I am in the second phase of a project where I need to extend my program into a menu driven application to query the database I have on a .txt file. So, my trouble is that I cannot get my loop to be perpetual. It always terminates when it initializes from one option to the next. Here is the snippet of my code that is my int main:
int main ()
{
char Q,q;
char S,s;
char task;
string pathname;
string z;
int count=0;
cout << "Welcome to Jason Rodriguez's Library Database." << endl;
cout << "Please enter the name of the backup file: ";
cin >> pathname;
ifstream inFile(pathname.c_str());
while(!inFile.eof())
{
getline(inFile,z);
count++;
}
while (task != 'Q' || task != 'q') {
cout << count << " records loaded successfully." << endl;
cout << "Enter Q to (Q)uit, Search (A)uthor, Search (T)itle, (S)how All: ";
cin >> task;
if ((task == 'Q')||(task =='q'))
{
cout << "Program will now terminate";
break;
}
else if ((task == 'S')||(task =='s'))
{
showAll (loadData (pathname));
cout << endl;
cout << "Enter Q to (Q)uit, Search (A)uthor, Search (T)itle, (S)how All: ";
cin >> task;
}
}
}
I need to add two more options into the loop on top of these two but I figured I should get the first two working correctly first. The other two should be plug & chug after that. Basically what I was trying to do is say if the user enters Q or q, terminate the program. Else, if user hits S or s, activate showall function and after ward, go back to the original query. It isn't working though. Assistance is welcome and appreciated.
Menus almost always require loops - especially ones that require the user to enter the correct choice input. The most applicable one in a case like this is the while loop - but essentially, any other loop variant can be used.
UPDATE:
int main ()
{
char task;//this is the only char needed. Your other chars were redundant
string pathname;
string temp;//I changed z to temp to better reflect its purpose
int count=0;
cout << "Welcome to Jason Rodriguez's Library Database." << endl;
cout << "Please enter the name of the backup file: ";
cin >> pathname;
ifstream inFile(pathname.c_str());//this is potentially a problem in that you aren't verifying that the pathname is a valid one
//you did not check to see that your file was open, otherwise there is no way to tell that you successfully opened the file
if (inFile.is_open()) {
//while(!inFile.eof()) is a character by character read and comparison
//made your life easier by shortening it down to this - which ensures
//that a line is read. (Much faster and more readable)
while(getline(inFile,temp))
{
count++;
}
inFile.close();//always close a file after you've used it
//At this point the entire file has been read. So, this is where this message SHOULD be
cout << count << " records loaded successfully." << endl;
}
else {
//if there was an error opening the file (i.e. wrong path, or it simply does not exist), this will be displayed
cout << "There was a problem opening your file" << endl;
exit(0);//and the program will terminate
}
while (task != 'Q' || task != 'q') {
cout << "Enter Q to (Q)uit, Search (A)uthor, Search (T)itle, (S)how All: ";
cin >> task;
if ((task == 'Q')||(task =='q'))
{
cout << "Program will now terminate";
break;
}
else if ((task == 'S')||(task =='s'))
{
string author;
//showAll (loadData (pathname));
cout << endl;
cout << "Search an Author" << endl;
cin >> author;//get the author name to search from the user
//write code to search an author here
}
}
}
There are a number of issues with the code that you posted which I will forgo for the sake of brevity. Hence, note the following:
Your code was printing the same message per option (except for quit). Of course it would appear that it didn't work. Each option is a different task. Print what each task does (similar to what I did).
You wish to search the file for an author, but you have not stored it. Look into a way of storing it that appeases your instructor.
It would be ideal for you to use switch in this case, considering the increasing complexity of your code.
Try breaking down each task into functions, and call them to make your main function readable. In fact, it is a good programming practice for your main function to be as small as possible.
And, as juanchopanza quite rightly pointed out: you have some fundamental issues with C++. Try doing some more exercises and do more examples from a good C++ book.
Related
I tried to make a an introduction to a "game", and in its functions I made some Yes/No, 1/2/3, situations.
Im new to this however it wasn't that difficult, worked perfectly. The problem appeared when handling with invalid inputs. So this is what the code looks like by now:
#include "Introduction.h"
#include "GameConstants.h"
#include "PlayerCharacter.h"
#include <iostream>
#include <windows.h>
using namespace std;
Introduction::Introduction()
{
}
/////////Function N.1///////////
void Introduction::presentation()
{
char confirm;
string enteredName;
cout << constants.line() << "Welcome traveler! What is the name?" << endl;
getline(cin,enteredName);// Gets the WHOLE LINE as the name.
while (confirm != 'Y') //If the player doesn't confirm the name with 'Y' in will run again until it does.
