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void main()
{
for(int i=0;i<5;i++);
printf("%d",i);
}

What is the output?..

Answers were Sorted based on User's Feedback



void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / kumaran

declaring int i inside for loop is not available in
traditional 'c'

Is This Answer Correct ?    80 Yes 30 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / mariaalex007

Answer is 5..........

Is This Answer Correct ?    114 Yes 77 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / mahfooz

we can not declare loop condition variable in c..we can do
in c++.

Is This Answer Correct ?    38 Yes 20 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / samir isakoski

why shoul have a cout if is c++

buth this is c (printf)

and why cannot daclare a int in the for ciclus

Is This Answer Correct ?    17 Yes 11 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / divya

no output because the for loop is terminated with a semicolon which means the loop will terminate at once.

Is This Answer Correct ?    15 Yes 12 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / radha raman

output=5

Is This Answer Correct ?    5 Yes 3 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / shilpi

it will give 6 as a answer because condition is true inside and then it will increase its value by 1 and the terminate from the loop.

Is This Answer Correct ?    2 Yes 0 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / ricky dobriyal

hi friend i am ricky dobriyal.
if this programe in c then this program produce error
expression syntex error.
if in c++ then it will produce o/p=5

Is This Answer Correct ?    10 Yes 9 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / pramod

if it is in C it displays an error "indicating that "i"
cannot be declared in loop and i as undefined symbol"

if it the case of C++ it displays "5" because C++ supports
dynamic declaration and once declared anywhere in the
program(even in loops) can be used anytime in that program

Is This Answer Correct ?    2 Yes 2 No

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?....

Answer / marius

if i have

$x=(11,22,33,44,55);
for($i=0;$i<5;$i++)
echo$x[$i]." ";

and we will print this:


11 22 33 44 55

$i=0 => 11
$i<5 => 44
$i++ => 55

ghigamarius@gmail.com

Is This Answer Correct ?    1 Yes 2 No

Post New Answer

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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); }

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