In my intro to programming course (c++), the instructor told us to:
"Write a program that inputs an unspecified number of integer ("int")
values and then outputs the minimum value and the maximum value that
were entered. At least one value will always be input. (read all input
until EOF). For extra credit make your program also work when the user
doesn't enter any ints at all (just the EOF.)"
I wanted to get fancy, so, in my solution, when just EOF is entered, the program responds with "Oops! You didn't enter anything. Please try again, this time entering at least one integer: " and prompts for input again.
My instructor is saying that this answer is wrong because
After the EOF, there should be no more input to a program (neither
expected by the user nor the program) — using the EOF to switch from
“one mode” of input to another mode of input isn’t supporting the
standards.
Every definition of EOF I've found on the internet doesn't seem to support my professor's definition. EOF, from what I can tell, is simply defined as the end of the current file. It seems perfectly valid to accept input from a user until EOF, do something with that input, and then ask for additional input until EOF again.
Because this is an online course, I was able to review everything we learned relating to EOF and we were only told that EOF meant "End of File" and could be 'used to signal an end to user input' (important, because, even if my professor was wrong, one could argue that I should have adopted his standards if he had specifically told us to. But he didn't tell us to).
What is the proper way to use EOF with user input? Is my professor's statement that "After the EOF, there should be no more input to a program" the standard
and expected way to use EOF? If a program accepts a variable amount of input, does something with it, and then accepts more variable input, is it not acceptable to use EOF with those inputs (aka don't use while(cin >> user_input) in that scenerio)? If so, is there a standard for what should be used to signal end of input in a scenario where you're accepting variable input multiple times?
My exact solution to the assignment is below. My solution to the main assignment "Write a program that inputs an unspecified number of integer ("int") values and then outputs the minimum value and the maximum value that were entered" was considered correct, by the second part of the assignment "make your program also work when the user doesn't enter any ints at all (just the EOF.)" was deemed incorrect ("make the program also work" is the only prompt we were given).
Thanks so much for any feedback!! Obviously, I'm skeptical of my professors feedback / decision, but, in general, I'm just trying to get a sense of C++ community standards.
#include <iostream>
#include <iomanip>
#include <string>
#include <stdlib.h>
using namespace std;
int main(){
string user_input;
int int_input, min_user_input, max_user_input;
bool do_it = true;
cout << "Hi John," << endl;
cout << "Please enter a few integers (signal EOF when finished): ";
while(do_it) {
cin.clear();
cin >> user_input;
if (user_input.empty()) {
cout << endl;
cout << "Oops! You didn't enter anything. Please try again, this time entering at least one integer: ";
}
else {
try {
int_input = atoi( user_input.c_str() );
min_user_input = int_input;
max_user_input = int_input;
while(cin >> int_input) {
if (min_user_input > int_input) {
min_user_input = int_input;
}
if (max_user_input < int_input) {
max_user_input = int_input;
}
}
cout << endl;
cout << "The max user input was: " << max_user_input << endl;
cout << "The min user input was: " << min_user_input << endl;
do_it = false;
}
catch (std::invalid_argument) {
cout << endl;
cout << "Oops! You didn't enter an integer. Please try again, this time only entering integers: ";
do_it = true;
}
}
}
return 0;
}
Note: additional feedback I got on this submission was: to not use c libraries (apparently stdlib.h is one) and that, on some computers (though, apparently, not mine), #include <stdexcept> will be needed to compile.
Answer
Short answer: my instructor is correct. When used with cin, no additional user input should follow an EOF signal. Apparently, in some cases the program won't let you enter more information, but, as #hvd points out
Although your system may let you continue reading from the same file
in the specific case that it is coming from a TTY, due to EOF being
faked there, you shouldn't generally rely on that.
Aka, because I'm using a terminal to enter user input, the program happens to work. In general, it won't work though.
As #RSahu answers, EOF just shouldn't be used to signal the end of variable length cin multiple times in a program. Importantly
There is no standard means, or commonly practiced coding standard, of
indicating when user input has ended for the time being. You'll have
to come up with your own mechanism. For example, if the user enters
"end", you can use it to deduce that the user has ended input for the
time being. However, you have to indicate to the user that that's what
they need to enter. Of course, you have to write code to deal with
such input.
Because this assignment required the use of EOF, what I was attempting to accomplish was, unintentionally, prohibited (aka receive input, check it, possibly receive more input).
Proper use of EOF (can it be used multiple times in a program?)
There is no single EOF. There is EOF associated with every input stream.
If you are reading from a file, you can reset the state of the std::ifstream when it reaches EOF to allow you to read the contents of the file again.
However, if you are reading data from std::cin, once EOF is reached, you can't read from std::cin any more.
In the context of your program, your professor is right. They are most likely talking about reading from std::cin.
EOF, from what I can tell, is simply defined as the end of the current file.
