State constitutions have been amended, but ______ affect the basic principles.
A.not so much as
B.not so as to
C.in order to
D.so long as to
A.not so much as
B.not so as to
C.in order to
D.so long as to
A.touch/var/state/dhcp/dhcpd.1eases
B.address/var/state/dhcp/dhcpd.1eases
C.nat/var/state/dhcP/dhcpd.1eases
D.resolve/var/state/dhcp/dhcpd.1eases
A.4
B.3
C.2
D.1
A.对于客户程序而言,State模式比Strategy模式更透明
B.Strategy模式用来处理算法变化,State模式用来处理状态变化
C.State模式的对象变化频率比Strategy模式的对象高
D.State模式的“状态”是在对象外部,Strategy模式的“策略”是在对象内部
(72)
A.function model,class model and state model
B.class model,interaction model and state model
C.class model,interaction model and sequence model
D.function model,interaction model and state model
● 针对以下程序段,对于变量 c 的取值,至少需要(61)个测试用例才能够满足语句覆盖的要求。
c = ((u8_t *)q->payload)[i];
switch (c)
{
case SLIP_END:
sio_send(SLIP_ESC, netif->state);
sio_send(SLIP_ESC_END, netif->state);
break;
case SLIP_ESC:
sio_send(SLIP_ESC, netif->state);
sio_send(SLIP_ESC_ESC, netif->state);
break;
default:
sio_send(c, netif->state);
break;
}
(61)A.4 B.3 C.2 D. 1
The boss could hardly () his temper when he saw the state of his office.
A.hold onto
B.hold up
C.hold off
D.hold in
● 在如下所示的一段 XML 代码中,根元素名为 (41 ) 。
<?xml version=”1.0” encoding=”GB2312” standalone=”yes”>
<state coursename=”课程”>
<courseid id=”0900”></courseid>
</state>
(41 )A. xml B. state C. coursename D. courseid
A.活动图(activity diagram)
B.状态图(state diagram)
C.序列图(sequence diagram)
D.协作图(collaboration diagram)
A.正确
B.错误
【C程序】
#include<stdio.h>
/*此处为栈类型及其基本操作的定义,省略*/
int main(){
STACK station;
int state[1000];
int n; /*车厢数*/
int begin, i, j, maxNo; /*maxNo为A端正待入栈的车厢编号*/
printf("请输入车厢数:");
scanf("%d",&n);
printf(“请输入需要判断的车厢编号序列(以空格分隔):”);
if(n<1)return-1;
for (i=0; i<n; i++) /*读入需要驶出的车厢编号序列,存入数组state[]*/
scanf("%d",&state[i]);
(1) ; /*初始化栈*/
maxNo=1;
for(i=0; i<n; ){ /*检查输出序列中的每个车厢号state[i]是否能从栈中获取*/
if((2) ){ /*当栈不为空时*/
if (state[i]=Top(station)) { /*栈顶车厢号等于被检查车厢号*/
printf("%d",Top(station));
Pop(&station);i++;
}
else
if ((3) ) {
printf(“error\n”);
return 1;
}
else{
begin= (4) ;
for(j=begin+l;j <=state [i];j++){
Push(&station, j);
}
}
}
else{ /*当栈为空时*/
begin=maxNo;
for(j=begin; j<=state[i];j++) {
Push(&station, j);
}
maxNo= (5) ;
}
}
printf("OK");
return 0;
}