L
larry_john
Unregistered / Unconfirmed
GUEST, unregistred user!
1.what packet switching provides assured delivery
a. X.25 b.Frame Relay c.ATM d. ISDN
2.which mthod can speed up the access memory disk virtual memory?
a. select a good replacement algorithm
b. make mapping from virtual address to real address fast
b. make memory accessing time shorten
d. all of above
3.In Unix, how many files(including directories) at maximum can be
contained in one directory
a. depends on the i-node space
b. depends on the file length
c. unlimited
d. 512
4.what limits the sensitivity of a receiver
a. thernal noise
b. nois figure of the receiver
c. phase noise in the receiver
d. all of the above
5.the use of SIM kard in GSM mobile phone
a. identification of the caller
b. authen
tification of the caller
c.
d.
6. y
=a1*y(n-1)+b0*x+b1*x(n-1)
w=a1*w(n-1)+x
求y与w两个信号的关系
a. b. c. d.
7. int i=5,j;
j=(++i)*(i--);
运行后i＝_______，j＝_________
补充几道
8. int a[] = {0,1,2,3,4};
int *p[] = {a,a+1,a+2,a+3};
int **pp = p;
假设a=8239150(记不清了，大概吧），p=8239168
*(pp++)-a = ?
9.int a[100],i,*p=a;
for(i=0;i<100;i++) a=i;
for(i=0;i<100;i++)
{
*p=a;
g(&p);
};
for(i=0;i<100;i++)
printf("%d",a);
};
void g(int **p);
{
**p++;
*p++;
}
10.下面函数的作用：
fun(char *a,char *b)
{
while (*a++=*b++)
｛｝;
return;
}
11.CDMA的分集方式
四个选项为空间分集、时间分集、频率分集、**分集等四选三的组合。
12.下面那项不属于PCB进程控制块的内容
13.抽象类的特点
a. X.25 b.Frame Relay c.ATM d. ISDN
2.which mthod can speed up the access memory disk virtual memory?
a. select a good replacement algorithm
b. make mapping from virtual address to real address fast
b. make memory accessing time shorten
d. all of above
3.In Unix, how many files(including directories) at maximum can be
contained in one directory
a. depends on the i-node space
b. depends on the file length
c. unlimited
d. 512
4.what limits the sensitivity of a receiver
a. thernal noise
b. nois figure of the receiver
c. phase noise in the receiver
d. all of the above
5.the use of SIM kard in GSM mobile phone
a. identification of the caller
b. authen
tification of the caller
c.
d.
6. y
=a1*y(n-1)+b0*x+b1*x(n-1)w
=a1*w(n-1)+x求y
与w两个信号的关系a. b. c. d.
7. int i=5,j;
j=(++i)*(i--);
运行后i＝_______，j＝_________
补充几道
8. int a[] = {0,1,2,3,4};
int *p[] = {a,a+1,a+2,a+3};
int **pp = p;
假设a=8239150(记不清了，大概吧），p=8239168
*(pp++)-a = ?
9.int a[100],i,*p=a;
for(i=0;i<100;i++) a=i;
for(i=0;i<100;i++)
{
*p=a;
g(&p);
};
for(i=0;i<100;i++)
printf("%d",a);
};
void g(int **p);
{
**p++;
*p++;
}
10.下面函数的作用：
fun(char *a,char *b)
{
while (*a++=*b++)
｛｝;
return;
}
11.CDMA的分集方式
四个选项为空间分集、时间分集、频率分集、**分集等四选三的组合。
12.下面那项不属于PCB进程控制块的内容
13.抽象类的特点