It is. Note that in particular, what it doesn't mean is the automatic start of a new file.
Although your system may let you continue reading from the same file in the specific case that it is coming from a TTY, due to EOF being faked there, you shouldn't generally rely on that. Try program </dev/null and see happens when you try to automate your program.
Related
My code is rather simple. I have a method called promptUserInput that is defined in UtilityFunctions.h; it is implemented in UtilityFunctions.cpp.
My main method includes UtilityFunctions.h. I've correctly written my makefile, and it compiles without issue.
My main method's first line reads:
string input = promptUserInput();
And here's the actual implementation of the promptUserInput function:
/* Prompts user to enter expression */
string promptUserInput()
{
string userInput;
cout << "> ";
getline(cin, userInput);
return userInput;
}
But when the program runs, it doesn't display the > symbol. Instead, there's an empty line waiting for user input. I enter some arbitrary character, hit Enter, and the > symbol appears afterwards. None of this makes sense to me. What's going on?
std::cout uses buffered output, which should always be flushed. You can achieve this by using std::cout.flush() or std::cout << std::flush.
You can also use std::cout << std::endl, which writes a line break and then flushes, but the line break might not satisfy your intention.
In C++, iostream automatically puts in a new line after cin. Is there a way to get rid of this?
I want use iomanip to format information into a table, like so:
cin cout
0.078125 3DA00000
-8.75 C10C0000
23.5 41BC0000
(random numbers)
example code:
#include <iostream>
using namespace std;
int main()
{
int num;
cin >> num; //now a new line.
cout << num << endl;
return 0;
}
You presumably pressed the return key to send your input from the command line to your program's standard input. That's where the newline is coming from. You can't read your number from cin before this newline appears in the console, because the newline is what causes the console to hand your input over to the program in the first place. You could, as a user, configure your console (or whatever is running your program) to act differently, but there's no way for the program itself to force such behavior.
If you really want to have your input and your output on the same line, you need to find a way to "write to the previous line". How that works depends on your console (see also How to rollback lines from cout?). There is no standard way to do this because cin and cout are in no way obligated to be attached to a console or anything resembling one, so it is not clear that "writing to the previous line" even means anything.
'endl' makes a new line just don't use it.
cout << num;
This question already has answers here:
Cin.Ignore() is not working
(4 answers)
Closed 9 years ago.
I am writing a C++ program and I need to program to end/terminate after the user hits enter.
This is what I have:
cout << "Press Enter to End" << endl;
cin.ignore(); //ends after the user hits enter
return 0;
But it doesn't work. Any advice?
I sense that you had some formatted input earlier in your program, i.e., it looks something like this:
std::cin >> value;
do_something(value);
std::cout << "press enter to end\n";
std::cin.ignore();
If that is the case, you have some character in the input buffer from the time you entered value. For example, there can be a '\n' but it can be any odd other characters, too. This character will be read when the program encounters std::cin.ignore().
What you probably want to is to get rid of the characters currently known to be in the buffer. This isn't quite what would be done as this can still miss characters which are already entered but not, yet, transferred but there is no portable approach to clear all potential characters (it is normally not a problem because hardly any user interface depends on character input from a terminal).
To ignore the characters which are known to be present you need to start off by breaking the connection between <stdio.h> and IOStreams using std::ios_base::sync_with_stdio() and later consume the known characters entered after your last read, e.g.,
#include <iostream>
int main()
{
std::ios_base::sync_with_stdio(false);
int x;
std::cin >> x;
std::cin.ignore(std::cin.rdbuf()->in_avail());
std::cout << "press enter\n";
std::cin.ignore();
}
The odd call to sync_with_stdio() is necessary to have the IOStreams actually buffer the characters received from the system rather than reading characters individually. As a nice side effect it dramatically improves the performance of using std:cin, std::cout, and the other standard stream objects.
maybe add system("pause") in the end of code; (include first). I guess you want enter to close termination.
Suppose we have a menu which presents the user with some options:
Welcome:
1) Do something
2) Do something else
3) Do something cool
4) Quit
The user can press 1 - 4 and then the enter key. The program performs this operation and then presents the menu back to the user. An invalid option should just display the menu again.
I have the following main() method:
int main()
{
while (true)
switch (menu())
{
case 1:
doSomething();
break;
case 2:
doSomethingElse();
break;
case 3:
doSomethingCool();
break;
case 4:
return 0;
default:
continue;
}
}
and the follwing menu():
int menu()
{
cout << "Welcome:" << endl
<< "1: Do something" << endl
<< "2: Do something else" << endl
<< "3: Do something cool" << endl
<< "4: Quit" << endl;
int result = 0;
scanf("%d", &result);
return result;
}
Entering numerical types works great. Entering 1 - 4 causes the program to perform the desired action, and afterwards the menu is displayed again. Entering a number outside this range such as -1 or 12 will display the menu again as expected.
