Fahrenheit to Celsius converter. Verify int not char - c++

Created a program to convert Fahrenheit to Celsius. Having trouble verifying if what the user input is an int. The problem I believe is the "if (!cin)" line.
#include <iomanip>
#include <iostream>
using namespace std;
int main()
{
float f, c;
string choice;
do{
cout << "Enter a Fahrenheit temperature to convert to Celsius:" << endl;
cin >> f ;
if ( !cin ) {
cout << "That is not a number..." << endl;
}
else if (f < -459.67) {
cout << "That is not a Fahrenheit temperature..." << endl;
}
if ( f >= -459.67) {
c = (( f - 32) * 5.0)/9.0 ;
cout << fixed ;
cout << setprecision(2) << "Celsius temperature is: " << showpos << c << endl;
}
cout << "Would you like to convert another? If so, enter Yes" << endl;
cin >> choice ;
}while ( choice == "Yes" || choice == "yes" );
return 0;
}

I'm not sure of what you mean about your "Continue yes or no statement", So I've written a code that asks the user to type yes to confirm the conversion, otherwise no to enter a new Fahrenheit value. After the conversion, The program also asks the user to type yes if he/she wants another conversion, If the user types anything except "yes", The program will close.
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
int main()
{
bool Continue = true;
while (Continue == true)
{
double f, c;
cout << endl << "Enter a Fahrenheit temperature to convert to Celsius:" << endl << endl;
while (!(cin >> f))
{
cout << endl << "Invalid Input. " << endl << endl;
cin.clear();
cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
while (f <= -459.67)
{
cout << "Invalid Input. " << endl;
cin.clear();
cin.ignore();
cin >> f;
}
cout << endl << "Continue? Type yes to proceed conversion, Otherwise type no." << endl << endl;
string confirmation;
cin >> confirmation;
while (confirmation != "yes" && confirmation != "no")
{
cout << endl << "Input not recognized. try again." << endl << endl;
cin >> confirmation;
}
if (confirmation == "yes")
{
c = ((f - 32) * 5.0) / 9.0;
cout << fixed;
cout << endl << setprecision(2) << "Celsius temperature is: " << showpos << c << endl;
cout << endl << "Another convertion? type yes to confirm. " << endl << endl;
string cont;
cin >> cont;
if (cont != "yes")
{
Continue = false;
}
}
}
return 0;
}
You should use while loop, So the program won't stop asking until the user enters the correct data. Use cin.clear(); to clear invalid data that the user inputted. And cin.ignore(); to ignore any succeeding erroneous data. For example, '25tg', 'tg' character is ignored since it's not valid. '25' will be accepted.
Edits in my answer and code provided are very welcome.

you can use a function to check if the input is numeric or not....
should be something like this:
bool isFloat( string myString ) {
std::istringstream iss(myString);
float f;
iss >> noskipws >> f; // noskipws considers leading whitespace invalid
// Check the entire string was consumed and if either failbit or badbit is set
return iss.eof() && !iss.fail();
}

