input validation while loop is infinite - c++

#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
double addition;
double subtraction;
double top, bottom;
double multiplication, multiplication2;
char variable;
double total = 0.0;
cout << "Type in:\n'A' For Addition\n"
<< "'S' For subtraction\n" << "'D' For division\n"
<< "'M' For multiplication\n";
cin >> variable;
switch (variable)
{
case 'A':
{
cout << "Enter 0 for input\n";
cin >> addition;
while(addition != 'Q' || addition != 'q')
{
cout << "Enter numbers for adding\nThen type in"
<< "Q or q to quit\n";
cin >> addition;
total += addition;
}
cout << "Your total is " << total << endl;
}
It loops infinitely starting out at the first cout statement in the while loop. I will type in numbers, then as soon as I type in q or Q and hit enter it will immediately loop infinitely. Thanks!

Your condition for the while loop uses a logical OR.
Let's say you try to quit the loop and enter the input 'Q'. The first part of the condition will be FALSE, but the second part of the condition will be TRUE. Since it is a logical OR, then the whole condition will be TRUE and the loop will execute. The converse is also true if you input 'q'.
So no matter what you enter, your loop will run.

There are two main problems in your program.
First, condition addition != 'Q' || addition != 'q' is always true, because for any value of addition, either addition != 'Q' or addition != 'q' is true (i.e. addition can never be both Q and q at the same time). You probably meant addition != 'Q' && addition != 'q'
Second, when you do cin >> addition with variable of type double, then you will either receive a valid number or - if somebody enters - 'Q', for example, "nothing" and an error flag is set. "Nothing" means that the value of addition remains unchanged.
To accomplish the "either a number or 'Q'"-thing, you need to read in a string and compare it to "Q" (or "q") and otherwise try to convert the string into a double.
The code fragment could look as follows:
int main() {
double sum = 0;
double toAdd;
std::string input;
bool end = false;
while (!end) {
cout << "enter a value to add (type Q or q to quit)" << endl;
cin >> input;
if (input == "Q" || input == "q") {
end = true;
}
else {
try {
toAdd = stod(input);
sum += toAdd;
} catch (out_of_range &e) {
cout << "input " << input << " is out of range." << endl;
} catch (invalid_argument &i) {
cout << "input " << input << " is not a valid number." << endl;
}
}
}
cout << "sum: " << sum << endl;
}

Related

How to evaluate number greater than and less than in the same while loop?

