Getting identifier choice is undefined - c++

I'm working on a calculator console app. I'm a beginner in C++ I want the program to ask a question after I'm done printing the previous result. I get the following error though and I have no idea how to fix it.
identifier "choice" is undefined
Here is my source code:
#include "stdafx.h"
#include <iostream>
int getUserInput() {
std::cout << "Please enter an integer: ";
int value;
std::cin >> value;
return value;
}
int getConversionFormula() {
std::cout << "What do you want to convert? (1 for Meters to feet, 2 for kilometers to miles, 3 for kilagrams to pounds.): ";
int op;
std::cin >> op;
//user might select an invalid operation that isnt there
//Implement a way to avoid this.
return op;
}
int getUserInput2() {
std::cout << "Do you want to convert anything else? (hit 1 for yes, hit 2 for no): ";
int choice; //for the user if they hit yes or no to the question above.
std::cin >> choice;
return choice;
}
int getUserChoice(int choice) {
//If user selects 1 show the converstion formula screen
if (choice == 1)
getConversionFormula();
if (choice == 2)
std::exit;
return -1;
}
int calculateResult(int x, int op) {
//we will use == to compare two variables to see if they are true or not
if (op == 1)
return x * 3.280839895; //meters to feet
if (op == 2)
return x / 1.609344; //km to miles
if (op == 3)
return x * 2.2046; //kg to pounds
return -1; //If the user entered an invalid operation
}
void printResult(int result) {
std::cout << "Your result is: " << result << std::endl;
}
int main()
{
int input1 = getUserInput(); //Gets the users input
int op = getConversionFormula();
int result = calculateResult(input1, op);
printResult(result);
int input2 = getUserInput2(); //Asks the user if they want to convert anything else
int input3 = getUserChoice(input2, choice);
std::cin.clear(); // reset any error flags
std::cin.ignore(32767, '\n'); // ignore any characters in the input buffer until we find an enter character
std::cin.get(); // get one more char from the user
}
Can someone please tell me how I can fix it? I get the error in the main function just to let everyone know.

choice is not declared in main, it's only in getUserInput2, which it returns. getUserChoice doesn't even take two arguments, so just do:
getUserChoice(input2);
You could start chaining your function calls a bit, e.g.: getUserChoice(getUserInput2()); and thus eliminate a few local variables.

Related

Whats a good way to get the program to end based on user input?

I did my "Hello World", I'm just getting started on my programming adventure with C++. Here is the first thing I've written, what are some ways to get it to end with user input? I'd like a yes or no option that would terminate the program. Also any feedback is welcome, thank you
#include <iostream>
using namespace std;
void Welcome();
void calculateNum();
void tryAgain();
int main() {
Welcome();
while (true) {
calculateNum();
tryAgain();
}
system("pause");
}
void calculateNum() {
float userNumber;
cin >> userNumber;
for (int i = 100; i >= 1; i--) {
float cNumber = i* userNumber;
cout << i << " >>>>> " << cNumber << endl;
}
}
void Welcome() {
cout << "Welcome \n Enter a number to see the first 100 multiples \n";
}
void tryAgain() {
cout << "Try again? Enter another number... ";
}
Here is one option:
Switch to do ... while loop, with the condition at the end.
Make your tryAgain() function return a boolean and put it in the while condition.
In tryAgain function read input from the user, and compare it to expected answers.
First, lets add a new header for string, it will make some things easier:
#include <string>
Second, lets rebuild the loop:
do {
calculateNum();
} while (tryAgain());
And finally, lets modify the function:
bool tryAgain() {
string answer;
cout << "Try again? (yes / no)\n";
cin >> answer;
if (answer == "yes") return true;
return false;
}
Now, there is a slightly shorter way to write that return, but it might be confusing for new learners:
return answer == "yes";
You don't need the if because == is an operator that returns bool type value.
You can change your calculateNum() in the following way:
Change the return value of your calculateNum() function into bool to indicate whether the program shall continue or stop
read the input into a std::string
check if the string is equal to your exit string like 'q' for quit
3.a in that case, your function returns false to indicate the caller that the program shall stop
3.b otherwise, create a stringstream with your string and read the content of the stream into your float variable and continue as you do like now
In your loop in your main function you break if calculateNum() returned false
Here is a simple solution:
#include <iostream>
// Here are two new Includes!
#include <sstream>
#include <string>
using namespace std;
void Welcome();
// Change return value of calculateNum()
bool calculateNum();
void tryAgain();
int main()
{
Welcome();
while (true)
{
if (!calculateNum())
break;
tryAgain();
}
system("pause");
}
bool calculateNum()
{
//Read input into string
string userInput;
cin >> userInput;
//Check for quit - string - here just simple q
if (userInput == "q")
return false;
//otherwise use a std::stringstream to read the string into a float as done before from cin.
float userNumber;
stringstream ss(userInput);
ss >> userNumber;
//and proces your numbers as before
for (int i = 100; i >= 1; i--)
{
float cNumber = i * userNumber;
cout << i << " >>>>> " << cNumber << endl;
}
return true;
}
void Welcome()
{
cout << "Welcome \n Enter a number to see the first 100 multiples \n";
}
void tryAgain()
{
cout << "Try again? Enter another number... ";
}
Having your users input in a string you can even do further checks like checking if the user entered a valid number, interpret localized numbers like . and , for decimal delimitters depending on your system settings and so on.

