对施工成本管理目标进行风险分析,并制定防范性对策,这是施工成本管理的()措施。
A.组织
B.技术
C.经济
D.合同
第1题
A.编译不通过
B.100
C.200
D.200 100
第2题
试图编译和运行以下代码的结果是什么? private class Base{ Base(){ int i = 100; System.out.println(i); } } public class PriBase extends Base{ static int i = 200; public static void main(String argv[]){ PriBase p = new PriBase(); System.out.println(i); } }
A、编译错误
B、200
C、100 200
D、100
第3题
下列程序的输出结果是非曲直【 】。
include<iostream, h>
class base
{
int x, y;
public:
base(int i, int j){x=i; y=j;}
virtual int add(){return x+ y;}
};
class three: public base
{
int z;
public:
three(int i, int j, int k) :base(i, j){z=k; }
int add() { return (base:: add()+z); }
};
void main()
{
three * q=new three(lO,20,30);
cout<<q->add()<<end1;
}
第4题
有如下程序:
include<iostream>
using namespace std
class Base{
int b;
public:
Base(int i) {b=i;}
Void disp () {cout<<"Base:b="<<b<<''; }
};
class Base1:virtual public Base{
public:
Base1(int i):Base(i){}
};
class Base2:virtual public Base{
public:
Base2(int i):Base(i){}
};
class Derived:public Basepublic Base1{
int d;
public:
Derived(int i ,int j):Base1(j),Base2(j),【 】
{ d=i; }
void disp() {cout<<"Derived:d="<<d<<' ';}
};
int main()
Derived objD(1,2);objD. disp()
objD. Base::disp();
objD. Base1::disp()
objD. Base2::disp();
return 0;
}
请将程序补充完整,使程序在运行时输出:
Derivd:d=1 Base:b=2 Base:b=2 Base:b=2
第5题
为完成下面的程序,应在划线处填入的语句是 #include<iostream> using namespace std; class Base { private: int x; public: Base(int i) { x=i; } ~Base(){} }; class Derived:public Base { public: ______ //完成类Derive构造函数的定义 }; int main() { Derived obj; return 0; }
A.Derived(int i):Base(i){}
B.Derived(){}
C.Void Derived (int i):Base(i){}
D.Denved(int i){Base(i);}
第6题
为完成下面的程序,应在划线处填入的语句是 #include<iostream> using namespace std; class Base { private: int x; public: Base (int i) { x=i; } ~Base(){} }; class Derived:public Base { public: ______________//完成类Derive构造函数的定义 }; int main() { Derived Obj; return 0; }
A.Derived(int i):Base(i){}
B.Derived(){}
C.void Derived (int i):Base(i){}
D.Derived(int i){Base(i);}
第7题
有以下程序
include <iostream>
using namespace std;
class Base
{
int a;
public:
Base(int x){ a=x; }
void show(){ cout<<a; }
class Derived : public Base
{
int b;
public:
Derived(int i) :Base(i+1),b(i){}
void show() { cout<<b;
};
int main ()
{
Base b(5),*pb;
Derived d(1);
pb=&d;
pb->show ();
return 0;
}
运行后的打印结果是______。
第8题
下列程序的输出结果是______。
include<iostream.h>
class base
{
int x,y;
public:
base(int i,int j){x=i;y=j;}
virtual int add(){return x+y;}
};
class three:public base
{
int z;
public:
three(int i,int j,int k):base(i,j){z=k;)
int add(){return(base::add()+z);}
};
void main()
{
three*q=new three(10,20,30);
cout<<q->add()<<endl;
}
第9题
为完成下面的程序,应在划线处填入的语句是()。 #include <iostream> using namespace std; class Base { private: int x; public: Base(int i) { x=i; } ~Base(){} }; class Derived : public Base { public: _______________ //完成类Derive构造函数的定义 }; int main() { Derived Obj; return 0; }
A.Derived(int i):Base(i){}
B.Derived(){}
C.voidDerived(int i):Base(0){}
D.Derived(int i){Base(i);}
第10题
下面程序的打印结果是【 】。
include <iostream>
using namespace std;
class Base
{
public:
Base(int x)
{
a=x;
}
void show()
{
cout<<a;
}
private:
int a;
};
class Derived : public Base
{
public:
Derived(int i) :Base(i+1) ,b(i) { }
void show()
{
cout<<b;
}
private:
int b;
};
int main ()
{
Base b(5) , *pb;
Derived d(1);
pb=&d;
pb->show();
return 0;
}
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