#include<stdio.h>
void main()
{
int i=1;
printf("%d%d%d",i++,++i,i);
}
Answers were Sorted based on User's Feedback
Answer / vinod
3 2 1.because the print f function print right to leftand
the compiler reads left to right.thus answer is 3 2 1
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / shruti
the answer will be:-
1,3,3
as i++ is postfix so first it will print the value then
increament..
after first increament the second preincreament comes and
the value becomes 3..
third time also it will 3..
| Is This Answer Correct ? | 4 Yes | 5 No |
Answer / aaradhana
1,2,2
since to print the postincremented value then i takes the
value 1 then gets preincremented by1 & takes i=2.To print i
value then it takes the updated value i=2.
| Is This Answer Correct ? | 1 Yes | 3 No |
Answer / ramya
Answer is 133.
first the compiler prints 'i' value and prints the value,
next it increments the 'i' value and then prints its value.
| Is This Answer Correct ? | 8 Yes | 14 No |
Answer / priyadarshan kasta
1 2 2
becoz, this line will execute frm right to left side.
that is, first i=1, then ++i will be 2 and then i++ will be
printed as 2. So , it will print as 1 2 2(i.e i++,++i,i)
| Is This Answer Correct ? | 1 Yes | 10 No |
what is run time error?
#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); }
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