I'm just learning C++ (1 week of experience) and was trying to write an input validation loop that ask the user to enter "Yes" or "No". I figured it out but have a feeling there's a better way of approaching this. Here's what I came up with:
{
char temp[5]; // to store the input in a string
int test; // to be tested in the while loop
cout << "Yes or No\n";
cin.getline(temp, 5);
if (!(strcmp(temp, "Yes")) || !(strcmp(temp, "No"))) // checks the string if says Yes or No
cout << "Acceptable input"; // displays if string is indeed Yes or No
else //if not, intiate input validation loop
{
test = 0;
while (test == 0) // loop
{
cout << "Invalid, try again.\n";
cin.getline(temp, 5); // attempts to get Yes or No again
if (!(strcmp(temp, "Yes")) || !(strcmp(temp, "No"))) // checks the string if says Yes or No
test = 1; // changes test to 1 so that the loop is canceled
else test = 0; // keeps test at 0 so that the loop iterates and ask for a valid input again
}
cout << "Acceptable input";
}
cin.ignore();
cin.get();
return 0;
}
I apologize for my poor notes, not sure what is relevant. I'm also using the cstring header.
Even better IMO:
std::string answer;
for(;;) {
std::cout << "Please, type Yes or No\n";
getline(std::cin, answer);
if (answer == "Yes" || answer == "No") break;
}
You also can transform answer into lower case that allows user to type not only "Yes", but also "yes", "yEs", etc. See this question
I think you want a do while loop:
bool test = false;
do
{
cout << "Yes or No\n";
cin.getline(temp, 5);
if (!(strcmp(temp, "Yes")) || !(strcmp(temp, "No"))) // checks the string if says Yes or No
{
cout << "Acceptable input"; // displays if string is indeed Yes or No
test = true;
}
else
{
cout << "Invalid, try again.\n";
}
} while (!test);
Related
Please note that I am a complete beginner at C++. I'm trying to write a simple program for an ATM and I have to account for all errors. User may use only integers for input so I need to check if input value is indeed an integer, and my program (this one is shortened) works for the most part.
The problem arises when I try to input a string value instead of an integer while choosing an operation. It works with invalid value integers, but with strings it creates an infinite loop until it eventually stops (unless I add system("cls"), then it doesn't even stop), when it should output the same result as it does for invalid integers:
Invalid choice of operation.
Please select an operation:
1 - Balance inquiry
7 - Return card
Enter your choice and press return:
Here is my code:
#include <iostream>
#include <string>
using namespace std;
bool isNumber(string s) //function to determine if input value is int
{
for (int i = 0; i < s.length(); i++)
if (isdigit(s[i]) == false)
return false;
return true;
}
int ReturnCard() //function to determine whether to continue running or end program
{
string rtrn;
cout << "\nDo you wish to continue? \n1 - Yes \n2 - No, return card" << endl;
cin >> rtrn;
if (rtrn == "1" and isNumber(rtrn)) { return false; }
else if (rtrn == "2" and isNumber(rtrn)) { return true; }
else {cout << "Invalid choice." << endl; ReturnCard(); };
return 0;
}
int menu() //function for operation choice and execution
{
int choice;
do
{
cout << "\nPlease select an operation:\n" << endl
<< " 1 - Balance inquiry\n"
<< " 7 - Return card\n"
<< "\nEnter your choice and press return: ";
int balance = 512;
cin >> choice;
if (choice == 1 and isNumber(to_string(choice))) { cout << "Your balance is $" << balance; "\n\n"; }
else if (choice == 7 and isNumber(to_string(choice))) { cout << "Please wait...\nHave a good day." << endl; return 0; }
else { cout << "Invalid choice of operation."; menu(); }
} while (ReturnCard()==false);
cout << "Please wait...\nHave a good day." << endl;
return 0;
}
int main()
{
string choice;
cout << "Insert debit card to get started." << endl;
menu();
return 0;
}
I've tried every possible solution I know, but nothing seems to work.
