Answer / roxy

this is my code:

#include<stdio.h>
#include<string.h>

main()
{
char cc1[]={"OTTAWA"},
{"MOSCOW"},
{"ROME"};
char cc2[]={"CANADA"}
{"RUSSIA"}
{"ITALY"};
char a;

clrscr();

printf("Enter Capital: ");
scanf("%s",&cc1);

a=strcmpi(cc1,cc2);
strncpy(cc2,cc1,[1]);

if(a==0)
{
printf("\nThe capital of %s is %s",cc1,cc2);
}
if(a==1)
{
printf("\nThe capital of %s is %s",cc1,cc2);
}
else(a==2)
printf("\nThe capital of %s is %s",cc1,cc2);

getch()
}

How to upgrade LOOP environment, I just mean, how can i make loop statement editable ? I just try some program using loop statement and checking it in multiple compilers. Every compiler showing different output, what's the wrong ? is it a compiler based problem, or loop based problem, tell me why ? and what will be the debugging process, for this kind of problem ?

1 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  

---

#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  

---

void main() { int i=1; printf("%d%d%d",i,++i,i++); } Cau u say the output....?

24 Answers   HCL,

---

what is syntax error?

3 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,

---

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

1 Answers  

---

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  

---

write the value of x and y after execution of the statements: int x=19,y; y=x++ + ++x; x++; y++;

0 Answers  

---

what is run time error?

7 Answers  

---

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

9 Answers  

---

full c programming error question based problem

3 Answers   HCL, TCS,

---