
irmovl $1, %edx
addl %edx, %edi #i++
irmovl $4, %edx
addl %edx, %esi #p++
rrmovl %ecx, %eax
subl %edi, %eax #last-i
jg Loop2
Test1:
irmovl $1, %edx
subl %edx, %ecx #last--
jg Loop1
End:
mrmovl 8(%ebp),%eax
popl %edx
popl %edx
ret
.pos 0x1300
stack:
Stopped in 190 steps at PC = 0x1c. Status 'HLT', CC Z=1 S=0 O=0
Changes to registers:
%eax: 0x00000000 0x0000001c
%edx: 0x00000000 0x00000005
%esp: 0x00000000 0x00001300
%ebp: 0x00000000 0x000012f4
%esi: 0x00000000 0x00000020
%edi: 0x00000000 0x00000001
Changes to memory:
0x001c: 0x00000005 0x00000001
0x0020: 0x00000004 0x00000002
0x0028: 0x00000002 0x00000004
0x002c: 0x00000001 0x00000005
0x12f4: 0x00000000 0x0000001b
0x12f8: 0x00000000 0x00000005
0x12fc: 0x00000000 0x0000001c
4.56
################ Fetch Stage ###################################
int f_pc = [
M_icode == IJXX && M_ifun != UNCOND && !M_Cnd && M_valE < M_valA : M_valA;
M_icode == IJXX && M_ifun != UNCOND && M_Cnd && M_valE >= M_valA : M_valE;
W_icode == IRET : W_valM;
1 : F_predPC;
];
int f_predPC = [
f_icode == ICALL: f_valC;
f_icode == IJXX && (f_ifun == UNCOND || f_valC < f_valP): f_valC;
f_icode ==
1 : f_valP;
];
################ Execute Stage #####################################
int aluA = [
E_icode in { IRRMOVL, IOPL } : E_valA;
E_icode in { IIRMOVL, IRMMOVL, IMRMOVL } : E_valC;
E_icode in { ICALL, IPUSHL } : -4;
E_icode in { IRET, IPOPL } : 4;
E_icode == IJXX && E_ifun != UNCOND : E_valC;
];
int aluB = [
E_icode in { IRMMOVL, IMRMOVL, IOPL, ICALL,
IPUSHL, IRET, IPOPL } : E_valB;
E_icode in { IRRMOVL, IIRMOVL } : 0;
E_icode == IJXX && E_ifun != UNCOND : 0;
];
bool D_bubble =
((E_icode == IJXX && E_ifun != UNCOND && e_Cnd && E_valC >= E_valA) ||
(E_icode == IJXX && E_ifun != UNCOND && !e_Cnd && E_valC < E_valA)) ||
!(E_icode in { IMRMOVL, IPOPL } && E_dstM in { d_srcA, d_srcB }) &&
IRET in { D_icode, E_icode, M_icode };
bool E_stall = 0;
bool E_bubble =
((E_icode == IJXX && E_ifun != UNCOND && e_Cnd && E_valC >= E_valA) ||
(E_icode == IJXX && E_ifun != UNCOND && !e_Cnd && E_valC < E_valA)) ||
E_icode in { IMRMOVL, IPOPL } &&
E_dstM in { d_srcA, d_srcB};
5.13
A.%xmm1 add %xmm1是关键路径
B.3（浮点加法的延迟为3）
C.1（整数加法的延迟为1）
D.需要相乘的元素都不是数据相关的，可以用流水线处理，理论上，每个时钟周期可以发射一个浮点乘法
所以CPA的限制在于具有数据相关性的sum,浮点加法的延迟为3个始终周期，所以，浮点版本的CPE都是3.00
5.14
A.加载两个标量数据需要至少1个时钟周期
B.即使展开成了sum1,sum2,...sum6,对于任意一个sumi，它所执行的浮点数加法，都是一个子关键路径
也就是每个并列执行的sumi都被浮点数加法的延迟所限制，所以展开后的CPA不会小于3
事实上，展开前，只有sum是数据相关的，所以展开前的限制也是浮点数加法
展开只是减少了不必要的循环操作，并没有降低CPA下界
第一题：正确
第二题：正确
第三题：正确
第四题：代码呢？？
HW4
3.68
A = 9, B = 5
3.69
A.CNT = 7
//288-8=280. 280/40=7.