Related

Verifying User Input as Integer: User has to enter input twice

I am prompting the user to enter an integer value. When the value is incorrect, the program works. However, when the user enters an integer input, the user needs to enter the input twice.
I looked at other tutorials on how to use the while loop to catch erroneous input, and that part worked for me. However, the integer values need to be entered twice in order for the program to run.
#include <iostream>
using namespace std;
int main() {
cout << "*************************************************" << endl;
cout << "******************|DVD Library|******************" << endl;
cout << "*************************************************" << endl;
cout << "1.\tAdd DVD" << endl;
cout << "2.\tDelete DVD" << endl;
cout << "3.\tSearch DVD" << endl;
cout << "4.\tList All DVDs in the Library" << endl;
cout << "5.\tAdd DVD to Favorites List" << endl;
cout << "6.\tDelete DVD from Favorites List" << endl;
cout << "7.\tSearch DVD in Favorites List" << endl;
cout << "8.\tList All DVDs in Favorites List" << endl;
cout << "9.\tQuit" << endl;
cout << "*************************************************" << endl;
int input;
cin >> input;
while (!(cin >> input)) {
cin.clear();
while (cin.get() != '\n')
continue;
cout << "Please enter an integer --> " << flush;
}
if (input < 1 || input > 9) {
cout << "Invalid input! Please try again!" << endl;
}
return 0;
}
You ask for the input twice:
cin >> input;
while(!(cin >> input )){
Removing the first line might make it work you intended.
'The user has to enter the input twice' Look at your code
int input;
cin >> input;
while(!(cin >> input )){
How many times do you ask the user for input?
You'd have more luck with this
int input;
while(!(cin >> input )){
Your error recovery code looks reasonable, haven't tested it though.
int input;
while (cout << "Your choice: ",
!(cin >> input) || input < 1 || 9 < input)
{
cin.clear();
while (cin.get() != '\n');
cerr << "Invalid input! Please try again!\n";
}
Thanks everyone! The "cin >> input;" line was unnecessary. At first, I left it there because it would actually tell the user the error message if the user entered a numeric input such as a double. So, if the user entered something like 3.3, the program would display an error message that I specified ("Please enter an integer" line). However, the program in this case (when there is a double) asks the user to prompt for the integer input twice and then continues the program. When I delete the said unnecessary line, the program accepts a double input, but what it does, it takes the numeric value before the decimal point and uses it as the integer. So, a value of 1.2 is recorded as 1 when I tested it. I'm unsure why this phenomenon happens, but the program works otherwise. Maybe it accounts for human error?
#include <iostream>
using namespace std;
int main() {
cout << "*************************************************" << endl;
cout << "******************|DVD Library|******************" << endl;
cout << "*************************************************" << endl;
cout << "1.\tAdd DVD" << endl;
cout << "2.\tDelete DVD" << endl;
cout << "3.\tSearch DVD" << endl;
cout << "4.\tList All DVDs in the Library" << endl;
cout << "5.\tAdd DVD to Favorites List" << endl;
cout << "6.\tDelete DVD from Favorites List" << endl;
cout << "7.\tSearch DVD in Favorites List" << endl;
cout << "8.\tList All DVDs in Favorites List" << endl;
cout << "9.\tQuit" << endl;
cout << "*************************************************" << endl;
int input;
while (!(cin >> input)) {
cin.clear();
while (cin.get() != '\n')
continue;
cout << "Please enter an integer --> " << flush;
}
if (input < 1 || input > 9) {
cout << "Invalid input! Please try again!" << endl;
}
return 0;
}

Why do I have this infinite loop

hey guys kind of new to the programming world and I know you should read in all input as a string but this is just a simple program and I'm having a brain fart I think but here is my question......why when I press q to quit am i getting an infinite loop and how would I condense the while loops because that looks gross
here is what I have so far
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int grade;
char quit = 'a';
cout << "Input your grade (0-100): ";
cout << endl;
cin >> grade;
while (grade < 0) {
cout << "If you have a negetive grade....drop out! otherwise enter another grade" << endl;
cin >> grade;
}
while (quit != 'q') {
while (grade < 0) {
cout << "If you have a negetive grade....drop out! otherwise enter another grade" << endl;
cin >> grade;
}
if (grade == 100) {
cout << "You got a perfect grade!" << endl;
cout << "Letter grade: A" << endl;
}
else if (grade >= 90 && grade <= 100) {
cout << "Letter grade: A" << endl << endl;
}
else if (grade >= 80 && grade <= 89) {
cout << "Letter grade: B" << endl << endl;
}
else if (grade >= 70 && grade <= 79) {
cout << "Letter grade: C" << endl << endl;
}
else if (grade >= 60 && grade <= 69) {
cout << "Letter grade: D" << endl << endl;
}
else if (grade < 60) {
cout << "Letter grade: F" << endl << endl;
}
else {
cout << "Invalid grade!" << endl;
}
cout << " would you like to enter another grade? or press q to quit" << endl;
cin >> grade;
}
system("pause");
return 0;`enter code here`
}
Because of grade var. You declared grade as int.
If you type correct int, it works well, but If you type another char ex:) q or f, function cin cannot recognize q or f as int type.
If char input, cin pass own process.
You have to change grade type into char to recognize char and int inputs both.
If you want to use only one input flow, this implementation code will help you.
int _tmain(int argc, _TCHAR* argv[])
{
char c_input[32] = {0};
cin>>c_input;
while(atoi(c_input) < 0)
{
cout<<"If you have a negative grade....drop out! otherwise enter another grade" << endl;
cin>>c_input;
}
while(c_input[0] != 'q')
{
while(atoi(c_input) < 0)
{
cout<<"If you have a negative grade....drop out! otherwise enter another grade" << endl;
cin>>c_input;
}
cout<<c_input;
cout<<"Would you like to enter another grade? or press q to quit" << endl;
cin>>c_input;
}
return 0;
}
Minimal, verifiable example:
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int grade;
char quit = 'a';
cout << "Input your grade (0-100): ";
cout << endl;
cin >> grade;
while (quit != 'q') {
cout << " would you like to enter another grade? or press q to quit" << endl;
cin >> grade;
}
system("pause");
return 0;`enter code here`
}
See the problem with quit?
edit
What I have done is remove (most of) the lines that have nothing to do with quit or the loop.
At this point you should notice that the loop never changes quit.
If you are having trouble with a program, one of the best ways to figure out what is wrong is to get rid of everything that doesn't have anything to do with the error. In time, you'll be able to do this using only your mind. Duuude!
while i'm at it
The correct way to handle user input is to get it as a string, then convert it to what you want.
For example:
#include <iostream>
#include <sstream>
#include <stdexcept>
#include <string>
template <typename T>
T string_to( const std::string& s )
{
T value;
std::istringstream ss( s );
ss >> value >> std::ws;
if (!ss.eof()) throw std::invalid_argument("T string_to()");
return value;
}
int main()
{
std::cout << "Enter a number or 'q': ";
std::string s;
getline( std::cin, s );
if (s == 'q')
{
std::cout << "Good job! You entered 'q'.\n";
}
else
{
try
{
double x = string_to <double> ( s );
std::cout << "Good job! You entered '" << x << "'.\n";
}
catch (const std::exception& e)
{
std::cout << "Foo, you didn't obey instructions and made me " << e.what() << ".\n";
}
}
}

