the following is an exercise from a book for practicing some class inheritance. But the problem is in the client, not with the class design. (BaseCore, baseDMA, lacksDMA and hasDMA are the classes BTW).
// usedma.cpp -- polymorphic example (compile with dma.cpp)
#include <iostream>
#include "dma.h" // includes <iostream>
const int ELEMENTS = 1;
const int LENGTH = 30;
int main()
{
using std::cin;
using std::cout;
BaseCore *pArr[ELEMENTS];
char tempDate[LENGTH];
char kind;
for (int i = 0; i < ELEMENTS; i++)
{
cout << "\nEntering data for element #" << i + 1 << "\n\n";
cout << "Enter the date it was created: ";
cin.getline(tempDate, LENGTH - 1);
cout << "Enter 1 for baseDMA, 2 for lacksDMA, or 3 for hasDMA: ";
while (cin >> kind && kind != '1' && kind != '2' && kind != '3')
cout <<"Wrong data. Please, try again: ";
while (cin.get() != '\n')
continue;
char tempLabel[LENGTH];
int tempRating;
cout << "Enter the label: ";
cin.getline(tempLabel, LENGTH - 1);
cout << "Enter the rating: ";
cin >> tempRating;
if (kind == '1') // baseDMA
pArr[i] = new baseDMA(tempDate, tempLabel, tempRating);
if (kind == '2') // lacksDMA
{
char tempColor[LENGTH];
cout << "Enter the color: ";
cin.getline(tempColor, LENGTH - 1);
pArr[i] = new lacksDMA(tempDate, tempLabel, tempColor, tempRating);
}
if (kind == '3') // hasDMA
{
char tempStyle[LENGTH];
cout << "Enter the style: ";
cin.getline(tempStyle, LENGTH - 1);
pArr[i] = new hasDMA(tempDate, tempLabel, tempStyle, tempRating);
}
while (cin.get() != '\n')
continue;
}
cout << "\n";
for (int i = 0; i < ELEMENTS; i++)
{
pArr[i]->View();
cout << "\n";
}
cout << "Done.\n";
std::cin.get();
return 0;
}
Sample execution:
Entering data for element #1
Enter the date it was created: 2012.01.01
Enter 1 for baseDMA, 2 for lacksDMA, or 3 for hasDMA: 2
Enter the label: lacksDMA
Enter the rating: 15
Enter the color: blue
Date of creation: 2012.01.01
Label: lacksDMA
Rating: 15
Color:
Done.
It seems the Color member gets assigned the null character. This behavior happens inside both the if (kind == '2') and if (kind == '3') statements (with the style member in this case).
If I put a cin.get(); just before cin.getline() it works fine but I have to press an extra key to make the program ask for input.
Why is this happening? If there was a '\n' pending in the input queue, cin.getline() would discard it and put '\0' in the variable, I can understand that. But the program asks me for the input for color and let's me enter it normally. Also, if I put a cin.get(), then the program shouldn't be waiting for an extra key stroke in the execution, it just should get rid of that extra '\n'. What am I missing here?
cout << "Enter the rating: ";
cin >> tempRating;
Unlike, istream::getline(), operator>> leaves the trailing \n in the stream. It causes the next call to getline inside one of your if statements to get empty input.
The stream is empty when control flow reaches while (cin.get() != '\n') statement at the end of for loop - it's waiting for input and it appears as if you're still inputing color.
Call cin.ignore() right after and it'll work.
Note that this kind of bug is immediately obvious if you put a "debugging cout" right after the input statement for color. There's one more issue with the way you're getting tempRating. If you enter invalid input, say "xy", the erros flags will be set on cin and the program will enter an infinite loop. Always check whether input operations suceeded.
If I put a cin.get(); just before cin.getline() it works fine but I have to press an extra key to make the program ask for input.
It seems to me that when you don't put the cin.get(), your getline is getting an empty character. Then, when you put the cin.get, you get that empty character and the your getline works fine..
But you should definitely go in debug to see exactly what's happening!
Related
I'm trying to only allow integer values into my program, so I've made the following function. The function is similar to other ones I've seen online, and mine seems to work just fine up until I add an ! in front of it to check if something is not an int.