***There is a different bug, which is that when I get to the "Do you wish to continue?" part and input any invalid value and follow it up with 2 (which is supposed to end the program) after it asks again, it outputs the result for 1 (continue running - menu etc.). I have already emailed my teacher about this and this is not my main question, but I would appreciate any help.
Thank you!
There are a few things mixed up in your code. Always try to compile your code with maximum warnings turned on, e.g., for GCC add at least the -Wall flag.
Then your compiler would warn you of some of the mistakes you made.
First, it seems like you are confusing string choice and int choice. Two different variables in different scopes. The string one is unused and completely redundant. You can delete it and nothing will change.
In menu, you say cin >> choice;, where choice is of type int. The stream operator >> works like this: It will try to read as many characters as it can, such that the characters match the requested type. So this will only read ints.
Then you convert your valid int into a string and call isNumber() - which will alway return true.
So if you wish to read any line of text and handle it, you can use getline():
string inp;
std::getline(std::cin, inp);
if (!isNumber(inp)) {
std::cout << "ERROR\n";
return 1;
}
int choice = std::stoi(inp); // May throw an exception if invalid range
See stoi
Your isNumber() implementation could look like this:
#include <algorithm>
bool is_number(const string &inp) {
return std::all_of(inp.cbegin(), inp.cend(),
[](unsigned char c){ return std::isdigit(c); });
}
If you are into that functional style, like I am ;)
EDIT:
Btw., another bug which the compiler warns about: cout << "Your balance is $" << balance; "\n\n"; - the newlines are separated by ;, so it's a new statement and this does nothing. You probably wanted the << operator instead.
Recursive call bug:
In { cout << "Invalid choice of operation."; menu(); } and same for ReturnCard(), the function calls itself (recursion).
This is not at all what you want! This will start the function over, but once that call has ended, you continue where that call happened.
What you want in menu() is to start the loop over. You can do that with the continue keyword.
You want the same for ReturnCard(). But you need a loop there.
And now, that I read that code, you don't even need to convert the input to an integer. All you do is compare it. So you can simply do:
string inp;
std::getline(std::cin, inp);
if (inp == "1" || inp == "2") {
// good
} else {
// Invalid
}
Unless that is part of your task.
It is always good to save console input in a string variable instead of another
type, e.g. int or double. This avoids trouble with input errors, e.g. if
characters instead of numbers are given by the program user. Afterwards the
string variable could by analyzed for further actions.
Therefore I changed the type of choice from int to string and adopted the
downstream code to it.
Please try the following program and consider my adaptations which are
written as comments starting with tag //CKE:. Thanks.
#include <iostream>
#include <string>
using namespace std;
bool isNumber(const string& s) //function to determine if input value is int
{
for (size_t i = 0; i < s.length(); i++) //CKE: keep same variable type, e.g. unsigned
if (isdigit(s[i]) == false)
return false;
return true;
}
bool ReturnCard() //function to determine whether to continue running or end program
{
string rtrn;
cout << "\nDo you wish to continue? \n1 - Yes \n2 - No, return card" << endl;
cin >> rtrn;
if (rtrn == "1" and isNumber(rtrn)) { return false; }
if (rtrn == "2" and isNumber(rtrn)) { return true; } //CKE: remove redundant else
cout << "Invalid choice." << endl; ReturnCard(); //CKE: remove redundant else + semicolon
return false;
}
int menu() //function for operation choice and execution
{
string choice; //CKE: change variable type here from int to string
do
{
cout << "\nPlease select an operation:\n" << endl
<< " 1 - Balance inquiry\n"
<< " 7 - Return card\n"
<< "\nEnter your choice and press return: ";
int balance = 512;
cin >> choice;
if (choice == "1" and isNumber(choice)) { cout << "Your balance is $" << balance << "\n\n"; } //CKE: semicolon replaced by output stream operator
else if (choice == "7" and isNumber(choice)) { cout << "Please wait...\nHave a good day." << endl; return 0; }
else { cout << "Invalid choice of operation."; } //CKE: remove recursion here as it isn't required
} while (!ReturnCard()); //CKE: negate result of ReturnCard function
cout << "Please wait...\nHave a good day." << endl;
return 0;
}
int main()
{
string choice;
cout << "Insert debit card to get started." << endl;
menu();
return 0;
}
This question already has answers here:
Read file line by line using ifstream in C++
(8 answers)
Closed 1 year ago.