However, entering something like 'q' will simply cause the menu to display over and over again infinitely, without even stopping to get the user input.
I don't understand how this could possibly be happening. Clearly, menu() is being called as the menu is displayed over and over again, however scanf() is part of menu(), so I don't understand how the program gets into this error state where the user is not prompted for their input.
I originally had cin >> result which did exactly the same thing.
Edit: There appears to be a related question, however the original source code has disappeared from pastebin and one of the answers links to an article which apparently once explained why this is happening, but is now a dead link. Maybe someone can reply with why this is happening rather than linking? :)
Edit: Using this example, here is how I solved the problem:
int getNumericalInput()
{
string input = "";
int result;
while (true)
{
getline(cin, input);
stringstream sStr(input);
if (sStr >> result)
return result;
cout << "Invalid Input. Try again: ";
}
}
and I simply replaced
int result = 0;
scanf("%d", &result);
with
int result = getNumericalInput();
When you try to convert the non-numeric input to a number, it fails and (the important part) leaves that data in the input buffer, so the next time you try to read an int, it's still there waiting, and fails again -- and again, and again, forever.
There are two basic ways to avoid this. The one I prefer is to read a string of data, then convert from that to a number and take the appropriate action. Typically you'll use std::getline to read all the data up to the new-line, and then attempt to convert it. Since it will read whatever data is waiting, you'll never get junk "stuck" in the input.
The alternative is (especially if a conversion fails) to use std::ignore to read data from the input up to (typically) the next new-line.
1) Say this to yourself 1000 times, or until you fall asleep:
I will never ever ever use I/O functions without checking the return value.
2) Repeat the above 50 times.
3) Re-read your code: Are you checking the result of scanf? What happens when scanf cannot convert the input into the desired format? How would you go about learning such information if you didn't know it? (Four letters come to mind.)
I would also question why you'd use scanf rather than the more appropriate iostreams operation, but that would suffer from exactly the same problem.
You need to verify if the read succeeded. Hint: it did not. Always test after reading that you successfully read the input:
if (std::cin >> result) { ... }
if (scanf("%d", result) == 1) { ... }
In C++ the failed state is sticky and stays around until it gets clear()ed. As long as the stream is in failed state it won't do anything useful. In either case, you want to ignore() the bad character or fgetc() it. Note, that failure may be due to having reached the end of the stream in which case eof() is set or EOF is returned for iostream or stdio, respectively.
I am new to C++ and am in a class. I am trying to finish the first project and so far I have everything working correctly, however, I need the user to input a number to select their level, and would like to validate that it is a number, and that the number isn't too large.
while(levelChoose > 10 || isalpha(levelChoose))
{
cout << "That is not a valid level" << endl;
cout << "Choose another level:";
cin >> levelChoose;
}
That is the loop I made, and it sometimes works. If I type in 11 it prints the error, and lets me choose another level. However if the number is large, or is any alpha character it floods the screen with the couts, and the loop won't end, and I have to force exit. Why does it sometimes stop at the cin and wait for user input, and sometimes not? Thanks for the help!
This is an annoying problem with cin (and istreams in general). cin is type safe so if you give it the wrong type it will fail. As you said a really large number or non-number input it gets stuck in an infinite loop. This is because those are incompatible with whatever type levelChoose may be. cin fails but the buffer is still filled with what you typed so cin keeps trying to read it. You end up in an infinite loop.
To fix this, you need to clear the fail bit and ignore all the characters in the buffer. The code below should do this (although I haven't tested it):
while(levelChoose > 10 || isalpha(levelChoose))
{
cout << "That is not a valid level" << endl;
cout << "Choose another level:";
if(!(cin >> levelChoose))
{
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
}
Edit: numeric_limits<> is located in the limits include:
#include<limits>
From your description, it seems likely (nearly certain) that levelChose is some sort of numeric type, probably an integer.
When you use operator>> to read a number, anything that couldn't be part of a number (e.g., most letters) will be left in the input buffer. What's happening is that you're trying to read the number, it's failing and leaving the non-digit in the buffer, printing out an error message, then trying to read exactly the same non-digit from the buffer again.
Generally, when an input like this fails, you want to do something like ignoring everything in the input buffer up to the next new-line.
levelChoose appears to be an integer type of some form (int, long, whatever).
It's not valid to input a character into an integer directly like that. The input fails, but leaves the character in the incoming buffer, so it's still there when the loop comes around again.
Here's a related question: Good input validation loop using cin - C++
I suspect the part while(levelChoose > 10..... This does not restrict level to less than 10 (assuming greater than 10 is a large number in your context). Instead it probably should be while(levelChoose < 10...
To check that an expression is not too large, the following could be a possibility to validate (brain compiled code!!)
const unsigned int MAX = 1000;
unsigned int x;
cin >> x;
while(x < MAX){}