有以下程序: #include <iostream> using namespace std; class Base{ public: Base(int x=0) {cout<<x;} }; class Derived : public Base{ public: Derived(int x=0) {cout<<x;} private: Base val; }; int main(){ Derived d(1); return 0; } 程序的输出结果是
有以下程序: #include <iostream> using namespace std; class Base{ public: Base(int x=0) {cout<<x;} }; class Derived : public Base{ public: Derived(int x=0) {cout<<x;} private: Base val; }; int main(){ Derived d(1); return 0; } 程序的输出结果是
有如下程序:class Base {public :int date ;} ;class Derived1 : public Base { };class Derived2 : protected Base { };int main(){Derived1 d1 ;Derived2 d2 ;d1. date = 0 ; //①d2.date = 0 ; //②return 0 ;}下列关于程序编译结果的描述中,正确的是()
有如下程序:class Base {public :int date ;} ;class Derived1 : public Base { };class Derived2 : protected Base { };int main(){Derived1 d1 ;Derived2 d2 ;d1. date = 0 ; //①d2.date = 0 ; //②return 0 ;}下列关于程序编译结果的描述中,正确的是()
把从键盘上输入的十进制数(long型)以二进制到十六进制形式输出。请填空。 int main(void) { char b[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F',}; int c[64],d,i=0,base; long n; printf("enter a number : "); scanf("%ld",&n); printf("enter new base : "); scanf("%d",&base); do{ c[i]=____; i++; n= n╱base; } while (n!=0 ); printf("transmite new base: "); for (--i;i>=0;--i){ d=c[i]; printf("%c",____);} return 0;}"
把从键盘上输入的十进制数(long型)以二进制到十六进制形式输出。请填空。 int main(void) { char b[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F',}; int c[64],d,i=0,base; long n; printf("enter a number : "); scanf("%ld",&n); printf("enter new base : "); scanf("%d",&base); do{ c[i]=____; i++; n= n╱base; } while (n!=0 ); printf("transmite new base: "); for (--i;i>=0;--i){ d=c[i]; printf("%c",____);} return 0;}"
下面程序是将十进制数转换成不超过十六进制的n进制。请分析程序填空。 #include main() { int i=0,base,n,j,num[20]; printf("Enter data that will be converted:\n"); scanf("%d",&n); printf("Enter base:\n"); scanf("%d",&base); do { i++; num[i]=n【1】base; n=n【2】base; } while(n); printf("The data %d has been converted into the %d base data:\n",n,base); for(【3】) if(num[j]>9) putchar(num[j]+'A'-10); else putchar(num[j]+'0'); }
下面程序是将十进制数转换成不超过十六进制的n进制。请分析程序填空。 #include main() { int i=0,base,n,j,num[20]; printf("Enter data that will be converted:\n"); scanf("%d",&n); printf("Enter base:\n"); scanf("%d",&base); do { i++; num[i]=n【1】base; n=n【2】base; } while(n); printf("The data %d has been converted into the %d base data:\n",n,base); for(【3】) if(num[j]>9) putchar(num[j]+'A'-10); else putchar(num[j]+'0'); }
已知:Option Base 0,Dim A() As Variant,则下列语句中,正确的语句是
已知:Option Base 0,Dim A() As Variant,则下列语句中,正确的语句是
标准以太网指的是 ( ) A: 10 BASE 5 B: 10 BASE 2 C: 10 BASE T D: 100 BASE T
标准以太网指的是 ( ) A: 10 BASE 5 B: 10 BASE 2 C: 10 BASE T D: 100 BASE T
(6-2)定义如下Base类,能在(1)处正确调用Base的构造方法。 class Base{ int x,y; Base(int x){} Base(int x,int y){ //(1)调用Base的构造方法 } }
(6-2)定义如下Base类,能在(1)处正确调用Base的构造方法。 class Base{ int x,y; Base(int x){} Base(int x,int y){ //(1)调用Base的构造方法 } }
如何能使程序调用Base类的构造方法输出"base constructor"; class Base{ Base(int i){ System.out.println("base constructor"); } Base(){ } } public class Sup extends Base{ public static void main(String argv[]){ Sup s= new Sup(); //One } Sup() { //Two } public void derived() { //Three } }
如何能使程序调用Base类的构造方法输出"base constructor"; class Base{ Base(int i){ System.out.println("base constructor"); } Base(){ } } public class Sup extends Base{ public static void main(String argv[]){ Sup s= new Sup(); //One } Sup() { //Two } public void derived() { //Three } }
已知:Option Base 0,Dim A(1, 2, 1, 2) As Integer,则数组A的元素个数是
已知:Option Base 0,Dim A(1, 2, 1, 2) As Integer,则数组A的元素个数是
对于视频中的代码,将 Derived 的构造函数改为 Derived(int var):Base0(var-1),Base1(var),Base2(var+1){}定义 Derived d(1),则 d.var0 的值为: A: 1 B: 0 C: 2 D: 不确定
对于视频中的代码,将 Derived 的构造函数改为 Derived(int var):Base0(var-1),Base1(var),Base2(var+1){}定义 Derived d(1),则 d.var0 的值为: A: 1 B: 0 C: 2 D: 不确定