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generic name
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（ 29 ）有如下程序：#includeiostreamusing namespace std;class Name{char name[20];public:Name(){strcpy(name,""); cout'?';}Name(char *fname)){strcpy(name,fname); cout'?';}};int main(){Name names[3]={Name(" 张三 "),Name(" 李四 ")};Return 0;}运行此程序输出符号？的个数是A ） 0B ） 1C ） 2D ） 3
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声明一个对象,给它加上name属性和show方法显示其name值,以下代码中正确 的是( )。
Avarobj = [name:"zhangsan",show:function(){alert(name);}];Bvarobj = {name:"zhangsan",show:”alert()”};Cvarobj = {name:"zhangsan",show:function(){alert(name);}};Dvarobj = {name:"zhangsan",show:function(){alert();}};
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Examine the structure of the EMPLOYEES table:Column name Data type RemarksEMPLOYEE_ID NUMBER NOT NULL, Primary KeyLAST_NAME VARCNAR2(30)FIRST_NAME VARCNAR2(30)JOB_ID NUMBERSAL NUMBERMGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBERYou need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task? （）A. CREATE INDEX NAME _IDX (first_name, last_name);B. CREATE INDEX NAME _IDX (first_name, AND last_name)C. CREATE INDEX NAME_IDX ON (First_name, last_name);D. CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);E. CREATE INDEX NAME_IDX ON employees (First_name, last_name);F. CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
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有如下程序：#includeiostreamusing flamespace std；class Name{char name[20]；public：Name(){strcpy(name，“”)； cout‘?’；}Name(char*fname){strcpy(name，fname)； cout‘?’；}}；int main(){Name names[3]={Name(”张三”)，Name(”李四”)}；return 0；}运行此程序输出符号?的个数是A．0B．1C．2D．3
考题
有如下程序:includeusing namespace std;class Name{char name[20];public:Name( ){s
有如下程序： #include＜iostream＞ using namespace std； class Name{ char name[20]； public： Name( ){strepy(name，" ")；tout＜＜'?'；} Name(char*fname){strcpy(name，fname)；cout＜＜'?'；} }； int main( ){ Name name[3]={Name("张三")，Name("李四")}； return 0； } 运行此程序输出符号“?”的个数是A．0B．1C．2D．3
考题
GRE 的英文全称是()
A. Generic Router EncapulationB. Generic Routing EncapulationC. General Routing EncapulationD. General Router Encapulation
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html语言中，创建一个位于文档内部的靶位的标记是？（）A、name="NAME" B、name="NAME"/name C、a name="NAME"/a D、a name="NAME"
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证件上同时出现“Name”、“Full name”和“First name”，且无护照资料页下方的机读码可供参照的，则（）。A、“Name”或“Full name”为其姓，“First name”为名B、“Name”或“Full name”为其名，“First name”为姓C、全输入英文姓中D、全输入英文名中
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Which syntax turns an existing constraint on?（）A、ALTER TABLE table_name ENABLE constraint_name;B、ALTER TABLE table_name STATUS = ENABLE CONSTRAINT constraint_name;C、ALTER TABLE table_name ENABLE CONSTRAINT constraint_name;D、ALTER TABLE table_name STATUS ENABLE CONSTRAINT constraint_name;E、ALTER TABLE table_name TURN ON CONSTRAINT constraint_name;
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已知httpClient连接网络的url，通过Post方式访问时要传递name参数，下列方式正确的是（）。A、httpGet（url，name）；B、url=url+"？name="+name；C、list.add（newBasicNameValuePair（"name"，name））；D、httpGet.addParams（"name"，name）；
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What describes the difference between NPIV and vSCSI in the way fibre-channel attached disk and tape are presented when VIO Server is used?（）A、NPIV presents Virtual HBA addresses and vendor specific information for disk and tape devices.vSCSI presents generic disk and tape device interfaces.B、NPIV presents generic disk and tape device interfaces on virtual HBA addresses.vSCSI presents HBA addresses as SCSI with vendor specific information for disk and tape devices.C、NPIV presents HBA addresses and vendor specific device information directly to the client partition.vSCSI presents vendor specific disk information and generic tape device interfaces through VIO server.D、NPIV presents virtual HBA addresses with generic device information for SCSI based disk and tape devices.vSCSI presents vendor specific information for disk and tape device devices attached via SCSI hardware interfaces.
考题
单选题What kind of object does the generic Dictionary enumerator return？（）A
ObjectB
GenericC
KeyD
Value
考题
单选题Which syntax turns an existing constraint on? （）A
ALTER TABLE table_name ENABLE constraint_name;B
ALTER TABLE table_name STATUS = ENABLE CONSTRAINT constraint _ name;C
ALTER TABLE table_name ENABLE CONSTRAINT constraint _ name;D
ALTER TABLE table_name TURN ON CONSTRAINT constraint _ name;
考题
单选题Many doctors direct their patients to name-brand drugs, but smart consumers know that generic drugs cost half as much as buying name-brand drugs.A
generic drugs cost half as much as buying name-brand drugsB
buying generic drugs costs half as much as name-brand drugsC
generic drugs cost half as much as name-brand drugsD
buying generic drugs cost half as much as buying name-brand drugsE
to buy generic drugs costs half as much as buying name-brand drugs
考题
单选题Which statement adds a constraint that ensures the CUSTOMER_NAME column of the CUSTOMERS table holds a value?（）A
ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;B
ALTER TABLE customers MODIFY CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;C
ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn NOT NULL;D
ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn IS NOT NULL;E
ALTER TABLE customers MODIFY name CONSTRAINT cust_name_nn NOT NULL;F
ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name NOT NULL;
考题
单选题Which syntax turns an existing constraint on?（）A
ALTER TABLE table_name ENABLE constraint_name;B
ALTER TABLE table_name STATUS = ENABLE CONSTRAINT constraint_name;C
ALTER TABLE table_name ENABLE CONSTRAINT constraint_name;D
ALTER TABLE table_name STATUS ENABLE CONSTRAINT constraint_name;E
ALTER TABLE table_name TURN ON CONSTRAINT constraint_name;
考题
单选题Examine the structure of the EMPLOYEES table: Column name Data type Remarks EMPLOYEE_ID NUMBER NOT NULL, Primary Key LAST_NAME VARCNAR2(30) FIRST_NAME VARCNAR2(30) JOB_ID NUMBER SAL NUMBER MGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBER You need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task?（）A
CREATE INDEX NAME _IDX (first_name, last_name);B
CREATE INDEX NAME _IDX (first_name, AND last_name)C
CREATE INDEX NAME_IDX ON (First_name, last_name);D
CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);E
CREATE INDEX NAME_IDX ON employees (First_name, last_name);F
CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
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