Answer / vivek

Its a very simple answer if u r thinking as a software
engineer . so these mimi things are very importants .
variable not found means . in this program
U HAVE NOT DECLARED THE VARIABLE CALLED iSum . AND U R
TRYING TO ASSIGN THE i + j TO iSum VARIABLE WHICH WAS NOT
DECLARED THATS Y THE COMPILER GIVING AN ERROR .

Answer / yogambika

when u have not declared variable in the main function or
any other function but used in the program.
example:
main()
{
int i,j;----------------> (iSum not declared)
printf("enter the value of i and j");
scanf("%d%d",&i,&j);
iSum = i + j;
printf("The Sum =",iSum);
getch();
}
In this case iSum will Show a compiler error "Variable not
found".

Answer / deepti khanna

in some of the programming language it is not required to
declare a variable before we use, but in c++ while using a
variable prior to that we need to declare the variable
first.

main()
{
i=25;
cout<<i;
}
above program may show error because we doesn't declare a
variable.

A sample program using data structure? what is file handling?

0 Answers   TCS,

---

Using string functions write a program that will accept the name of the capital as input value and will display the corresponding country. ------------------------ Capitals Countries ------------------------ Capitals Countries Ottawa Canada Moscow Russia Rome Italy I can't not get it to run properly

1 Answers   AMA,

---

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

9 Answers  

---

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

50 Answers   Apple, Aptech, Arwen Tech, BCS, C2D Software, CEC,

---

when i use cout or cin call & then either << or >> .....it shows declaration syntax error...what should i do? cout<<"anything"; int a; cin>>a; return 0;

2 Answers  

---

---

How to create a program that lists countries capitals when country is entered? (Terribly sorry, I'm a complete novist to coding with C, am looking for inspiration and general tips on how to code and create this program.)

0 Answers  

---

Assume that the int variables i and j have been declared, and that n has been declared and initialized. Write code that causes a "triangle" of asterisks of size n to be output to the screen. Specifically, n lines should be printed out, the first consisting of a single asterisk, the second consisting of two asterisks, the third consistings of three, etc. The last line should consist of n asterisks. Thus, for example, if n has value 3, the output of your code should be * ** *** You should not output any space characters. Hint: Use a for loop nested inside another for loop.

2 Answers   HCL,

---

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.....

2 Answers  

---

what is macro in c? Difference between single linked list & double linked list what is fifo & lifo? what is stack & queue?

2 Answers   TCS,

---

#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  

---

Given that two int variables, total and amount, have been declared, write a loop that reads integers into amount and adds all the non-negative values into total. The loop terminates when a value less than 0 is read into amount. Don't forget to initialize total to 0. Instructor's notes: This problem requires either a while or a do-while loop.

3 Answers  

---

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

12 Answers   HCL,

---