Function to check if input is an integer:
bool isInteger(std::string s)
{
for (int i = 0; i < s.length(); i++)
{
if (isdigit(s[i]) == false)
{
return false;
}
return true;
}
}
Function being put to use:
int getLevel()
{
int level;
std::cout << "Level One\n";
std::cout << "Level Two\n";
std::cout << "Level Three\n";
std::cout << "Level Four\n";
std::cout << "Level Five\n";
std::cout << "Enter your level (1-5): ";
std::cin >> level;
while (!isInteger(std::to_string(level)) || level < 1 || level > 5)
{
std::cout << "Enter an integer value between 1-5 inclusive: ";
std::cin >> level;
}
clrscr();
return level;;
}
I believe the function works just fine until I put the ! in front of it. I am trying to only allow integer input into my program, and when I enter a double or string, the console becomes flooded with the message "Enter an integer value between 1-5 inclusive: " and doesn't give any time to enter an input. I am fairly new to c++ programming and could use some advice. Thank you!
std::cin >> level;
will try to read an integer and it will never read anything other than an integer. If this fails std::cin's failbit is set and further input operations (like std::cin >> level; inside the loop) are skipped.
You need to check if the reading succeeded and ignore the current input if not. Like this for example:
std::cout << "Enter your level (1-5): ";
while(!(std::cin >> level) || level < 1 || level > 5) {
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
std::cout << "Enter an integer value between 1-5 inclusive: ";
}
As little semi-related hint: level will always be an integer. Converting it to a string will always be the string-representation of an integer, so isInteger(std::to_string(level)) will always be true, unless level is negative, because you don't check for the sign.
Also that return true; in isInteger must be outside the loop, else you only check the first character.
Thanks to all the replies and clarification, I've managed to come up with a solution of my own.
New isInteger function that now checks for everything that is needed including inputs like "0004" that a user suggested above:
bool errorCheck(std::string s)
{
int intLevel;
std::stringstream tempLvl(s);
tempLvl >> intLevel;
for (int i = 0; i < s.length(); i++)
{
if (isdigit(s[i]) == false || s[0] == '0' || intLevel < 1 || intLevel > 5)
{
return false;
}
}
return true;
}
The method in action:
std::cout << "Enter your level (1-5): ";
std::cin >> stringLevel;
while (!errorCheck(stringLevel))
{
std::cout << "Enter an integer value between 1-5 inclusive: ";
std::cin >> stringLevel;
}
std::stringstream lvl(stringLevel);
lvl >> level;
clrscr();
return level;
}
Please let me know if you spot any problems with the code or have any easier solutions. Thanks for all the help!
ok i am gonna tell u the fact that console input extracts the input from console so if u ever tried to do something like that
i.e read string in place of integer the cin is going to be in bad state you can check this fact by putting an if like this
if(!cin>>level) break;
and u will find it working actually stream takes input from the console and convert it to boolean value so u can always check it's state bad state return false else true...... ..
SO,finally the bug is in cin>>level...
I hope u understood.... also check out that return true statement..
i am gonna put u reference link for more answer on this bug...
user enters String instead of Int
I'm learning c++ and reading c++ primer plus, but I don't understand why this code need two "cin >> ch". I know the first cin will read character that was user input.but then I delete first "cin >> ch" and run code ,the program have no error.So the fist cin is necessary? why the second cin needn't user to input?
#include <iostream>
int main()
{
using namespace std;
char ch;
int count = 0;
cout << "Enter characters; enter # to quit:\n";
cin >> ch; //get a character
while (ch != '#')
{
cout << ch;
++count;
cin >> ch; // get the next character
}
cout << endl << count << " characters read\n";
return 0;
}
You can evaluate your input right inside condition of while loop.
#include <iostream>
int main()
{
char ch;
int count = 0;
std::cout << "Enter characters; enter # to quit:\n";
while (std::cin >> ch && ch != '#')
{
std::cout << "entered: " << ch << std::endl;
++count;
}
std::cout << std::endl << count << " characters read" << std::endl;
return 0;
}
When while condition is entered it will wait for you to enter anything first. Once input is received it will check if the input is not #. If input is not # the loop is entered, input printed out, counter increased, and back to waiting for another input. If # is entered, condition becomes false, and loop is aborted.