how do you do a integer validation in C++?

ok, I have been looking for days now but I cant find anything that will work.
I have a program and I want to make sure that the user enters a integer and not a double.
this program works fine but I need to validate the numOne and numTwo to make sure they are integers and not doubles, (5.5)
int main()
{ //This is where my variables are stored
int numOne, numTwo, answer, rightAnswer, ranNumOne, ranNumTwo;
//this will display to the user to enter a range of numbers to be used
cout << "Please enter a set of numbers to be the range for the problems." << endl;
cout << "Please enter the beginning number." << endl;
cin >> numOne;
cout << "please enter the ending number." << endl;
cin >> numTwo;
//this makes sure that the user entered a integer(if not the program will close)
if (!(cin >> numOne))
{
cout << "You did not enter a integer PLEASE RE-RUN THE PROGRAM AND TRY AGAIN!" << endl;
cin.clear();
cin.ignore(100, '\n');
exit(0);
}
cout << "please enter the ending number." << endl;
cin >> numTwo;
//this makes sure that the user entered a number(if not the program will close)
if (!(cin >> numTwo))
{
cout << "You did not enter a integer PLEASE RE-RUN THE PROGRAM AND TRY AGAIN!" << endl;
cin.clear();
cin.ignore(100, '\n');
exit(0);
}
//this is where the first number is generated
srand(time(0));
ranNumOne = rand() % (numOne - numTwo) + 1;
system("PAUSE");
//this is where the second number is generated
srand(time(0));
ranNumTwo = rand() % (numOne - numTwo) + 1;
//this is where the calculations are done
rightAnswer = ranNumOne + ranNumTwo;
//this displays the problem that was generated
cout << "What is: " << endl;
cout << setw(11) << ranNumOne << endl;
cout << setw(6) << "+" << setw(3) << ranNumTwo << endl;
cout << " -------\n";
cin >> answer;
//this checks to see if the answer is right or not and displays the result
if (answer == rightAnswer)
{
cout << "Your answer was correct! " << endl;
}
else
cout << "The correct answer is: " << rightAnswer << endl;
return 0;
}
Use std:n:ci.fail() to see if it failed.
int numOne;
cin >> numOne;
if(cin.fail())
cout << "Not a number...")
Maybe even a nice template function.
template<typename T>
T inline input(const std::string &errmsg = "") {
T var;
std::cin >> var;
while (std::cin.fail()) {
std::cin.clear();
std::cin.ignore(256, '\n');
std::cout << errmsg;
std::cin >> var;
}
return var;
}
Or not:
#include <iostream>
#include <cstdlib>
#include <iomanip>
#include <string>
#include <ctime>
#define DIFF(n1, n2) (n1 > n2 ? n1 - n2 : n2 - n1)
using namespace std;
int input(const string &firstmsg = "", const string &errmsg = "") {
int var;
std::cout << firstmsg;
std::cin >> var;
while (cin.fail()) {
cin.clear();
cin.ignore(256, '\n');
cout << errmsg;
cin >> var;
}
return var;
}
int main(){
//This is where my variables are stored
int numOne, numTwo, answer, rightAnswer, ranNumOne, ranNumTwo;
//this will display to the user to enter a range of numbers to be used
cout << "Please enter a set of numbers to be the range for the problems." << endl << endl;
numOne = input("Please enter the beginning number: ", "Invalid. Enter again: ");
//this asks the user for the second number
numTwo = input("Please enter the ending number: ", "Invalid. Enter again: ");
//this is where the first number is generated
srand(time(0));
ranNumOne = rand() % (DIFF(numOne, numTwo)) + 1; // ensures it will always be positive
system("PAUSE");
//this is where the second number is generated
srand(time(0));
ranNumTwo = rand() % (DIFF(numOne, numTwo)) + 1;
//this is where the calculations are done
rightAnswer = ranNumOne + ranNumTwo;
//this displays the problem that was generated
cout << "What is: " << endl;
cout << setw(11) << ranNumOne << endl;
cout << setw(6) << "+" << setw(3) << ranNumTwo << endl;
cout << " -------\n";
cin >> answer;
//this checks to see if the answer is right or not and displays the result
if (answer == rightAnswer){
cout << "Your answer was correct! " << endl;
}
else
cout << "The correct answer is: " << rightAnswer << endl;
return 0;
}
why not, get the number into a double and then see if that double is an int. ie
double d;
cin>>d;
if (ceil(d) != d)
cout >> " not an integer";