// DiceRollProject.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <time.h>
using namespace std;
int diceRoll(int max); // function definition
int getValidInteger();// function definition
int main() {
srand(time(0)); // seed the random number generator
int exitProgram = 0;
int guess, rollValue;
int maxRollValue = 6;
cout << "Hello! Let's play a dice game. Let me do the first roll for you.\n" << endl;
rollValue = diceRoll(maxRollValue);
cout << "In this roll, you got: " << rollValue << "\n" << endl;
do {
rollValue = diceRoll(maxRollValue);
cout << "What's your guess for the next roll? Enter an integer between 1 and " << maxRollValue << ": ";
guess = getValidInteger();
// TODO: Validate input
if (guess > rollValue)
{
cout << "The guess was too high!";
}
if (guess < rollValue)
{
cout << "The guess was too low!";
}
if (guess == rollValue)
{
cout << "You guessed correctly, congrats!";
}
cout << "In this roll, you got: " << rollValue << "\n" << endl;
// TODO: Evaluate result
cout << "Enter 1 to exit or any other integer to continue rolling ";
exitProgram = getValidInteger();
cout << "\n";
if (exitProgram == 1)
{
cout << "Sorry to see you go. Have a wonderful day!\n" << endl;
}
} while (exitProgram != 1);
return 0;
}
// Roll the die
int diceRoll(int max) {
int rollValue;
rollValue = (rand() % max) + 1;
return rollValue;
}
// Check if user entered an integer
int getValidInteger() {
int userInput;
cin >> userInput;
while (userInput < 1) {
if (userInput < 1)
{
cout << "Please enter a number greater than or equal to 1\n";
}
if (userInput > 6)
{
cout << "Please enter a number less than or equal to 6\n";
}
}
if (cin.fail()) {
cin.clear();
cin.ignore();
cout << "Please enter an Integer only ";
cin >> userInput;
cout << "\n";
}
return userInput;
}
I have a dice roll guessing game, I'm trying to evaluate the users input, to make sure that they can't enter a number less than 1 and greater than 6, unfortunately, with just my if statements, they can still enter these numbers, although a string is displayed that the input is not valid, I want to make a while loop that keeps asking them to enter a valid number equal or greater than 1 and equal to and less than 6, if the user keeps inputting an incorrect number, the while loop will keep asking them for a valid number, until they do enter one, which will then run the program as normally.
First of all, inside the while loop you have dead code.
while (userInput < 1) {
if (userInput < 1)
{
cout << "Please enter a number greater than or equal to 1\n";
}
if (userInput > 6)
{
cout << "Please enter a number less than or equal to 6\n";
}
}
Within the loop body, the first if is always true and the second one is always false. You should enter in a loop when the user writes an invalid input. This happens when (userInput < 1 or userInput > 6)
After the evaluation of the while's condition, you should ask the user to write input
do {
cout << "Please enter an Integer only ";
cin >> userInput;
if (userInput < 1)
{
cout << "Please enter a number greater than or equal to 1\n";
}
if (userInput > 6)
{
cout << "Please enter a number less than or equal to 6\n";
}
}while(userInput < 1 || userInput > 6);
So your condition that will keep you in the while loop is if the person guesses too high or too low. Inside the while loop I would add the updating condition or statement that you would like to repeat. So in your case, "your guess is too high" or " your guess is too low" and ask for their input again. I am not a pro but I would keep it simple by constructing 2 while loops, one for too high and one for too low just like your if statements. literally you can just change your first two if statements to while loops and adding an few extra lines of cout to ask the person to guess again and validate their input. I hope this helped.
from what I've understood you are looking for something like this:
int main (){
int my_magic_number=(rand()%6)+1,usernumber=-1;
bool state;
while (usernumber!=my_magic_number){
cin>>usernumber;
state = (usernumber<1||usernumber>6);
while (state) {
cout<<"You entered a number outside the range [1,6] please try again\n";}
cin>>usernumber;
state = (usernumber<1||usernumber>6);
}
if (usernumber!=my_magic_number) {/* do whatever you want */}
} //while loop
} // main

Programming Principles and Practice: chapter 4 drill part 1

I just can't seem to get this program to work properly. I can get it to accept two integers and print them to the screen. But I can't get the program to terminate when the '|' is used. Once that its entered it loops infinitely. Here is the code that I have so far:
#include "../../std_lib_facilities.h"
int main()
{
int num1 = 0;
int num2 = 0;
char counter = '\0';
cout << "Please enter two integers and press enter. \n";
bool test = true;
while (counter != '|')
{
cin >> num1 >> num2;
cout << "Your numbers are: " << num1 << " " << num2 << endl;
if (cin.fail())
{
cout << "Goodbye!\n";
test = false;
}
else (counter != '|');
cout << "Enter more numbers or press '|' to exit.\n";
}
system("pause");
}
You are using the wrong condition in your while loop. You are never changing counter so the loop will never end. However you do change test to false in the while loop if the input fails. You can change the condition of the while loop to use test instead like
while(test)
{
//...
}
Since counter is no longer being used you can get rid of it completely.
Please note that unless you change to taking in string and parsing the input any input that will cause cin to fail will end the loop not just a |.

