Я написав програму, яка передбачає використання операторів переключення ... Однак при компіляції це показує:
Помилка: перейти до регістру.
Чому це робиться?
#include <iostream>
#include <cstdlib>
#include <fstream>
#include <string>
using namespace std;
class contact
{
public:
string name;
int phonenumber;
string address;
contact() {
name= "Noname";
phonenumber= 0;
address= "Noaddress";
}
};
int main() {
contact *d;
d = new contact[200];
string name,add;
int choice,modchoice,t;//Variable for switch statement
int phno,phno1;
int i=0;
int initsize=0, i1=0;//i is declared as a static int variable
bool flag=false,flag_no_blank=false;
//TAKE DATA FROM FILES.....
//We create 3 files names, phone numbers, Address and then abstract the data from these files first!
fstream f1;
fstream f2;
fstream f3;
string file_input_name;
string file_input_address;
int file_input_number;
f1.open("./names");
while(f1>>file_input_name){
d[i].name=file_input_name;
i++;
}
initsize=i;
f2.open("./numbers");
while(f2>>file_input_number){
d[i1].phonenumber=file_input_number;
i1++;
}
i1=0;
f3.open("./address");
while(f3>>file_input_address){
d[i1].address=file_input_address;
i1++;
}
cout<<"\tWelcome to the phone Directory\n";//Welcome Message
do{
//do-While Loop Starts
cout<<"Select :\n1.Add New Contact\n2.Update Existing Contact\n3.Display All Contacts\n4.Search for a Contact\n5.Delete a Contact\n6.Exit PhoneBook\n\n\n";//Display all options
cin>>choice;//Input Choice from user
switch(choice){//Switch Loop Starts
case 1:
i++;//increment i so that values are now taken from the program and stored as different variables
i1++;
do{
cout<<"\nEnter The Name\n";
cin>>name;
if(name==" "){cout<<"Blank Entries are not allowed";
flag_no_blank=true;
}
}while(flag_no_blank==true);
flag_no_blank=false;
d[i].name=name;
cout<<"\nEnter the Phone Number\n";
cin>>phno;
d[i1].phonenumber=phno;
cout<<"\nEnter the address\n";
cin>>add;
d[i1].address=add;
i1++;
i++;
break;//Exit Case 1 to the main menu
case 2:
cout<<"\nEnter the name\n";//Here it is assumed that no two contacts can have same contact number or address but may have the same name.
cin>>name;
int k=0,val;
cout<<"\n\nSearching.........\n\n";
for(int j=0;j<=i;j++){
if(d[j].name==name){
k++;
cout<<k<<".\t"<<d[j].name<<"\t"<<d[j].phonenumber<<"\t"<<d[j].address<<"\n\n";
val=j;
}
}
char ch;
cout<<"\nTotal of "<<k<<" Entries were found....Do you wish to edit?\n";
string staticname;
staticname=d[val].name;
cin>>ch;
if(ch=='y'|| ch=='Y'){
cout<<"Which entry do you wish to modify ?(enter the old telephone number)\n";
cin>>phno;
for(int j=0;j<=i;j++){
if(d[j].phonenumber==phno && staticname==d[j].name){
cout<<"Do you wish to change the name?\n";
cin>>ch;
if(ch=='y'||ch=='Y'){
cout<<"Enter new name\n";
cin>>name;
d[j].name=name;
}
cout<<"Do you wish to change the number?\n";
cin>>ch;
if(ch=='y'||ch=='Y'){
cout<<"Enter the new number\n";
cin>>phno1;
d[j].phonenumber=phno1;
}
cout<<"Do you wish to change the address?\n";
cin>>ch;
if(ch=='y'||ch=='Y'){
cout<<"Enter the new address\n";
cin>>add;
d[j].address=add;
}
}
}
}
break;
case 3 : {
cout<<"\n\tContents of PhoneBook:\n\n\tNames\tPhone-Numbers\tAddresses";
for(int t=0;t<=i;t++){
cout<<t+1<<".\t"<<d[t].name<<"\t"<<d[t].phonenumber<<"\t"<<d[t].address;
}
break;
}
}
}
while(flag==false);
return 0;
}
ось одне вражаюче повідомлення про помилку
—
jozxyqk,
caseблок в дужки?