C++ loop error that my professor can't figure out

I am currently taking a C++ programming class and am working on a project in which I have to create a fairly simple movie database. My code essentially works as intended yet in certain cases it causes the main menu to loop infinitely and I cannot figure out why. I brought this to my teacher and he cannot explain it either. He gave me a workaround but I would like to know if anyone can see the cause of the problem. Full code is as follows:
#include <cstdlib>
#include <iostream>
#include <vector>
#include <string>
using namespace std;
struct MovieType
{
string title;
string director;
int year;
int length;
string rating;
};
MovieType addMovie() {
MovieType newMovie;
cout << "Movie title :";
getline(cin, newMovie.title);
cout << "Director :";
getline(cin, newMovie.director);
cout << "Year :";
cin >> newMovie.year;
cout << "Length(in minutes) :";
cin >> newMovie.length;
cout << "Rating :";
cin >> newMovie.rating;
cout << endl;
return newMovie;
}
void listMovie(MovieType movie) {
cout << "______________________________________" << endl;
cout << "Title : " << movie.title << endl;
cout << "Director : " << movie.director << endl;
cout << "Released : " << movie.year << endl;
cout << "MPAA Rating : " << movie.rating << endl;
cout << "Running time : " << movie.length << " minutes" << endl;
cout << "______________________________________" << endl;
}
void search(vector<MovieType> movieVector) {
string strSearch;
cout << endl << "Search title: ";
getline(cin, strSearch);
for (int c = 0; c < movieVector.size(); c++) {
if (movieVector.at(c).title == strSearch)
listMovie(movieVector.at(c));
}
}
int main() {
bool quit = 0;
vector<MovieType> movieVector;
while (quit == 0) {
char selection = 'f';
cout << "Main Menu:" << endl;
cout << "'a' - Add movie" << endl;
cout << "'l' - List movies" << endl;
cout << "'s' - Search by movie title" << endl;
cout << "'q' - Quit" << endl;
cout << "Please enter one of the listed commands:";
cin >> selection;
cin.ignore();
cout << endl;
if (selection == 'a')
movieVector.push_back(addMovie());
else if (selection == 'l') {
for (int c = 0; c < movieVector.size(); c++) {
listMovie(movieVector.at(c));
}
}
else if (selection == 's') {
search(movieVector);
}
else if (selection == 'q')
quit = 1;
}
return 0;
}
When an unexpected input type is entered during the addMovie function(like entering text for the int type year), it just runs through the function then loops through the menu infinitely. It appears to me that the code just stops even looking at the input stream. I have tried using cin.ignore() in many different places but it doesn't matter if there is nothing left in the stream it just keeps going.
I am using NetBeans to compile my code.
I really have no idea why it behaves like this otherwise I would offer more information but I am just curious as to why this happens, because as I said before, my professor doesn't even know why this is happening.
Any help or insight is greatly appreciated.
cin enters an error state where cin.fail() is true. In this state it just ignores all input operations. One fix is to clear the error state, but better, only use getline operations on cin, not formatted input.
E.g., instead of
cin >> newMovie.year;
… do
newMovie.year = stoi( line_from( cin ) );
… where line_from can be defined as
auto line_from( std::istream& stream )
-> std::string
{
std::string result;
if( not getline( stream, result ) )
{
// Throw an exception or call exit(EXIT_FAILURE).0
}
return result;
}
Disclaimer: code untouched by compiler.

