Answer / erdem

{
int num, sum ;
printf("Enter digit 1\n");
scanf("%d",num);
sum=sum+num;
for(i=1;i<4;i++)
{
printf("Enter digit %d\n",i);
scanf("%d",num);
sum=sum+num;
}
printf("the sum of digits is %d",sum);
getche();

Answer / greeshmaanair

{
int sum,digit1,digit2,digit3,digit4;
printf("enter the four digit number\n");
scanf("%d",digit1,digit2,digit3,digit4);
digit1=(digit1%1000)/10;
digit2=(digit2%1000)/10;
digit3=(digit3%1000)/10;
digit4=(digit4%1000)/10;
sum=digit1+digit2+digit3+digit4;
printf("the sum of digits is %d",sum);
return(0);
}

I'm having trouble with coming up with the correct code. Thank You!! The assignment was to write a program using string functions that accepts a price of an item and displays its coded value. The base of the keys: X C O M P U T E R S 0 1 2 3 4 5 6 7 8 9 Sample I/O Dialogue: Enter Price: 489.50 Coded Value: PRS.UX

0 Answers  

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which typw of errors ? & how to solve it ?

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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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Given that two int variables, total and amount , have been declared, write a sequence of statements that: initializes total to 0 reads three values into amount , one at a time. After each value is read in to amount , it is added to the value in total (that is, total is incremented by the value in amount ). Instructor's notes: If you use a loop, it must be a for loop. And if you use a loop control variable for counting, you must declare it.

1 Answers   Google,

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

1 Answers  

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printy(a=3,a=2)

3 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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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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How to develop a program using C language to convert 8-bit binary values to decimals. TQ

1 Answers   Amazon,

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

6 Answers   HCL,

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How to convert hexadecimal to binary using c language..

1 Answers   Bajaj, GAIL, Satyam, Zenqa,

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what is the error in the following code: main() { int i=400,j; j=(i*i)/i; }

4 Answers  

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