{
cout << constants.xline() << "Your name is " << enteredName << " right? (Y/N)" << endl;
cin >> confirm; //The player's answer
cin.sync(); //Only takes the first character
confirm = toupper(confirm); //Turns player message into CAPS for easier detection in the "if" statements
if (confirm == 'N'){ //If not the correct name, gives another chance
cout << constants.xline() << "Please, tell me your name again..." << endl;
cin >> enteredName;
cin.sync();}
if ((confirm != 'Y')&&(confirm != 'N')){ //If an invalid input is entered, gives another chance. And insults you.
cout << constants.xline() << "Fool Go ahead, just enter your name again." << endl;
cin >> enteredName;
cin.sync();}
}
if (confirm == 'Y'){ //When the answer is yes ('Y') /* Uneeded line */
PC.setName(enteredName); //Saves the name
cout << constants.xline() << "Excellent! I have a few more questions for you " << PC.name() << "..." << endl;
}
}
//////////Function N.2///////////
void Introduction::difSelection(){
int selectedDif = 0; //Variable to store selected difficulty whitin this function.
Sleep(2500);
cout << constants.xline() << "What kind of adventure do you want to take part in?" << endl;
Sleep(2500); //Wait 2,5 s
cout << "\n1= Easy\n2= Normal\n3= Hard" << endl;
while(selectedDif != 1&&2&&3){ //Selected option must be 1/2/3 or will run again
cin >> selectedDif; //Sets the user selected difficulty
cin.sync(); //Gets only first character
if((selectedDif != 1||2||3)&&(!(selectedDif))){ //If the input isn't 1/2/3 AND is an invalid character, this will run. And it'll start again
cout << constants.xline() << "Criminal scum. Go again." << endl;
cin.clear();
cin.ignore();
}
if(selectedDif != 1&&2&&3){ //If selected option isn't 1/2/3, this will run and will loop again. However I know this conflicts with the previous statement since this will run anyways.
cout << constants.xline() << "Wrong input, please try again." << endl;
}
else if(selectedDif == 1){
constants.setDiff(1);
constants.setStatPoints(15);
} else if(selectedDif == 2){
constants.setDiff(2);
constants.setStatPoints(10);
} else if (selectedDif == 3){
constants.setDiff(3);
constants.setStatPoints(5);}
}
}
The first function works perfectly you can type "aaa" or "a a a" and will work. However I'd like to know if there's a simpler way to do it. (Understandable for a beginner, just started 3 days ago lol; if it includes some advanced or less known code prefer to stay like this by now).
Now, the second one, I really have no idea how to fix it. I need something that if the user's input was an invalid character type, throw certain message, and if it's an int type, but out of the range, another message. And of course, run again if it fails. Did a lot of search and couldn't find anything that meet this requirements.
To check if the user input is an int, you could use the good() function.
int val;
cin >> val;
if( cin.good() ) {
// user input was a valid int
} else {
// otherwise
}
As for the range check, the syntax is a bit different.
This returns true if the number is not equal to 1 nor 2 nor 3:
selectedDif != 1 && selectedDif != 2 && selectedDif != 3
Another shorter way would be to use:
selectedDif < 1 || selectedDif > 3
Another thing, in c++, there are two keywords break and continue which will allow to reduce the code in the loops.
Disclaimer: I am a beginner to programming, so what I say might sound really stupid
I have to make a "Telephone Directory" for school. The program isn't complete, but there are some things that I need to fix before moving on. The array TelephoneNumbers either isn't storing the numbers from the file correctly, or isn't displaying them. For the SeaerchRecords function, the first number in the file is displayed correctly, the second is displayed as "2147483647," and the rest of the numbers display as "0." The modify function also doesn't change the number, and I confirmed this with the while in the function. The string array works perfectly fine, however. May someone explain what I'm doing incorrectly?
