I am attempting to build a small program that includes a while loop:
int main() {
int x = 0;
int y = 0;
while (x != '|' || y != '|') {
cout << "Please enter two numbers: \n";
cin >> x >> y;
cout << "You entered: " << x << " and " << y << "\n\n" << endl;
}
return 0;
}
I compiled and ran on both my windows box with visual studio and my linux box, and when the break requirement is filled my terminal/cmd fills up infinitely like an infinite loop. I've searched in quite a few places and couldn't find a clear answer. I have tried making the while infinite and putting the break parameter inside to no avail. I have a feeling it has to do with putting the vertical symbol in an integer, but I have had for loops running well with it. The lesson I'm in asks for it and I have workarounds that are 'okay', but would like to know a clean/standard way of doing this.
The infinite loop is because you never check for input failure. After the line cin >> x >> y add:
if ( !cin )
break;
Bear in mind that x and y are ints, so they can only store numbers. When you write x != '|' you are testing whether the person successfully entered the number matching the ASCII character code of |. If you want to be able to inspect each character the person entered, input into a string instead.
Related
I am learning C++, and I am doing some exercises in the book I am using. One of them asks to write a program that asks a user how many numbers they want to add up. Then prompt for the numbers the user wants to add or to enter '|' once finished. The numbers are then pushed into a vector. Part of the program asks to check if the size of the vector is equal to the original number of input items and that is where I keep getting an error.
cout << "Please enter the numbers and | once you are done: ";
while(true)
{
for(int num; cin >> num; )
{
if(num == '|')
{
break;
}
ints.push_back(num);
}
if(ints.size() != n)
{
cout << "There are more or less numbers in the vector than originally specified\n"
<< "Vector will be cleared; please re-enter the values: ";
ints.clear();
continue;
}
else
{
break;
}
}
The problem is that if the number of input is off, the message goes into an infinite loop and I am not sure how to fix it.
EDIT: n is the variable that holds in the number of values user wanted to enter.
Thanks!
num is an integer and cin >> num won't extract | symbol. Comparison num == '|' may not work as expected because num could have the numeric value of | ascii symbol even when user did not input any | symbol. You should properly handle end marker reading:
// loop will break when user enters `|` because it is not an integer
// setting failbit of cin
for(int num; cin >> num;)
{
ints.push_back(num);
}
cin.clear(); // reset failbit making cin able to read again
// check the end marker entered by user
{
string end_marker;
cin >> end_marker;
if("|" != end_marker)
{
// use of endl will flush the stream ensuring that
// printed text won't stuck in the buffer
cout << "Please use | as end marker" << endl;
continue;
}
}
Here is how I implemented it. I am worried about the logic in your while loop. I had been taught to avoid while(true) whenever possible. You know the logic behind how your code should work. With more practice you'll start to recognize the conditions you need to use. I am sure there are better ways to do it. But this is the way I tried it.
But to answer your question, the main reason it is failing is because integers cannot compare themselves with characters.
if(num == '|')
That does not work since num is an integer and not a character.
Normally I would implement this in a class and since global variables are not highly looked upon I created my own namespace. You'll have to finish the rest of the logic yourself however:
#include <iostream>
#include <vector>
#include <string>
namespace global
{
std::vector<std::string> strings;
std::vector<int> ints;
std::string a = " ";
int num = 0;
}
void doWork()
{
std::cout << "Please enter the number of integers you would like to add up: ";
std::cin >> global::num;
std::cout << "Please enter the numbers and | once you are done: ";
while (global::a != "|")
{
std::cin >> global::a;
global::strings.push_back(global::a);
}
global::strings.pop_back();
for(auto &e : global::strings)
{
global::ints.push_back(std::stoi(e));
}
}
int main()
{
doWork();
if(global::ints.size() != global::num)
{
std::cout << "Size of vector does not match the size specified. Clearing vector" << std::endl;
global::ints.clear();
global::strings.clear();
global::num = 0;
global::a = " ";
doWork();
}
}
I made a vector of char's and converted those into integers so that way you could add them up. The while loop should be checking for | rather than always running true. It then will check the size of the vector in the end, clear it if it does not match, and ask you to do it again. This is the best way that I could think of doing it.
EDIT: as VTT pointed out, char can only do one character at a time. I have converted it into a string in order to handle the conversion.
EDIT 2: reset the values of global::num and global::a to their default at the end of the failure in order to prevent crashing.
I'm very new to C++. I started teaching myself C++ last week, and up until now, I haven't had any real problems.
I use Microsoft Visual Studio 2017, and I'm having problems with if and else statements. You see, I'm making this very limited calculator. Basically, the program gives you the 4 basic mathematical operations, and you choose which operation you want to use to calculate by entering either 1, 2, 3, or 4. Then, it runs another program which you can then calculate. (e.g.: 2 is for subtraction, if you enter 2, it will run the subtraction calculator)
Here's the code for the program.
