Answer / unknown

#include<stdio.h>

void main()
{
int i,j,k,num = 5;

for(i = num;i > 0;i--){
for(j = 1;j <= i;j++){
printf(" ");
}
for(k = (j - 1);k < num;k++){
if(k==j-1 || k==num-1)
printf("*");
else
printf(" ");
printf(" ");
}
printf(" \n");
}


for(i = num;i > 0;i--){
for(j = (num - i);j > 0;j--){
printf(" ");
}
for(k = 0;k < i;k++){
if(k==0|| k==i-1)
printf("*");
else
printf(" ");
printf(" ");
}
printf(" \n");
}
}

class test { int a; public: test(int b):a(b){} void show(){ cout<<a; } }; void main() { test t1; test t2(5); t1.show(); t2.show(); } }

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Answering Yes or No in C++...using only stdio.h and conio.h..........help me please...? here's must be the output of the program: Screen A Exam No. items Score 1 20 20 2 35 35 Another Entry? [Y] or [N] : Screen B: Record No. Student's Name: 1 Fernando Torres 2 Chuck Norris Note: if you press Y, the program must repeat the procedure in screen A, then if N, the program must proceed to the screen B....Please Help me out............

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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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how to convert decimal to hexadecimal without using arrays just loops

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What is probability to guarantee that the task a programmer is going to create will be created and be able to run on a particular system (RTOS/GPOS).

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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.

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How to convert hexadecimal to binary using c language..

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2 Answers  

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