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
string TelephoneNames[100];
int TelephoneNumbers[100];
void ModifyRecords(); //Function to Modify Records
void SearchRecords(); //Function to Search Records
void DeleteRecords(); //Function to Delete Records
int main()
{
fstream inputFile;
fstream outputFile;
char choice;
inputFile.open("Telephone Names.txt"); //To store
for (int count=0;count<100;count++) //file names
{ //into a
inputFile >> TelephoneNames[count]; //string
}
inputFile.close();
inputFile.open("Telephone Numbers.txt");//To store
for (int count=0;count<100;count++) //file #'s
{ //into a
inputFile >> TelephoneNumbers[count];//string
}
inputFile.close();
//Display options available
cout << " Hello, do you want to:\n";
cout << " ======================\n";
cout << "-Modify Records|Enter M\n";
cout << "-Search Records|Enter S\n";
cout << "-Delete Records|Enter D\n";
//Store choice
cin >> choice;
//Send to different function
if (choice=='M'||choice=='m')
{
ModifyRecords();
}
if (choice=='S'||choice=='s')
{
SearchRecords();
}
return 0;
}
void ModifyRecords()
{
string name;
string newname;
int newnumber;
int count=0;
cout << "Enter the name of the person: ";
cin >> name;
for (count=0;TelephoneNames[count]!=name;count++)//To determine where in the strings the new numbers need to be
{
}
cout << "Enter the new name of the person: ";
cin >> newname;
cout << "Enter the new number of the person: ";
cin >> newnumber;
TelephoneNames[count]={newname};
TelephoneNumbers[count]={newnumber};
count=0;
while (count<6)
{
cout << TelephoneNames[count] << endl;
cout << TelephoneNumbers[count] << endl;
cout << endl;
count++;
}
}
void SearchRecords()
{
string name;
int count=0;
cout << "Enter the name of the person you would like to find: ";
cin >> name;
for (count=0;TelephoneNames[count]!=name;count++)//To determine where in the strings the new numbers need to be
{
}
cout << "Name: " << TelephoneNames[count] << endl;
cout << "Number: " << TelephoneNumbers[count] << endl;
}
Since there is no any answer still and I don't see exactly the problem at this point I'll provide some suggestions how you can find a problem in your code.
In any programming situation when you can't find a bug, first task is to locate it as much precisely as you can and check all input data and assumptions. Usually, debugger is used for such purposes, but you can just output text in console before creating final version of your program.
To start with, you must check that you really received names and telephones from your file:
inputFile.open("Telephone Names.txt"); //To store
for (int count=0;count<100;count++) //file names
{ //into a
inputFile >> TelephoneNames[count]; //string
cout << TelephoneNames[count] << endl; //WE MUST SEE WHAT IS REALLY STORED IN TelephoneNames
}
inputFile.close();
inputFile.open("Telephone Numbers.txt");//To store
for (int count=0;count<100;count++) //file #'s
{ //into a
inputFile >> TelephoneNumbers[count];//string
cout << TelephoneNumbers[count] << endl; //WE MUST SEE WHAT IS REALLY STORED IN TelephoneNumbers
}
inputFile.close();
Ok, when it is checked and you are defenitely sure there is no problem in your data we can move to SeaerchRecords function doing the same procedure. We must check what is happening while you are searching:
for (count=0;TelephoneNames[count]!=name;count++)//To determine where in the strings the new numbers need to be
{
cout << "Search step: " << count << " name " << name << " found name " << TelephoneNames[count] << " number " << TelephoneNumbers[count] << endl;
}
Doing so you will locate your bug rather quickly. The problem can be in input files format, in difference of "name" and stored names format etc.
I'll provide several additional suggestion how you can improve your code.
1) Try to use const declarations for such commonly used things as number of records (const int NUMBER_OF_RECORDS = 100; insted of just putting '100' everywhere), it will reduce the amout of work and possible bugs. 2) Try to check all possible problems that you program can encounter if someting is wrong with data. What will happen if you have less than 100 records in your files now? Program crush or silent reading of unappropriate data which is even worse. Check that you haven't reach file end on any step of reading along with current check that you've reached you number of records and do something in case of unappropriate data.
3) Check the possible problems with conditions in your cycles not to run them infinite number of times. Now your condition for(count=0;TelephoneNames[count]!=name;count++)
will execute forever if there is no such name or just crush the program on count 100 or more. You should check that count doesn't exceed that value. Good luck!
