71 Your sponsor announce to you and your team that his/her door is open to everyone wishing to discuss project problems. Yet, the sponsor seems to be annoyed whenever someone does stop by, even when discussing a serious problems. This is an example of.A.
题目
71 Your sponsor announce to you and your team that his/her door is open to everyone wishing to discuss project problems. Yet, the sponsor seems to be annoyed whenever someone does stop by, even when discussing a serious problems. This is an example of.
A. semantics
B. credibility
C. inconsistent signals
D. perceptual differences
E. misunderstood role
相似考题
更多“71Your sponsor announce to you and your team that his／her door is open to everyone wishing ”相关问题
-
第1题：
阅读以下说明和C++代码。
【说明】
传输门是传输系统中的重要装置。传输门具有Open(打开)、Closed(关闭)、Opening (正在打开)、StayOpen(保持打开)和Closing(正在关闭)五种状态。触发传输门状态转换的事件有click、complete和timeout三种。事件与其相应的状态转换如下图所示。
下面的C++代码1与C++代码2分别用两种不同的设计思路对传输门进行状态模拟，请填补代码中的空缺。
【C++代码1】
const int CLOSED=1； const int PENING=2；
const int PEN=3； const int CLOSING=4；
const int STAYOPEN=5； //定义状态变量，用不同整数表示不同状态
class Door {
Private：
int state； //传输门当前状态
void setState(int state){ this-＞state=state；} //设置当前状态
public：
Door()：state(CLOSED){}；
void getState(){ //根据当前状态输出相应的字符串
switch(state){
case OPENING： cout＜＜"OPENING"＜＜endl； break；
case CLOSED： cout＜＜"CLOSED"＜＜endl； break；
case OPEN： cout＜＜"OPEN"＜＜endl； break；
case CLOSING： cout＜＜"CLOSING"＜＜endl； break；
case STAYOPEN：cout＜＜"STAYOPEN"＜＜endl； break；
}
};
void click() { //发生click事件时进行状态转换
if ((1)) setState(OPENING)；
else if ((2)) setState(CLOSING)；
else if ((3)) setState(STAYOPEN)；
}
void timeout(){ //发生timeout事件时进行状态转换
if (state == OPEN) setState(CLOSING)；
}
void complete(){ //发生complete事件时进行状态转换
if (state == OPENING) setState(OPEN)；
else if (state == CLOSING) setState(CLOSED)；
}
};
int main(){
Door aDoor；
aDoor.getState()；aDoor.click()； aDoor.getState()；
aDoor.complete()；aDoor.getState()； aDoor.click()；
aDoor.getState()；aDoor.click()； aDoor.getState()； return 0；
}
【C++代码2】
class Door {
public：
DoorState *CLOSED，*OPENING，*OPEN，*CLOSING，*STAYOPEN，*state；
Door()；
virtual～Door(){……//释放申请的内存，此处代码省略)；
void setState(DoorState *state) { this-＞state = state；}
void getState(){
//此处代码省略，本方法输出状态字符串，
//例如，当前状态为CLOSED时，输出字符串为“CLOSED”
}；
void click()；
void timeout()；
void complete()；
}；
Door::Door(){
CLOSED = new DoorClosed(this)； OPENING = new DoorOpening(this)；
PEN = new DoorOpen(this)； CLOSING = new DoorClosing(this)；
STAYOPEN = new DoorStayOpen(this)；state = CLOSED；
}
void Door :: click() {(4)；)
void Door :: timeout() {(5)；)
void Door :: complete() {(6)；}
class DoorState//定义一个抽象的状态，它是所有状态类的基类
{
protected：Door *door；
public：
DoorState(Door *door) {this-＞door = door；}
virtual～DoorState(void)；
virtual void click() {}
virtual void complete(
正确答案：(1)state == CLOSED || state == CLOSING (2)state == OPENING || state == STAYOPEN (3)state == OPEN (4)state-＞click() (5)state-＞timeout() (6)state-＞complete() (7)door-＞setState(door-＞OPENING)