If you remove the first cin then count will never be incremented. The user can enter # character before entering the loop and the program can never enter it therefore.
The first cin>>ch is obviously used to take input from user but you
have again accepting data in while loop using the same variable name "ch" ,
So when you run the program it will not give u error but accept only first value that you have accept before the while loop not in while loop.
In while loop you can assign new value to variable "ch" but not accept the new value again.
I am writing a library program that displays a menu of options letting the user add new books to the library, but in my add statement it accepts the title and then gets caught in an infinite loop. I wrote a book class that mainly uses pointers to assign things, if I need to post that I will. But when you run the program it compiles, displays the menu, and when you choose add a book it accepts the title but as soon as you hit enter it starts an a infinite loop.
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <string>
using namespace std;
int main()
{
int bookCounter = 0;
Book library[25];
int menuOption = 0;
char tempt[50] = "\0";
char tempauth[50] = "\0";
char search[50] = "\0";
unsigned int tempp = 0;
do
{
menuOption = 0;
cout << endl << "1. Show the Library" << endl;
cout << "2. Add a Book" << endl;
cout << "3. Search the Library by Title" << endl;
cout << "4. Exit Library" << endl;
cout << "Select a menu option (e.g. 1, 2, etc.): ";
cin >> menuOption;
if(menuOption == 1)
{
for(int i = 0; i < bookCounter; i++)
{
library[i].displayBook();
}
}
else if(menuOption == 2)
{
cout << "Enter the Title: ";
cin >> tempt[50];
cout << endl << "Enter the Author's name: " ;
cin >> tempauth[50];
cout << endl << "How many pages does the book have? (just enter a number, e.g. 675, 300): ";
cin >> tempp;
library[bookCounter].setAuthor(tempauth);
library[bookCounter].setTitle(tempt);
library[bookCounter].setPages(tempp);
bookCounter++;
menuOption = 0;
}
else if(menuOption == 3)
{
cout << "Enter a title you would like search for (will return partial matches): ";
cin >> search[50];
for (int i = 0; i < bookCounter; i++)
{
int temp = strcmp(search, library[i].getTitle());
if (temp == 1)
{
library[i].displayBook();
}
}
}
}while(menuOption != 4);
system("pause");
return 0;
}
The problem is caused by the way you are trying to read into the arrays:
cin >> tempt[50];
This tries to read a single character into the character at index 50 of the array tempt, which is outside the bounds of the array (which has valid indices in the range [0,49]).
This means only the first character of the entered title will be consumed from the output. Similarly for author. Hence, only the first two characters which you have entered are actually read. Then, this line will be encountered:
cin >> menuOption;
Here, what is left in the buffer (the remainder of the title) will be read, expecting a number. As this does not match a valid format for a number, you will get an error flag in cin. This will mean that all resulting inputs will also fail, menuOption will never change and your program gets stuck in a loop.
A solution to your problem would be to read into tempt without index. You can also check if a read has failed using if(cin.fail()) which should only trigger if there's been an error. If so, handle it and then call cin.clear() to reset the error flags.
I think that this line cause the problem,
cin >> search[50];
You're accessing out bound of search array.
One error is when you type in the menu option, the 'return' stays in the input buffer. The next read of char[] in your tempt variable, will be skipped.
Type cin.ignore(); after cin >> menuOption;
Also, you should read tempt instead instead of tempt[50].
This
cin >> tempt[50];
accesses a non-existent entry in the array. You probably meant to code
cin >> tempt;
Or, better, use std::string instead of raw char array.
