Answer / sampath alagappan

3,3,0

Answer / manikanta

x=3,y=3,z=1

Answer / arunkumar

it will be 3,3,1
no doubt it is the correcct one..
if anybody have a doubt compile and execute it with turbo c.

Answer / monika thakur

3,3,1

Answer / jasmeet singh

hey i typed the output of x=4 but its 3. coz after the
decrement now its value is 3.

Answer / ramya b.m

The output will be 3 3 1. initially the value of x is 4 and
subtracting it with value of y which is 3 the difference is
1. while printing the value of x is decremented by 1 and
hence the output 3 3 1.

Answer / venkat

The output will be 331
this is so because while assigning x -y is assinged to z..
so z=4-3=1
then the value of x will be decremented { post decrement }
now the value of x will be 4-1 = 3
so while printing
x = 3
y = 3{no change}
z= 1

Answer / tommy tom

331

x = 4-- = 3
y = 3
z = 1

then, when printing, there are no spaces in the "%d%d%d" format string, thus there are no spaces in the printing of the numbers. Also, at the end of the run, the command prompt will be tacked onto the end of the output, as the format string contains no newlines, line returns, or other reasons to assume a nice printing.

Answer / guru1985

#include<stdio.h>
#include<conio.h>
void main()
{
int x=4,y=3,z;
z=x-- -y;
printf("x=%d\ny=%d\nz=%d",x,y,z);
}
Output=>
x=3
y=3
z=1

Answer / geetha.s.b

4,3,1

wap for bubble sort

3 Answers  

#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  

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

6 Answers   HCL,

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what is exceptions?

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#include"stdio.h" #include"conio.h" void main() { int a; printf("\n enter a number:"); scanf("%c\n"); getch(); }

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void main() { int i=5; printf("%d",i+++++i); }

14 Answers   HCL,

Given an int variable n that has already been declared and initialized to a positive value, and another int variable j that has already been declared, use a do...while loop to print a single line consisting of n asterisks. Thus if n contains 5, five asterisks will be printed. Use no variables other than n and j .

2 Answers  

Answering Yes or No in C++...using only stdio.h and conio.h..........help me please...? here's must be the output of the program: Screen A Exam No. items Score 1 20 20 2 35 35 Another Entry? [Y] or [N] : Screen B: Record No. Student's Name: 1 Fernando Torres 2 Chuck Norris Note: if you press Y, the program must repeat the procedure in screen A, then if N, the program must proceed to the screen B....Please Help me out............

1 Answers  

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.

1 Answers  

printy(a=3,a=2)

3 Answers  

quoroum of computer languages?

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