During the closing process, some balance sheet accounts are closed and end the period with a zero balance.()此题为判断题(对,错)。

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During the closing process, some balance sheet accounts are closed and end the period with a zero balance.()

此题为判断题(对,错)。


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2.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(

3.请认真阅读以下关于某传输系统的技术说明、状态转换图及C++代码,根据要求回答问题1和问题2。【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)5种状态。触发状态的转换事件有click、complete和timeout 3种,事件与其相应的状态转换如图6-18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,请填补代码段中的空缺语句。【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:private:int state; //传输门当前状态void setState(int state) { this->state = stale; } //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 s

更多“During the closing process, some balance sheet accounts are closed and end the period with a zero balance.() ”相关问题
  • 第1题:

    The boy lay in the street,his eyes_______and his hands______.

    A.closing;trembling
    B.closed;trembled
    C.closing;trembled
    D.closed;trembling

    答案:D
    解析:

  • 第2题:

    8、以下代码描述的是Moore型状态机: always @(in or state) case (state) zero: begin out = 0; if (in) next_state = one; else next_state = zero; end …………


    通常用于比较复杂的过程控制

  • 第3题:

    以下代码用于产生固定数目为clk_cnt的时钟信号。 parameter clk_cnt = 5, clk_period = 2; reg clk; initial begin clk = 0; repeat(clk_cnt) #(clk_period/2) clk = ~clk; End


    正确

  • 第4题:

    The boy lay in the street,his eyes_______and his hands______.

    A.closing;trembling
    B.closed;trembled
    C.closing;trembled
    D.closed;tremblin

    答案:D
    解析:

  • 第5题:

    以下代码描述的是Moore型状态机: always @(in or state) case (state) zero: begin out = 0; if (in) next_state = one; else next_state = zero; end …………


    通常用于比较复杂的过程控制

  • 第6题:

    以下关于VDB函数的语法正确的是()。

    A.VDB(cost,salvage,life,period,factor)

    B.VDB(cost,salvage,life,period,month)

    C.VDB(cost,salvage,life,per)

    D.VDB(cost,salvage,life,start-period,end-period,factor,no-switch)


    VDB(cost,salvage,life,start-period,end-period,factor,no-switch)