#include "stdafx.h"
#include "iostream"
using namespace std;
int main(){
int add = 1;
int sub = 2;
int mul = 3;
int div = 4;
cout << "Extremely Limited C++ Calculator: Enter a number between one and
four to start calculating and press enter."
<< '\n'
<< "LEGEND"
<< '\n' << "1 = Addition"
<< '\n' << "2 = Subtraction"
<< "\n" << "3 = Multiplication"
<< '\n' << "4 = Division"
<< '\n' << "Operation: ";
if (cin >> add) {
system("start C:\\CalculatorApps\\addition.exe");
return 0;
}
if (cin >> sub) {
system("start C:\\CalculatorApps\\subtraction.exe");
return 0;
}
if (cin >> mul) {
system("start C:\\CalculatorApps\\multiplication.exe");
return 0;
}
if (cin >> div) {
system("start C:\\CalculatorApps\\division.exe");
return 0;
}
}
So I have the addition.exe and subtraction.exe done, but the problem is that no matter what number I enter, it will always run addition.exe. In the subtraction calculator, I experimented having the user choose if they wanted to subtract more than 2 numbers, but that also didn't end up working because it was ignoring if statements. I also at one point had an else statement on both the subtraction calculator and the main program that takes you to the calculators that displayed text reading that the number they entered was not a valid choice and to enter a valid choice, but even that was ignored by the program. Now, maybe I didn't look hard enough on the internet, but I couldn't find one that helped me out. If you know the answer, please tell me but in words that I can understand (I am new after all), or please link me to another question that has been answered that will solve my question.
Thank you in advanced!
cin is an object of class istream that represents the standard input stream. It corresponds to the cstdio stream stdin. The operator >>overload for streams return a reference to the same stream. The stream itself can be evaluated in a boolean condition to true or false through a conversion operator.
cin provides formatted stream extraction. The operation cin >> x;
where "x" is an int will fail if a non-numeric value is entered. So:
if(cin>>x)
will return false if you enter a letter rather than a digit.
As you are always entering a digit, your first statement if(cin>>sum) is always true. So, your other statements are skipped.
/**
Write a program that reads a series of numbers and stores them in a vector. After the user inputs all the numbers he or she wishes to, ask how many of the numbers the user wants to sum. For an answer N. print the sum of the first N elements of the vector. For example: "Please enter some numbers (press 'I' at prompt to stop ) : " 12 23 13 24 15 "Please enter how many of the numbers you wish to sum, starting from the first:" 3 "The sum of the first 3 numbers : 12, 23, and 13 is 48." Handle all inputs. For exam ple, make sure to give an error message if the user asks for a sum of more numbers than there are in the vector.
**/
#include<iostream>
#include<vector>
using namespace std;
int main()
{
try
{
vector<int> numbers;
int num;
cout<<"Now enter the numbers";
while(cin>>num)
numbers.push_back(num);
int n,sum=0;
cout << "Enter the nth number to find sum of elements till n : ";
cin>>n;
if(n >numbers.size())
throw 66;
for(int i=0;i<n;i++)
sum+=numbers[i];
cout << "sum is "<<sum;
return 0;
}
catch(int k)
{
cerr<<"Error "<<k;
return -1;
}
}
So , when I enter EOF , CTRL+D , the program terminates. I am not sure where it is going wrong. I even tried to debug using gdb(with the help from an online tutorial) . It didn't just work out . can someone tell me what's wrong with the code ?
You are not checking if you actually read anything.
Consider this little test program:
#include <iostream>
int main()
{
std::cout << "std::cin is " << (std::cin ? "ready" : "done") << "\n";
int n = -42;
std::cin >> n;
std::cout << n << "\n";
std::cout << "std::cin is " << (std::cin ? "ready" : "done") << "\n";
n = -42;
std::cin >> n;
std::cout << n << "\n";
std::cout << "std::cin is " << (std::cin ? "ready" : "done") << "\n";
}
The output, when fed with an empty standard input (which is equivalent to immediately declaring its end with ctrl+d) is:
std::cin is ready
-42
std::cin is done
-42
std::cin is done
As you can see, n is never changed, as there is never a new value to change it to! Also, you can easily spot that the state of std::cin reflects if the previous read went past the end.
Since you are only checking the value of your integers without ensuring that they have a sane default (just check what happens to n if it is not set by reading the input), this can easily lead to your program exhibiting unexpected behavior.
Note: The behavior of the test program is different when fed input that simply is not a number.
When you supply cin with EOF it causes cin.failbit to become true. With the failbit set to true, all subsequent cin reads will be ignored. Since n has no default value execution becomes unpredictable from here. In my case the program was crashing because it was throwing 66. Adding cin.clear() after the while loop will fix this, but is not advisable. Two simple solutions would to stop on a magic number/prompt the user after every input if they want to continue.
hi i am new to c++ and i dont understand why my while statement doesnt work now. it was working when i tried to do it earlier.