C++ cin gets skipped, even after I use cin.ignore()

Just to make clear, I am very new to C++.
But I wrote I very small program to test my skill with arrays and ran into a problem with cin.
If the user enters number, like the program expects them to, all is well. But if a string gets entered, all input is skipped and the program ends.
I set up all of my inputs like this: cin >> x;cin.clear();cin.ignore();
So what is awry??
Here is the full code:
#include <iostream>
#include <cstdlib>
using namespace std;
int main()
{
system("cls");
int create = 1;
int entry;
int x;
string chc;
cout << "How long should the array be?" << endl;
cout << ":";
cin >> x;cin.clear();cin.ignore();
if(x<1){x=1;}
int myArray[x];
string askcontinue;
for(int x=0;x<sizeof(myArray)/sizeof(myArray[0]);x++){
system("cls");
cout << "Enter value #" << x+1 << endl;
cout << ":";
cin >> entry;cin.clear();cin.ignore();
myArray[x]=entry;
}
system("cls");
cout << "Index - Value" << endl;
for(int x=0;x<sizeof(myArray)/sizeof(myArray[0]);x++){
cout << x << " ------ " << myArray[x] <<endl;
}
system("cls");
cout << "Restart? [Y/N]" << endl;
cout << ":";
cin >> chc;cin.clear();cin.ignore();
if(chc=="y" || chc=="Y"){main();}
}
cin >> x;cin.clear();cin.ignore();
This thing that you're doing throughout your program is part of the problem. If the user enters something that doesn't meet the formatting requirements for an integer, the stream goes into a failure state. Directly after that happens you clear the stream and discard the next character. If the user entered in more than one character as part of the invalid input, the ignore() call is simply discarding the next character, but not all of the invalid input.
You need to check if the input did not succeed, and then discard the input using the overload of ignore() that takes the number of characters you wish to discard. Do the following if you wish to consistently ask the user for input if he does not provide valid characters:
while (!(std::cin >> x)) {
std::cout << "How long should the array be?" << std::endl;
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
But judging from your code, it doesn't look like you want to repeatedly ask the user for input. In that case, you should check for valid input instead and do nothing in the invalid case:
if (std::cin >> x) {
...
}
Also, VLAs (or variable-length arrays) are a non-standard feature of C++, provided as extentions in some compilers. Don't use them. Instead, allocate dynamically by using std::vector:
std::vector<int> myArray(x);
NOTE: you should also change the fact that you defining the variable 'x' three times
the problem you are having, is that c input does not type checking, so it does not care what was entered, so this is up to you. You should input everything as a string, and then make sure that the string contains nothing but numbers, THEN you can use std::stoi, or whatever the appropriate conversion method is. if they DO NOT enter a valid number, then you can just say INVALID, and tell the user to enter a valid number, and go back to the input, you could use something such as:
system("cls");
cout << "Enter value #" << x + 1 << endl;
cout << ":";
cin >> entry; cin.clear(); cin.ignore();
while(!is_valid_integer(entry))
{
system("cls");
cout << "INVALID NUMBER \n Enter value #" << x + 1 << endl;
cout << ":";
cin >> entry; cin.clear(); cin.ignore();
}
myArray[x] = std::stoi(entry);
And then entry is a string.
is_valid_integerwould be defined as:
bool is_valid_integer(std::string str)
{
for(auto it : str)
{
if(!(ch == '0' || ch == '1' || ch == '2' || ch == '3' || ch == '4' || ch == '5' || ch == '6' || ch == '7' || ch == '8' || ch == '9'))
return false;
//OR: this is more efficient, but is reliant on using ascii codes (which in this case we are)
//if(!(ch >=48 && ch <= 57)) return false;
}
return true;//all numbers
}

