Answer / vera

012345
not
0
1
2
3
4
cz their is no printf("\n");
first time i=0 it print 0
scnd time i=1 it print 1
last time i=4 it print 4
but not 5 cz when i=5 :i is not <5
so it will not print 5
10x

Answer / ghiga

0 1 2 3 4 5

Answer / sanjeevi

Guys remember things
1) we cant declare loop variable inside for
2) here i is local to for loop alone..

Answer / chuimaya

It gives an error coz we cannot declare a variable inside the for loop.

Answer / krishna krish

0 1 2 3 4

Answer / avi

Error will occur

Answer / ttt

0
1
2
3
4

Answer / ramveer

for the first loop i is 0 so 0 is printed
for 2nd time i++ i becomes 1 so 1 is printed
for 3rd time i becomes 2 so 2 gets printed
for 4th time i becomes 3 so it gets prited
for 5th time i becomes 4 so it gets printed
now for this time i has became 5 and this isn't in loop boudary as the condition is i<5 so its breaks out of loop.

so the ans is "0 1 2 3 4" .

Answer / erdem

Ofcourse

01234

(logic for start from 0 to 4 and no blank or tab)

Answer / sansiri

the answer is...........0




because in c,c++ program when the semicolon is present then it is considered as one statement
here for loop has a semicolon so it consider one statement and execute only one so that the initial value o is printed

void main() { int i=5,y=3,z=2,ans; clrscr(); printf("%d",++i + --z + i++ + --i * ++y); i=5,y=3,z=2; ans=++i + --z + i++ + --i * ++y; printf("\n%d",ans); getch(); } Its output is 37 and 31.... Please explain me why its different How it works.....

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what are the techniques for reducing the fragility of a memory bug?

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2. A student studying Information Technology at Polytechnic of Namibia is examined by coursework and written examination. Both components of assessment carry a maximum of 50 marks. The following rules are used by examiners in order to pass or fail students. a. A student must score a total of 40% or more in order to pass (total = coursework marks + examination marks) b. A total mark of 39% is moderated to 40% c. Each component must be passed with a minimum mark of 20/50. If a student scores a total of 40% or more but does not achieve the minimum mark in either component he/she is given a technical fail of 39% (this mark is not moderated to 40%) d. Grades are awarded on marks that fall into the following categories. Mark 100-70 69-60 59-50 49-40 39-0 Grade A B C D E Write a program to input the marks for both components (coursework marks out of 50 and examination marks out of 50), out put the final mark and grade after any moderation. [30]

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main() { char c; for(c='A';c<='Z';c++) getch(); }

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how to convert decimal to hexadecimal without using arrays just loops

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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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