class test
{
int a;
public:
test(int b):a(b){}
void show(){
cout<<a;
}
};
void main()
{
test t1;
test t2(5);
t1.show();
t2.show();
}
}
void main() { int i=1; printf("%d%d%d",i,++i,i++); } Cau u say the output....?
char* f() return "hello:"; void main() {char *str=f(); }
Assume that the int variables i and j have been declared, and that n has been declared and initialized. Write code that causes a "triangle" of asterisks of size n to be output to the screen. Specifically, n lines should be printed out, the first consisting of a single asterisk, the second consisting of two asterisks, the third consistings of three, etc. The last line should consist of n asterisks. Thus, for example, if n has value 3, the output of your code should be * ** *** You should not output any space characters. Hint: Use a for loop nested inside another for loop.
#include"stdio.h" #include"conio.h" void main() { int a; printf("\n enter a number:"); scanf("%c\n"); getch(); }
class test { int a; public: test(int b):a(b){} void show(){ cout<<a; } }; void main() { test t1; test t2(5); t1.show(); t2.show(); } }
what is the large sustained error signal that eventually cause the controller output to drive to its limit
#include<iostream.h> #include<stdlib.h> static int n=0; class account { int age,accno; float amt; char name[20]; public: friend void accinfo(account [] ,int); void create(); void balenq(); void deposite(); void withdrawal(); void transaction(account []); }; void account :: create() { static int acc=1231; accno=acc+n; cout<<"\n\tENTER THE CUSTOMER NAME : "; cin>>name; cout<<"\n\t ENTER THE AGE : "; cin>>age; cout<<"\n\t ENTER THE AMOUNT : "; cin>>amt; // if(amt<=500) // cout<<"\n\tAMOUNT IS NOT SUFFICIENT TO CREATE AN ACCOUNT..."; cout<<"\n\t YOUR ACCOUNT NUMBER : "<<accno<<endl; n++; } void accinfo(account cus[],int ch) { int no,flag=0; cout<<"\n\t\tENTER YOUR ACCOUNT NUMBER : "; cin>>no; for(int i=0;i<=n&&flag==0;i++) if(no==cus[i].accno) { flag=1; switch(ch) { case 2: cus[i].balenq(); break; case 3: cus[i].deposite(); break; case 4: cus[i].withdrawal(); break; case 5: cus[i].transaction(cus); break; default: cout<<"\n\t\tEND OF THE OPERATION"; exit(1); } } if(flag==0) cout<<"\n\t\tYOUR ACCOUNT DOES NOT EXIST..."<<endl; } void account :: balenq() { cout<<"\n\t\tCUSTOMER NAME : "<< name << endl; cout<<"\n\t\tBALANCE : "<< amt << endl; } void account :: deposite() { int damt; cout<<"\n\t\tCUSTOMER NAME : "<< name <<endl; cout<<"\n\t\tBALANCE : "<< amt <<endl; cout<<"\n\tENTER THE AMOUNT TO BE DEPOSITED : "; cin>>damt; amt+=damt; cout<<"\n\t\tYOUR CURRENT BALANCE : "<<amt<<endl; } void account :: withdrawal() { int wamt; cout<<"\n\t\tCUSTOMER NAME : "<< name; cout<<"\n\t\tBALANCE : "<< amt; cout<<"\n\tENTER THE AMOUNT TO BE WITHDRAWN : "; cin>>wamt; if(amt-wamt>=500) { amt-=wamt; cout<<"\n\t\tYOUR CURRENT BALANCE : "<<amt; } else cout<<"\n\tYOUR BALANCE IS TOO LOW FOR WITHDRAWAL..."<<endl; } void account :: transaction (account cus[]) { int no,tamt,flag=0; cout<<"\n\tENTER THE RECEIVER'S ACCOUNT NUMBER : "; cin>>no; cout<<"\n\t\t ENTER THE AMOUNT : "; cin>>tamt; for(int i=0;i<=n&&flag==0;i++) if(cus[i].accno==no) { flag=1; cus[i].amt+=tamt; amt-=tamt; cout<<"\n\t\tYOUR CURRENT BALANCE : "<<amt<<endl; cout<<"\n\t\t RECEIVER'S BALANCE : "<<cus[i].amt<<endl; } if(flag==0) cout<<"\n\tRECEIVER'S ACCOUNT NUMBER IS NOT AVALIABLE..."<<endl; } void main() { account cus[10]; int ch; do { cout<<"\n\t\t BANK ACCOUNT"; cout<<"\n\t\t ************\n"; cout<<"\n\t\t1.CREATE AN ACCOUNT"; cout<<"\n\t\t2.BALANCE ENQUIRY"; cout<<"\n\t\t3.DEPOSITE"; cout<<"\n\t\t4.WITHDRAWAL"; cout<<"\n\t\t5.TRANSACTION"; cout<<"\n\t\t6.EXIT\n\n"; cout<<"\n\t\tENTER YOUR CHOICE : "; cin>>ch; if(ch==1) cus[n].create(); else accinfo(cus,ch); }while(1); }
How to create a program that lists the capital country when told what the original country is? (Terribly sorry, I'm a novice programmer and would appreciate any help ;). Cheers, Alexxis
Write a c-programe that input one number of four digits and find digits sum?
What is probability to guarantee that the task a programmer is going to create will be created and be able to run on a particular system (RTOS/GPOS).
What is the code for following o/p * * * * * * * * * * * * * * * *
what is meant for variable not found?