I am trying to get the user to key in data and store them inside a vector so when -1 is being entered it will break the entire loop and start displaying the vector. However it is currently not working even if -1 has been entered the loop will still continue to go on.
void cargo::addCargo(vector<cargo> &userCargo)
{
fstream file;
file.open("cargo.txt", ios::out | ios::app);
int i = 0;
bool checker = true;
cargo newCargo;
while (checker == true)
{
cout << "Enter id" << endl;
if (cin.peek() == '-1')
{
checker = false;
cout << "Hello";
}
else
{
cin >> idCon;
newCargo.setId(idCon);
}
cout << "Enter Destination Country" << endl;
if (cin.peek() == '-1')
{
checker = false;
}
else
{
cin >> destCountryCon;
newCargo.setDestCountry(destCountryCon);
}
cout << "Enter Time" << endl;
if (cin.peek() == '-1')
{
checker = false;
}
else
{
cin >> timeCon;
newCargo.setTime(timeCon);
newCargo.setIndex(i + 1);
userCargo.push_back(newCargo);
}
cout << endl;
i++;
}
//loop to display the entire vector
displayCargo(userCargo);
}
You are code is not self-contained so I can't fix it for you. The pragmatic answer is break or continue with a flag like checker. peek() looks at the next character and when the user enters "-1" it will return '-' and not the multi-character constant '-1'. Alternative, user can press ctrl-d to trigger eof. If you want to use "-1" (who came up with that terrible ux?), read that as a string or int.
So, I'm triying to learn c++ (coming from python), and I wanted to make a program just to see if i could do it with what i've learned, here's the code
#include <iostream>
using namespace std;
int response(string i) {
if (i == "yes" or i == "Yes") {
cout << "\nHello, sad, I'm dad\n";
return(0);
}
else if (i == "no" or i == "No") {
cout << "Good for you pal\n";
return(0);
}
else {
cout << "Answer properly you overgrown flatworm\n";
response(i);
};
};
int main() {
string i;
cout << "Are you sad?";
cin >> i;
response(i);
};
Pretty simple huh? No. For some reason, yes and no answers work fine, but when I try something different I get insulted infinitely and the program crashes from exceeding it's memory limit. How do I solve this?
(English is not my native language, so feel free to correct any ortography mistakes)
At no point do you request further input. For bad input 'i', the response routine prints out an insult, and then calls itself with exactly the same string.
The response routine prints out an insult, and then calls itself with exactly the same string.
The response routine prints out an insult, and then calls itself with exactly the same string.
…
You need to allow the user to enter a new string, and then (if you want to use recursion) make the recursive call to validate the new input.
But as mentioned in the comment, this is not really a problem that needs a recursive solution.
This can be solved by eliminating recursion ad it involves moving the input routine inside of a function that's more self-contained:
int getResponse(string i) {
for(;;) {
string i;
cout << "Are you sad?";
cin >> i;
if (i == "yes" or i == "Yes") {
cout << "\nHello, sad, I'm dad\n";
return(0);
}
else if (i == "no" or i == "No") {
cout << "Good for you pal\n";
return(0);
}
else {
cout << "Answer properly you overgrown flatworm\n";
}
}
}
You have 2 issues:
In the else case, you are not asking for new user input.
You need to return the result of calling response(i), otherwise the code invokes undefined behavior.
else {
cout << "Answer properly you overgrown flatworm\n";
cin >> i;
return response(i);
};
Alternatively, since you never use the return value from response, you can just remove all the return statements, and make it a void function.