How can I avoid bad input from a user?

I am a very newbie programmer, so I don't really know much about writing code to protect the application.. Basically, I created a basicMath.h file and created a do while loop to make a very basic console calculator (only two floats are passed through the functions). I use a series of if and else if statements to determine what the users wants to do. (1.add, 2.subtract, 3.multiply, 4.divide) I used a else { cout << "invalid input" << endl;} to protect against any other values, but then I tried to actually write a letter, and the program entered a infinite loop. Is there anyway to protect against users who accidentally hit a character instead of a number?
`#include <iostream>
#include "basicMath.h"
using namespace std;
char tryAgain = 'y';
float numOne = 0, numTwo = 0;
int options = 0;
int main()
{
cout << "welcome to my calculator program." << endl;
cout << "This will be a basic calculator." << endl;
do{
cout << "What would you like to do?" << endl;
cout << "1. Addition." << endl;
cout << "2. Subtraction." << endl;
cout << "3. Multiplication" << endl;
cout << "4. Division." << endl;
cin >> options;
if (options == 1){
cout << "Enter your first number." << endl;
cin >> numOne;
cout << "Enter your second number." << endl;
cin >> numTwo;
cout << numOne << " + " << numTwo << " = " << add(numOne, numTwo) << endl;
}
else if (options == 2){
cout << "Enter your first number." << endl;
cin >> numOne;
cout << "Enter your second number." << endl;
cin >> numTwo;
cout << numOne << " - " << numTwo << " = " << subtract(numOne, numTwo) << endl;
}
else if (options == 3){
cout << "Enter your first number." << endl;
cin >> numOne;
cout << "Enter your second number." << endl;
cin >> numTwo;
cout << numOne << " * " << numTwo << " = " << multiply(numOne, numTwo) << endl;
}
else if (options == 4){
cout << "Enter your first number." << endl;
cin >> numOne;
cout << "Enter your second number." << endl;
cin >> numTwo;
cout << numOne << " / " << numTwo << " = " << divide(numOne, numTwo) << endl;
}
else {
cout << "Error, invalid option input." << endl;
}
cout << "Would you like to use this calculator again? (y/n)" << endl;
cin >> tryAgain;
}while (tryAgain == 'y');
cout << "Thank you for using my basic calculator!" << endl;
return 0;
}
`
One way would be to use exception handling, but as a newbie you're probably far from learning that.
Instead use the cin.fail() which returns 1 after a bad or unexpected input. Note that you need to clear the "bad" status using cin.clear().
A simple way would be to implement a function:
int GetNumber ()
{
int n;
cin >> n;
while (cin.fail())
{
cin.clear();
cin.ignore();
cout << "Not a valid number. Please reenter: ";
cin >> n;
}
return n;
}
Now in your main function wherever you are taking input, just call GetNumber and store the returned value in your variable. For example, instead of cin >> numOne;, do numOne = GetNumber();
When you input to cin, it is expecting a specific type, such as an integer. If it receives something that it does not expect, such as a letter, it sets a bad flag.
You can usually catch that by looking for fail, and if you find it, flush your input as well as the bad bit (using clear), and try again.
Read a whole line of text first, then convert the line of text to a number and handle any errors in the string-to-number conversion.
Reading a whole line of text from std::cin is done with the std::getline function (not to be confused with the stream's member function):
std::string line;
std::getline(std::cin, line);
if (!std::cin) {
// some catastrophic failure
}
String-to-number conversion is done with std::istringstream (pre-C++11) or with std::stoi (C++11). Here is the pre-C++11 version:
std::istringstream is(line);
int number = 0;
is >> number;
if (!is) {
// line is not a number, e.g. "abc" or "abc123", or the number is too big
// to fit in an int, e.g. "11111111111111111111111111111111111"
} else if (!is.eof()) {
// line is a number, but ends with a non-number, e.g. "123abc",
// whether that's an error depends on your requirements
} else {
// number is OK
}
And here the C++11 version:
try {
std::cout << std::stoi(line) << "\n";
} catch (std::exception const &exc) {
// line is not a number, e.g. "abc" or "abc123", or the number is too big
// to fit in an int, e.g. "11111111111111111111111111111111111"
std::cout << exc.what() << "\n";
}