I like to get up early so that I can get plenty of work ________.
A. to do
B. doing
C. done
D. being done
以下PHP代码的运行结果是()。?phpob_start();for($i=0;$i10;$i++){echo $i;}$output = ob_get_contents();ob_end_clean();echo $ouput;?A.12345678910B.1234567890C.0123456789D.Notice提示信息
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阅读以下说明和C++程序,将应填入(n)处的字句写在对应栏内。【说明】字符串在程序设计中扮演着重要角色。现需要设计字符串基类string,包含设置字 符串、返回字符串长度及内容等功能。另有一个具有编辑功能的串类edlt_string,派生于string,在其中设置一个光标,使其能支持在光标处的插入、删除操作。【程序】include <iostream.h>include <stdio.h>include <string.h>class string{int length;char *data;public:int get_length() {return length;}char *get_data() {return data;}~string() {delete data;}int set data(int in_length, char *in_data);int set_data(char *data);void print() {cout<<data<<endl;}};class edit_string: public string{int cursor;public:int get_cursor() {return cursor;}void move_cursor(int dis) {cursor=dis;}int add_data(string *new_data);void delete_data(int num);};int string::set_data(int in_length,char *in_data){length=in_length;if(!data)delete data;(1)strcpy(data,in_data);return length;}int string::set data(char *in_data){(2)if(!data)delete data;(1)strcpy(data,in_data);return length;}int edit_string::add_data(string *new_data){int n,k,m;char *cp,*pt;n=new_data->get_length();pt=new_data->get_data();cp=this->get_data();m=this->get_length();char *news=new char[n+m+1];for(int i=0; i<cursor; i++)news[i]=cp[i];k=i;for(int j=0; j<n; i++,j++)news[i]=pt[j];cursor=i;for(j=k; j<m; j++,i++)(3)news[i]='\0';(4)delete news;return cursor;}void edit string::delete_data( int num){int m;char *cp;cp=this->get_data();m=this->get_length();for(int i=cursor; i<m; i++)(5)cp[i]='\0';}
[A] get[B] take[C] work[D] try
I() packing in wooden cases.A、thinkB、preferC、get
向量类vector中的get(i)方法不能够返回向量中下标为i的元素值。()此题为判断题(对,错)。
A:I prefer watching TV. B:So( ) I.A. haveB. amC. do
ArraryLista=newArrayList();a.add(Alpha”);a.add(Bravo”):a.add(Charlie”);a.add(Delta”);Iteratoriter=a.iterator();Whichtwo,addedatline17,printthenamesintheArrayListinalphabeticalorder?()A.for(inti=0;i<a.size();i++) System.out.println(a.get(i)));B.for(inti=0;i<a.size();i++) System.out.println(a[i]);C.while(iter.hasNext()) System.out.println(iter.next());D.for(inti=0,i<a.size();i++) System.out.println(iter[i]);E.for(inti=0;i<a.size();i++) System.out.println(iter.get(i));