I'm new to C++. I decided to not watch the next tutorial and put my skills to use, by making a funny Mind Reader application. I'm pleased with myself, however, even though I've ironed out most bugs, I still have one concerning the exit function. I read the C++ documentation for it, and I'm not sure what I did wrong. I did exit(0);. I have a very weird error, which is:
no match for call to '(std::string {aka std::basic_string<char>}) (int)
I have searched online, however I am still unaware of what the problem is. My error is on line 59 (marked in the code):
#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
int main()
{
//declaring variables to be used later
string name;
string country;
int age;
//header goes below
cout << "#######################################";
" ############ MIND READER ############"
"#######################################\n\n";
//asks if the user would like to continue and in not, terminates
cout << "Would like you to have your mind read? Enter y for yes and n for no." << endl;
cout << "If you do not choose to proceed, this program will terminate." << endl;
string exitOrNot;
//receives user's input
cin >> exitOrNot;
//deals with input if it is 'y'
if (exitOrNot == "y"){
cout << "Okay, first you will need to sync your mind with this program. You will have to answer the following questions to synchronise.\n\n";
//asks questions
cout << "Firstly, please enter your full name, with correct capitalisation:\n\n";
cin >> name;
cout << "Now please enter the country you are in at the moment:\n\n";
cin >> country;
cout << "This will be the final question; please provide your age:\n\n";
cin >> age;
//asks the user to start the sync
cout << "There is enough information to start synchronisation. Enter p to start the sync...\n\n";
string proceed;
cin >> proceed;
//checks to see if to proceed and does so
if (proceed == "p"){
//provides results of mind read
cout << "Sync complete." << endl;
cout << "Your mind has been synced and read.\n\n";
cout << "However, due to too much interference, only limited data was aquired from your mind." << endl;
cout << "Here is what was read from your mind:\n\n";
//puts variables in sentence
cout << "Your name is " << name << " and you are " << age << " years old. You are based in " << country << "." << endl << "\n\n";
cout << "Thanks for using Mind Reader, have a nice day. Enter e to exit." << endl;
//terminates the program the program
string exit;
cin >> exit;
if (exit == "e"){
exit(0); // <------------- LINE 59
}
}
}
//terminates the program if the input is 'n'
if (exitOrNot == "n"){
exit(0);
}
return 0;
}
Thanks
The local variable exit shadows other identifiers from outer scopes with the same name.
To illustrate with a smaller example:
int main()
{
int i;
{
int i;
i = 0; // assign to the "nearest" i
// the other i cannot be reached from this scope
}
}
Since the only exit visible is an object of type std::string, the compiler sees exit(0) as a call to operator()(int) and throws a hissy fit when it doesn't find one among std::string members.
You can either qualify the name (std::exit(0);) or rename the variable. And since all of your code is in main you can simply say return 0; instead.
Try using return 0; or return EXIT_SUCCESS;. It's the exact same thing. Also, you can only input one word into a cin. Instead, use getline(cin, string name); If it still doesn't work, add a cin.ignore(); before your getline(cin, string name);, like this:
//stuff
string country;
cout << "Now please enter the country you are in at the moment:\n\n";
cin.ignore();
getline(cin, country);
//stuff
return 0;
The problem is arrising because you declared a standard keyword as the name of a local variable.
Now as the local variable is of type sting it is not able to take it as its value.
I have a pretty basic ofstream() question. I have an app that matches data that was inputted by a user in a text document. Can i skip lines with ofstream and not modify text already there? If possible, how? Please forgive me my English isn't too great.
#include <fstream>
#include <iostream>
#include <string>
using namespace std;
int main()
{
int count = 0;
int num;
int numcopy;
string clientNames[3000];
string caseNumbers[3000];
int userInp = 1;
string confirm = "2";
cout << "Do you have a file already started (y/n)?"<<endl;
cin >> confirm;
if(confirm == "y")
{
goto input;
}
if(confirm == "n")
{
goto postinput;
}
input:
cout << "What is the number of the query last entered?";
cin >> userInp;
num = userInp;
numcopy = userInp;
postinput:
for(int i = 1; i <3000; i++)
{
userInp ++;
repeat:
cout <<"Enter Client's Name:";
cin >> clientNames[userInp];
cout << " " <<endl;
cout <<"Enter Case Number:";
cin>> caseNumbers[userInp];
cout <<"Client Name "<< i << " "<<clientNames[userInp]<<endl;
cout << "Case Number" << i << " "<<caseNumbers[userInp]<<endl;
cout <<"Is This Correct?"<<endl;
confirm == " ";
cin >> confirm;
if(confirm == "y")
{
cout <<"Confirmed"<<endl;
}
if(confirm == "n")
{
goto repeat;
}
if(confirm == "/end")
{
break;
}
}
ofstream file;
file.open("caseData.txt");
for(int w = 0; w <3000;w++)
{
num++;
file <<
}
}
"skip lines with ofstream and not modify text already there" is not possible.
But you can store all lines of the file in your program first. And while you are processing the file, output the stored line when you want to leave that line unchanged.