B.typedef struct{ long idx; long x[4]; }a_struct; 或者 typedef struct{ int idx; long x[4]; }a_struct;
//idx:bp+40i+8 x:bp+40i+16 bp代表结构b的起始地址
3.70
A. 0 8 0 8
B.16
C.up->e2.x = *((up->e2.next)->e1.p) - (up->e2.next)->e1.y;
第一题：正确
第二题：有点小问题（idx）
第三题：正确
HW3
3.60
A.x %rdi,n %esi 和 %ecx, result %rax, mask %rdx
B.result = 0; mask = 1;
C.mask != 0;
D.mask = (mask << n);
E.result |= (x & mask);
F. long loop(long x,int n)
{
long result = 0;long mask;
for (mask = 1; mask != 0; mask = (mask << n){
result |= (x & mask);
}
return result;
}
3.62
typedef enum {MODE_A, MODE_B, MODE_C, MODE_D, MODE_E} mode_t;
long switch3(long * p1, long * p2, mode_t action)
{
long result = 0;
switch(action){
case MODE_A:
result = * p2;
* p2 = * p1;
break;
case MODE_B：
result = * p1 + * p2;
* p1 = result;
break;
case MODE_C:
* p1 = 59;
result = * p2;
break;
case MODE_D:
* p1 = * p2;
result = 27;
break;
case MODE_E:
result = 27;
break;
default:
result = 12;
break;
}
return result;
}
3.64
A. 对于 T D[R][S][T], & D[i][j][k] = Xd + L(S * T * i + T * j + k); L是数据类型T以字节为单位的大小，Xd是数组D的起始位置
B. 由8 * R * S * T = 3640，8 * S * T =520, 8 * T = 104
得 R = 7, S = 5, T = 13
第一题：正确(括号匹配上有问题)
第二题：正确
第三题：正确
2.61 A. !~x B. !x C. !((x & 0xFF)^(0xFF)) D. !(x >> ((sizeof(int) - 1) << 3))
2.62
int int_shifts_are_arithmetic()
{
int x = -1;
return -1 == (x >> 1);
}
2.65 int odd_ones(unsigned x) { x ^= (x >> 16); x ^= (x >> 8); x ^= (x >> 4); x ^= (x >> 2); x ^= (x >> 1); return x & 1; }
第一题：正确
第二题：正确
第三题：正确
HW2 2.88 line 1:A 1 01110 001 -9/16 B 1 0110 0010 -9/16 line 2:A 0 10110 101 13*(2^4) B 0 1110 1010 13*(2^4) line 3:A 1 00111 110 -7/(2^10) B 1 0000 0111 -7/(2^10) line 4:A 0 00000 101 5/(2^17) B 0 0000 0001 1/(2^10) line 5:A 1 11011 000 -2^12 B 1 1110 1111 -31*(2^3) line 6:A 0 11000 100 3*(2^8) B 0 1111 0000 +∞ 2.92 float_bits float_negate(float_bits f) { if ((f & 0x7F800000) == 0x7F800000); return f; return f ^ 0x80000000; } 2.96 int float_f2i(float_bits f)// int [-2^31,2^31-1] { unsigned sign = f >> 31; unsigned exp = (f >> 23) & 0xFF; unsigned frac = f & 0x7FFFFF; unsigned bias = 127; int res; unsigned E; unsigned M; if (exp >= 31 + bias) // 数太大超范围，或为TMin res = 0x80000000; else if (exp < 0 + bias) // exp>0即使是1.111...除以2也小于1;exp=0显然 res = 0;//下面考虑exp在[bias,30+bias] else // >=1且能用int表示 { E = exp-bias; // E在[0,30]; M = frac | 800000; if (E >= 23) // 精确表示 res = M << (E - 23); else res = M >> (23 - E); // 向0舍入 } return (sign? -res : res); }
第一题：正确
第二题：有误
第三题：正确