If you insist on using recursion then move the input and the check in the same function response() - that function doesn't need to return int at all. In main you can just call response().
void response()
{
string i;
cout << "Are you sad?";
cin >> i;
if (i == "yes" or i == "Yes")
{
cout << "\nHello, sad, I'm dad\n";
}
else if (i == "no" or i == "No")
{
cout << "Good for you pal\n";
return;
}
else
{
cout << "Answer properly you overgrown flatworm\n";
response();
}
}
int main()
{
response();
}
I know for a fact similar questions have been asked before but I really can't figure out what's wrong with my code specifically. For some reason if I input "n" I have to press enter twice. But if I input "y", everything works fine and the code moves to the next section. My code is as follows:
do{
try {
if (test) cout << " Re-enter: ";
test = false;
getline(cin, choice);
checkinput(choice);
}
catch (int flag) {
if (flag == 1){ cout << "Error: Input must be y or n."; test = true; }
}
} while (test);
and the checkinput function is as follows:
// function for checking the input of y/n
string checkinput(string c) {
if (c != "Y" && c != "y" && c != "N" && c != "n") {
throw 1;
}
if (cin.fail()) throw 1;
return c;
}
I think you are trying to do too much here. You can simplify this.
There is no need to throw and catch exceptions inside checkinput. Since there is only two cases you can use a boolean. Secondly, you are returning c. I don't know why you are doing that, it isn't being used. You should instead, return a boolean.
checkinput becomes:
bool checkInput(string c) {
if (c.length() > 1)
return false;
return c == "Y" || c == "y" || c == "N" || c == "n";
}
Now you can simplify the do-while and remove the try statement. Additionally, there is no need for the test variable now since we are successfully grabbing any input:
int main() {
string choice = "";
do {
cout << "Enter yes or no (y/n): ";
getline(cin, choice); // or cin >> choice;
bool check = checkInput(choice);
if (!check)
cout << "Error: Input must be y or n." << endl;
} while (true);
}
You may also simplify this further but that will be at the cost of readability. Good luck!
I'm making a small program that uses a if else statement, but instead of using numbers to control the flow i want to be able to make the control work with with yes and no;
for example:
cout << "would you like to continue?" << endl;
cout << "\nYES or NO" << endl;
int input =0;
cin >> input;
string Yes = "YES";
string No = "NO";
if (input == no)
{
cout << "testone" << endl;
}
if (input == yes)
{
cout << "test two" << endl;
//the rest of the program goes here i guess?
}
else
{
cout << "you entered the wrong thing, start again" << endl;
//maybe some type of loop structure to go back
}
but I can't seem to get any variations of this to work, i could make the user type a 0 or 1 instead but that seems really stupid, i'd rather it be as natural as possible, users don't speak numbers do they?
also i need to be able to simply add more words, for example "no NO No noo no n" all would have to mean no
hopefully that makes some sense
also i would love to make this using a window but i've only learned basic c++ so far not even that and i cant find any good resources online about basic windows programming.
You're not reading in a string, you're reading in an int.
Try this:
string input;
instead of
int input = 0;
Also, C++ is case-sensitive, so you can't define a variable called Yes and then try to use it as yes. They need to be in the same case.
btw, your second if statement should be an else if, otherwise if you type in "NO" then it will still go into that last else block.
First of all, input must be std::string, not int.
Also, you've written yes and no wrong:
v
if (input == No)
// ..
// v
else if (input == Yes)
^^^^
If you want your program to work with "no no no ..", you could use std::string::find:
if( std::string::npos != input.find( "no" ) )
// ..
The same with "Yes".
Also, you could do this to be almost case-insensitive - transform the input to upper-case letters (or lower, whatever ), and then use find.This way, yEs will be still a valid answer.
bool yesno(char const* prompt, bool default_yes=true) {
using namespace std;
if (prompt && cin.tie()) {
*cin.tie() << prompt << (default_yes ? " [Yn] " : " [yN] ");
}
string line;
if (!getline(cin, line)) {
throw std::runtime_error("yesno: unexpected input error");
}
else if (line.size() == 0) {
return default_yes;
}
else {
return line[0] == 'Y' || line[0] == 'y';
}
}
string input;
cin >> input;
if (input == "yes"){
}
else if (input == "no"{
}
else {
//blah
}