This effectively does what you want.
Also, you really should get rid of the gotos in your code. They should be used only in rare cases. And for beginners, I always feel they should not use it until they are very familiar with programming.
I am going to take a guess that you want to do something about the circumstance when the user has inputted the same clientName or same caseNumber multiple times. It is actually not entirely clear from your question that this is what you want to do, but you asked:
I have an app that matches data that was inputted by a user in a text document. Can i skip lines with ofstream and not modify text already there? If possible, how?
However, I did not see any matching logic in your program. You are simply recording up to 2999 entries (since you don't use the 0 entry of your arrays), or wait until the user enters /end as a confirmation message.
If you had actual matching logic, you could detect on input whether the user has typed in the same clientName or same caseNumber, and you can prompt the user for what to do about it (for example, keep the old existing entry, keep newly entered entry). If you had such logic, you would know that you would only be outputting unique lines of data, and so your output loop would be fairly simple.
There were some comments made on your use of goto. Instead of the repeat: label, you could start another loop:
//repeat:
do {
// ...read in input, echo it back, wait for confirmation...
cin >> confirm;
if (confirm == "y") {
cout << "Confirmed" << endl;
break;
}
} while (confirm != "/end");
if (confirm == "/end") {
break;
}
In this code, anything other than a y or /end is treated the same as n.
I'm writing a file matching program for a project for school. The idea is that one program allows you to enter info as follows: 1000 (acct number) Jane Doe 54.50 (balance). Then allow you to enter the account number and a transaction amount for the second program to combine and update a new master file.
The programs are working together just fine (the second one takes information from the first, including any transactions and updates the new balance - searching by account number) but the problem I am running into is with the name.
---Wasn't clear here. When I ask for a name and I put in a single string of characters, the program works fine, if I try to put in a full name, like Jane Doe I go into the loop mentioned below.
I've tried char name[20] which puts me into an infinite loop and I have to 'x' out of the program and I've tried assigning first and lastName to string. That worked for the writing but the program that takes the input file oldMaster and the transaction file inTransaction then outputs a new file newMaster, doesn't recognize the name.
I've tried getline also which isn't working for me, probably programmer error.
Should this be done as an array, if that's possible for this? I think I'm getting hung up on the fact that I am editing files. Answers are fine - but I like to figure it out on my own, just looking for a little guidance on where to go from here.
Hopefully this was fairly clear - if not I'll be happy to explain again in a different way. Just frustrated that I'm this close and can't solve it.
Thanks in advance!
#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <iomanip>
using namespace std;
void createOldMaster()
{
ofstream oldMaster;
int accountNum;
double balance;
char name[15];
oldMaster.open("oldmast.dat", ios::out);
if(!oldMaster)
{
cout << "Unable to open the file." << endl;
exit(1);
} // end if
cout << "Enter the account number (0 to exit)." << endl;
while(true)
{
cout << "Account Number: ";
cin >> accountNum;
if(accountNum == 0)
break;
else
{
\\ This is where it hangs up if I use a first and last name
cout << "\nName: ";
cin >> name;
cout << "\nBalance : " << endl;
cin >> balance;
oldMaster << accountNum << " " << name << " " << balance << endl;
}
}
} //end createOldMaster
void createTransaction()
{
ofstream inTransaction;
int accountNum;
double balance;
inTransaction.open("trans.dat");
if(!inTransaction)
{
cout << "Unable to open the transaction file." << endl;
exit(1);
}
cout << "Enter the account number and balance (0 to exit): " << endl;
while(true)
{
cout << "Account Number: " << endl;
cin >> accountNum;
if(accountNum == 0)
break;
else
{
cout << "Balance: " << endl;
cin >> balance;
inTransaction << accountNum << " " << balance << endl;
}
}
} //end createTransaction
int main()
{
createOldMaster();
createTransaction();
return 0;
}
Your best bet is to use as much of the standard C++ library as you can. Have a reference handy, maybe even a copy of the C++ standard if you're so inclined, and look for shortcuts to make your work easier and your code shorter.
Avoid primitive arrays and primitive strings wherever possible. Instead of primitive arrays try to use std::vector. Instead of primitive strings try to use std::string. Instead of C's FILE* try to use std::ofstream and std::ifstream. If you need to prohibit two accounts with the same account number then choose a C++ container that guarantees unique elements. If you need to find an element in a container try to use a member function of the container for the search, and if that doesn't exist then a standard search function from the standard C++ algorithms.
Reuse and steal mercilessly.