Answer / nipun

You can use while loop for this. Insert your code for screen
A in the while loop and just let it run if the user enters
a 'Y' and if the user choses 'N' then go to screen B out of
the loop

Or
you can use label:
ex:
main()
{
code....
......

label:
code for screen A...
......
......
printf("Another Entry? [Y] or [N] :");
scanf("%s", &op);
if(choice=='y' || choice=='Y')
{
goto label;
}
}

how to convert decimal to binary in c using while loop without using array

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which typw of errors ? & how to solve it ?

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errors are known as?

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void main() { int i=1; printf("%d%d%d",i,++i,i++); } Cau u say the output....?

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quoroum of computer languages?

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wap for bubble sort

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How to upgrade LOOP environment, I just mean, how can i make loop statement editable ? I just try some program using loop statement and checking it in multiple compilers. Every compiler showing different output, what's the wrong ? is it a compiler based problem, or loop based problem, tell me why ? and what will be the debugging process, for this kind of problem ?

1 Answers  

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#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); }

1 Answers  

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I can not get my C++ program to work right. It is supposed to tell if a word is a palindrome or not, but it only tells thet the word is not a palindrome. And I can't fix it.

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Find the error (2.5*2=5) (a) X=y=z=0.5,2.0-5.75 (b) s=15;

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void main() { int i=7; printf("N= %*d",i,i); }

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char* f() return "hello:"; void main() {char *str=f(); }

1 Answers  

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