So, this program I am working on is not handling incorrect user input the way I want it to. The user should only be able to enter a 3-digit number for use later in a HotelRoom object constructor. Unfortunately, my instructor doesn't allow the use of string objects in his class (otherwise, I wouldn't have any problems, I think). Also, I am passing the roomNumBuffer to the constructor to create a const char pointer. I am currently using the iostream, iomanip, string.h, and limits preprocessor directives. The problem occurs after trying to enter too many chars for the roomNumBuffer. The following screenshot shows what happens:
The relevant code for this problem follows:
cout << endl << "Please enter the 3-digit room number: ";
do { //loop to check user input
badInput = false;
cin.width(4);
cin >> roomNumBuffer;
for(int x = 0; x < 3; x++) {
if(!isdigit(roomNumBuffer[x])) { //check all chars entered are digits
badInput = true;
}
}
if(badInput) {
cout << endl << "You did not enter a valid room number. Please try again: ";
}
cin.get(); //Trying to dum- any extra chars the user might enter
} while(badInput);
for(;;) { //Infinite loop broken when correct input obtained
cin.get(); //Same as above
cout << "Please enter the room capacity: ";
if(cin >> roomCap) {
break;
} else {
cout << "Please enter a valid integer" << endl;
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}
}
for(;;) { //Infinite loop broken when correct input obtained
cout << "Please enter the nightly room rate: ";
if(cin >> roomRt) {
break;
} else {
cout << "Please enter a valid rate" << endl;
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}
}
Any ideas would be greatly appreciated. Thanks in advance.
Read an integer and test whether it's in the desired range:
int n;
if (!(std::cin >> n && n >= 100 && n < 1000))
{
/* input error! */
}
Although Kerrek SB provide an approach how to address the problem, just to explain what when wrong with your approach: the integer array could successfully be read. The stream was in good state but you didn't reach a space. That is, to use your approach, you'd need to also test that the character following the last digit, i.e., the next character in the stream, is a whitespace of some sort:
if (std::isspace(std::cin.peek())) {
// deal with funny input
}
It seems the error recovery for the first value isn't quite right, though. You probably also want to ignore() all characters until the end of the line.
I have a while loop here that only takes in 1 and 2 as the number, if i insert and number that is not these my else statement will keep asking for the correct one, which works correctly. But if i insert a letter my else statement loops forever. How can i fix this?
#include <iostream>
using namespace std;
int main()
{
int myChoice;
cin >> myChoice;
while ( myChoice >= 2 || myChoice <= 1)
{
if (myChoice == 1)
{
cout <<"food1";
break;
}
else if (myChoice == 2)
{
cout <<"food2";
break;
}
else
{
cout << " " << endl;
cout << "Please select the proper choices" << endl;
cout << "Try again: ";
cin >> myChoice;
}
}
return 0;
}
If you enter a non-number, then cin >> myChoice fails. That means that it leaves the input intact in the input buffer and when you get there again it tries to parse it and fails, and so on... You must clear the error state and ignore the non-digits. The simplest way is something like this:
cout << "Try again: ";
cin.clear(); // clear error state
cin.ignore(std::numeric_limits<streamsize>::max(), '\n'); // ignore till the end of line
cin >> myChoice;
The problem here is that the cin >> operator expects to receive an int input and receives a char input.
The istream module, of which cin is an instance, is using buffered I/O. This means that the user input is first stored in a buffer, and then read from that buffer when the user program accesses the >> operator. Ordinarily, if the >> operator succeeds in reading and parsing the user input, the read data is extracted from the buffer and the next invocation of the >> operator would continue where the last call left off. In you case, however, the >> operator attempts to parse the user input as a number and fails since it contains illegal chars which are not digits. The >> operator doesn't extract the read data from the buffer in this case and this same data is being referred to over and over again in the following calls to the >> operator.
You should empty the buffer on failure, the way ybungalobill suggested, for instance.
Your while condition is always true, then you use break to exit the loop. You could simplify things a bit like this:
#include <iostream>
using namespace std;
int main()
{
int myChoice;
cin >> myChoice;
while( myChoice != 1 || myChoice != 2 ) {
cout << endl;
cout << "Please select the proper choices" << endl;
cout << "Try again: ";
cin.clear();
cin.ignore(std::numeric_limits<streamsize>::max(), '\n');
cin >> myChoice;
}
// At this point myChoice is 1 or 2
if (myChoice == 1)
cout << "food1";
else if (myChoice == 2)
cout << "food2";
}