How can you make input take strings and int? c++

is it possible, say your trying to do calculations so the primary variable type may be int... but as a part of the program you decide to do a while loop and throw an if statement for existing purposes.
you have one cin >> and that is to take in a number to run calculations, but you also need an input incase they want to exit:
Here's some code to work with
#include <iostream>
using namespace std;
int func1(int x)
{
int sum = 0;
sum = x * x * x;
return sum;
}
int main()
{
bool repeat = true;
cout << "Enter a value to cube: " << endl;
cout << "Type leave to quit" << endl;
while (repeat)
{
int input = 0;
cin >> input;
cout << input << " cubed is: " << func1(input) << endl;
if (input = "leave" || input = "Leave")
{
repeat = false;
}
}
}
I'm aware they wont take leave cause input is set to int, but is it possible to use a conversion or something...
another thing is there a better way to break the loop or is that the most common way?
One way to do this is read a string from cin. Check its value. If it satisfies the exit condition, exit. If not, extract the integer from the string and proceed to procss the integer.
while (repeat)
{
string input;
cin >> input;
if (input == "leave" || input == "Leave")
{
repeat = false;
}
else
{
int intInput = atoi(input.c_str());
cout << input << " cubed is: " << func1(intInput) << endl;
}
}
You can read the input as a string from the input stream. Check if it is 'leave' and quit.. and If it is not try to convert it to a number and call func1.. look at atoi or boost::lexical_cast<>
also it is input == "leave" == is the equal operator. = is an assignment operator.
int main() {
cout << "Enter a value to cube: " << endl;
cout << "Type leave to quit" << endl;
while (true)
{
string input;
cin >> input;
if (input == "leave" || input == "Leave")
{
break;
}
cout << input << " cubed is: " << func1(atoi(input.c_str())) << endl;
}
}
you can use like
int input;
string s;
cint>>s; //read string from user
stringstream ss(s);
ss>>input; //try to convert to an int
if(ss==0) //not an integer
{
if(s == "leave") {//user don't want to enter further input
//exit
}
else
{
//invalid data some string other than leave and not an integer
}
}
else
{
cout<<"Input:"<<input<<endl;
//input holds an int data
}

Comparing Doubles and Chars in C++

I am trying to write a simple program in C++ that reads in an unspecified number of marks, then once the user inputs the character 'q', the program must calculate and display the average mark. However I am having some trouble. The approach I am taking is to save each value as a double, the I want to compare the double to the character 'q' and if they are the same character, end the loop, calculate and display the average.
However I think that the comparison between the char value 'q' and double value for the mark seem to be incomparable. This worked for me when I did the same using integer values for the mark but not doubles it seems. Any help would be appreciated.
Here is the code:
int main()
{
cout << "Please enter any number of marks. Enter 'q' to stop." << endl;
double total = 0;
int counter = 0;
bool repeat = true;
do
{
double mark;
cin >> mark;
if (mark != 'q')
{
total += mark;
counter++;
}
else
{
repeat = false;
}
}
while (repeat == true);
double average = total/counter;
cout << "Average: " << average << endl;
return 0;
}
you'll need to change the mark variable to string, and then compare it to 'q', else try to parse is as a number.
otherwise this entire code, does not make a lot of sense, because 'q' in ASCII is 113, which I guess is a possible value
Typecast double to int and then compare, It must work because it compares ASCII value of character
Here is the code:
int main()
{
cout << "Please enter any number of marks. Enter 'q' to stop." << endl;
double total = 0;
int counter = 0;
bool repeat = true;
do
{
double mark;
cin >> mark;
if ((int)mark != 'q')
{
total += mark;
counter++;
}
else
{
repeat = false;
}
}
while (repeat == true);
double average = total/counter;
cout << "Average: " << average << endl;
return 0;
}
you can not cast a double to char. You may need to use additional c++ library functions which convert string (char*) to double. There are different ways to do this.
Try this :
#include <iostream>
#include <sstream>
using namespace std;
int main()
{
cout << "Please enter any number of marks. Enter 'q' to stop." << endl;
double total = 0;
int counter = 0;
bool repeat = true;
do
{
char userinput[8];
cin >> userinput;
std::stringstream ss;
double mark;
if (userinput[0] != 'q')
{
ss << userinput;
ss >> mark;
total += mark;
counter++;
}
else
{
repeat = false;
}
}
while (repeat == true);
double average = total/counter;
cout << "total : " << total << " count : " << counter << endl;
cout << "Average: " << average << endl;
return 0;
}
You are doing it wrong. If you try to cin a char into a double variable, the input char stays in the input buffer and the double variable remains the same. So this will end in an infinite loop.
If you really want the user to enter a char to end input, you need to take the whole input in a string variable. Check the string for q. If not present, use atof() to convert it to double.