Answer / susheel kumar

bug

Answer / anii

It's a human mistake associated with the program

Answer / rekha

Issue

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

0 Answers   TCS,

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Why are memory errors hard to debug?

1 Answers  

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quoroum of computer languages?

0 Answers   Infosys,

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how to convert decimal to hexadecimal without using arrays just loops

2 Answers  

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which typw of errors ? & how to solve it ?

0 Answers  

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Write a program to accept two strings of Odd lengths. Then take all odd characters from one string and even characters from the other and concatenate and produce a string.

1 Answers  

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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]

0 Answers  

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loop1: { x=i<n?(i++):0; printf("%d",i); exit(x); continue; } Error- misplaced continue. Doubt-1.will the exit(x) be executed for all values of x 2.will this statement go out of the program.

5 Answers   CMC,

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what is the large sustained error signal that eventually cause the controller output to drive to its limit

1 Answers   TCS,

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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  

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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); }

1 Answers  

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

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

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