Full code is available at: http://pastebin.com/aeH5fKwh
basically here is the while loop (i excluded all the unnecessary parts, i left the inside of the while loop intact for viewing purpose)
int main()
{
unsigned int seed;
char input;
bool done;
for (int round = 0; round < 5; round++)
{
done = false;
cout << "\nEnter seed: ";
cin >> seed;
cout << "\nRound 1" << endl;
while(!done)
{
cout << "\nDo you wish to draw another card [y][n]: ";
cin >> input;
while (input != 'y' && input != 'n')
{
cout << "Invalid input! Please enter [y][n]!" << endl;
cin >> input;
}
if (input == 'y')
{
dealExtra(playerHand, deck, gameInfo);
cout << "Your cards are ";
printHand(playerHand, gameInfo.playerCardCount);
}
else
done = true;
}
}
cout << endl;
return 0;
}
when i try entering anything that is not 'y', 'n', it will tell me that my input is invalid. But when i try to enter 'y' or 'n', it kinda just ignored it and nothing else happened.. i checked with cout statement and found that it manage to get into the if (input == 'y') statement, but it doesnt seem like it is doing anything else. Everything was fine till 20 minutes ago and i really couldnt figure out whats wrong.
Edit: i ran another test using "cout << '[' << input << ']' << endl;".. it seems like the program is able to get my first input, but then it just hangs there afterwards.. what i get is something like:
Do you wish to draw another card [y][n]: y
[y]
y
y
y
y
I compiled this on linux terminal using g++
if extra codes is needed, i'll edit and add them.. thanks!
When you ask for input from the console, most implementations buffer characters until a newline key is pressed.
After the newline is received, the first character of the buffer is returned. The newline still remains in the buffer as well as any extra characters.
In your case, the second cin >> input statement will read the newline from the buffer.
As an experiment, try entering "frog" and single step through your program. This should illustrate the case of residual characters in the buffer.
Try cin.ignore(1000, '\n') after the first cin >> input. The ignore method will eat up any remaining characters in the buffer until the newline is found.
Make below statements inactive
dealExtra(playerHand, deck, gameInfo);
printHand(playerHand, gameInfo.playerCardCount);
and check if it works, then try making one of the above statements active alternately to find out in which function the flow is getting lost. And so on.
If you feel lazy to run a debugger, and plan to use cout<< statements to find a hanging call, you should flush you cout:
( cout << "I am here and going to hang" ).flush() ;
Otherwise you can't see recent output just because it's still in the output buffer. Try this and you well might see what call hangs your program.
You have an infinite loop inside checkComputerHand:
bool done = false;
while(!done)
{
if(sum == 11 && checkAce == true)
{
computerHand[aceLocation].value = 11;
done = true;
}
if(sum > 11 && checkAce == true)
{
computerHand[aceLocation].value = 1;
done = true;
}
// What if checkAce wasn't true? Infinite loop!
}
Also, the first two lines of newGame do not make any sense:
void newGame(Card playerHand[], Card computerHand[], Statistics &gameInfo)
{
playerHand = '\0';
computerHand = '\0';
// ...
}
Array parameters are silently rewritten by the compiler as pointer parameters. So all you're doing is assigning the null pointer to those local pointers. Probably not what you intended...
Title probably sounds confusing so first I'll show you my code, I made this simple program to get two input values and multiply them, and another thing, but that's not important, It works correctly:
#include <iostream>
using namespace std;
main()
{
int a,b,c,d,e;
char j = 4;
cout << "Welcome to Momentum Calculator\n\n";
cout << "------------------------------\n";
cout << "Please Enter Mass in KG (if the mass in in grams, put \"9999\" and hit enter): \n\n";
cin >> a;
if (a==9999) {
cout << "\nPlease Enter Mass in grams: \n\n";
cin >> d;
}
else {
d = 0;
}
cout << "\nPlease Enter Velocity \n\n";
cin >> e;
if (d == 0)
{
c = (a*e);
}
else {
c = (e*d)/100;
}
cout << "\nMomentum = " << c;
cin.get();
cin.ignore();
while (j == 4)
{
cout << "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n";
main();
}
}
Now as you can see, my variable is an int (integer) and my problem is If I enter an English letter (a-z) or anything that is not a number will cause it to repeat my program unlimited times at an unlimited speed. I want a string/char to see if my var "a" is a letter or anything but don't know how to. I can do it, however, I want user to input only one time in "a" and mine makes him to enter again. Please Help :)
There is a function called isalpha in ctype library, checks whether your variable is an alphabetic letter so you can do using isalpha function.
Will isdigit or isalpha from standard library help you?
P.S.
1KG contains 1000 grams, so you should divide by 1000, not by 100;
UPDATE:
Seems I understood your question...
You need cin.clear(); before cin.get() and cin.ignore().
Otherwise the these calls won't do anything, as cin is in an error state.
I think you can get a as an String, and see if it contains English letter or not, if it contains, again ask for the input ( you can do it in a while loop ). And when a correct input entered, parse it and find what is it's number.