(1)state == CLOSED || state == CLOSING (2)state == OPENING || state == STAYOPEN (3)state == OPEN (4)state-＞click() (5)state-＞timeout() (6)state-＞complete() (7)door-＞setState(door-＞OPENING) 解析：本题考查的是状态转换图的程序设计与实现。
空(1)、(2)和(3)需要根据状态转换图来填写，空(1)、(2)和(3)所在的方法为click，表示当发生click事件时应该发生什么状态转换。根据代码可知，发生click事件时，状态分别跳转到OPENING，CLOSING和STAYOPEN，则发生click前的状态由状态转换图可以得到，分别为CLOSED或CLOSING、STAYOPEN或OPENING以及OPEN。
代码2中空(4)、(5)和(6)考查当发生click、timeout以及complete事件的时候，状态应该如何迁移。类Door的state成员变量用于记录类Door所处的状态，而state变量的类型为DoorState *，DoorState中分别具有click、timeout和complete方法用来响应对应的事件，因此，空(4)、(5)和(6)分别为：state-＞click()、state-＞timeout()和 state-＞complete()。
空(7)主要考查门的当前状态为CLOSED时，发生click事件时状态的迁移。根据状态图可知，CLOSED状态的在click事件下将迁移到OPENING，因此，此处应该将传输门状态设置为OPENING，DoorState变量存储了当前其存储的传输门的实例，因此，可直接调用其方法setState来设置状态，由于传输门状态采用类的实例变量表示，所以此处应该填写door-＞setState(door-＞OPENING)。
代码1和代码2的区别是：代码2将状态间的转换规则封装到具体的类中，当状态转换图的转换规则发生变化时，只需更改部分对应类中的状态迁移规则，而代码1中的迁移规则散落在程序中，维护起来较为困难。 -
第2题：
I wish to propose a toast to our friendship.A:state
B:say
C:suggest
D:announce答案：C解析：本句意思：我提议为我们的友谊干杯。propose提议，（为……）祝酒。suggest建议，提 议；state陈述；say说；announce宣布。第2部分：阅读判断 -
第3题：
材料题
D
With the large number of dogs roaring through our communities，people need to know the facts about rabies（狂犬病），a fatal disease caused by animal bites．Despite vaccination（接種疫苗）programs，rabies is still very prevalent，and will continue to be a serious public health problem for many years to come．
Rabies strikes the central nervous system and brings on choking，convulsions（抽搐）and inability to swallow liquids．It can even cause death．If you or anyone in your family is bitten by dog，cat or other animal，you should not panic，but thoroughly wash the wound with plenty of soap and water and rush to nearby hospital for immediate treatment．If you own the animal which did the biting，you should immediately call a veterinarian for advice and make sure the public health authorities know when and where the biting took place and who was bitten．
（）Rabies is a kind of disease which________．
f a person is bitten by some kind of animal，you________．A.should be panic
B.should take him（her）to a big hospital right away
C.should help to clean the wound and ask the patient to have a good rest at home
D.should help to clean the wound and then take him（her）to a nearby hospital quickly as possibl答案：D解析： -
第4题：
（ ） – Is Miss White __________ English teacher, Maria?
– No, she teaches __________ geography.
A．your；my
B．you；mine
C．you；us
D．your；us
正确答案：D
-
第5题：
If a person is bitten by some kind of animal，you________.A.should be panic
B.should take him（her）to a big hospital right away
C.should help to clean the wound and ask the patient to have a good rest at home
D.should help to clean the wound and then take him（her）to a nearby hospital quickly as possible.答案：D解析： -
第6题：
【单选题】下面程序的输出是。 main（) {enum team {my,your=4,his,her=his+10}; printf（"%d%d%d%dn",my,your,his,her);}
A．0 1 2 3
B．0 4 0 10
C．0 4 5 15
D．l 4 5 15
DDBBCC C语言对枚举的定义规定：在枚举中声明的各个枚举元素，如果没有明确指出某个枚举元素的值，它的上一个元素存在并有明确值的情况下，这个枚举元素的值为其上一个元素的值+1。在本题中，没有明确说明枚举元素em3的值，则em3=em2+1=1+1=2，进而可知，在printf()打印函数中，要打印的数组元素是aa[3]、aa[1]、aa[2]，因此最后的打印结果应当为“DDBBCC”。