C++ program stuck in an infinite loop

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;
}

Undefined behaviour when entering `char` for integer variable with `std::cin`

I have this while loop, just to check if the entered number is 2. If the user entered by accident a letter instead of a number the loop goes to infinity even though I've added isdigit, but didn't fix the loop from going crazy if the input is a character. This is code:
int num1;
bool cond {false};
while(!cond){
cout<<"enter 2:";
cin>>num1;
if (!isdigit(num1)){
cout<<"not a digit:";
cin>>num1;
}
//
if(num1 == 2)
cond = true;
}
I would suggest trying something a little more straightforward instead of your current code:
#include <iostream>
#include <limits>
using namespace std;
int main()
{
int num1;
cout << "Please enter the number 2:";
cin >> num1;
while (num1 != 2)
{
cin.clear(); //Clears the error flag on cin.
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "You did not enter the number 2, Please try again:";
cin >> num1;
}
return 0;
}
Now, cin.ignore(numeric_limits<streamsize>::max(), '\n'); is when it ignores up until '\n' or EOF \n is the delimiter meaning that, that is the character at which cin will stop ignoring.
Furthermore, numeric_limits<streamsize>::max() is basically saying there is no limit to the number of characters to ignore.
You need to use the header file #include<limits> to use this.
I recommend separating the reading of input data from converting it to a number. A good method for this is to use stringstream. Here's a working example:
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
int num1;
string input;
bool cond{ false };
cout << "enter 2:";
while (!cond) {
getline(cin, input);
stringstream ss(input);
ss >> num1;
if( ss.fail() ){
cout << "not a digit:";
continue;
}
//
if (num1 == 2)
cond = true;
else
cout << "enter 2:";
}
return 0;
}
int num1;
bool cond {false};
do{
cout<<"enter 2:";
cin>>num1;
if (cin.good()) {
cond = true;
}else {
cin.clear();
cin.ignore();
cout << "Invalid, please enter 2." << endl;
}
}while(!cond);
While false, execute statements. Also, if you want the user to re-enter a number, cin must be flushed.
Try declaring the variable num1 as char because isdigit(ch) works for char and not for int.
I hope this solves your problem
Why does the loop iterate infinitely?
Let's take this example
int a;
std::cin >> a;
std::cout << "Entered: " << a;
Now let's test it with different inputs
with int
5
Entered: 5
10
Entered: 10
Yes just as we would expect, but what happens when you enter a char?
with char
r
Entered: 0
f
Entered: 0
Why does this happen?
When you declare the variable int, and then do std::cin >> , you are telling the input method that the user will enter an integer, but when it doesn't get what it expected, it will fail. C++ will not implicitly convert the value of char into int. Hence, you get strange results.
How to solve this?
As I have mentioned earlier, it fails. When this fails you can catch it this way
if (!(std::cin >> a))
{
std::cout << "Invalid input ! \n";
}
We're saying, if the input fails, show the message.
let's test this new code.
int a;
if (!(std::cin >> a))
{
std::cout << "Invalid input !\n";
}
else
{
std::cout << "Entered: " << a;
}
5
Entered: 5
r
Invalid input !
How to print char value of the int?
If you want to just print the ASCII value of the entered number, you need to cast the value into char.
You can do
int num = 48;
char character_value = num;
Here C++ will implicitly convert char into int.
But if you need a safer type of conversion, prefer using static_cast<>.
The syntax looks like this
int a = 5;
char character_value = static_cast<char>(a);
Static cast in C++
Type casting in C++
Dealing with invalid input in C++

How to print out a double value without losing first digit

When I run my code, it only prints the decimal parts of the double. On another page, I took a inputted double and printed out the double the way it was inputted.
But for my following code, it only prints out the decimals. For example, when I input 1.95 it only prints out 0.95. Why is it removing the first digit? I see nothing in my code that points to this.
I have already tried it in a more simple way and it worked. And I dont see any problems that would mess with the double in my code.
#include <iostream>
using namespace std;
int main()
{
double price;
char user_input;
do
{
cout << "Enter the purchase price (xx.xx) or `q' to quit: ";
cin >> user_input;
if (user_input == 'q')
{
return 0;
}
else
{
cin >> price;
int multiple = price * 100;
if (multiple % 5 == 0)
{
break;
}
else
{
cout << "Illegal price: Must be a non-negative multiple of 5 cents.\n" << endl;
}
}
} while (user_input != 'q');
cout << price << endl;
}
When I input 1.95, I get 0.95. But the output should be 1.95.
Problem covered in other answer: Reading for the 'q' removed the first character from the stream before it could be parsed into a double.
A solution: Read the double first. If the read fails, check to see if the input is a 'q'.
#include <iostream>
#include <limits>
using namespace std;
int main()
{
double price;
while (true)
{
cout << "Enter the purchase price (xx.xx) or `q' to quit: ";
if (cin >> price)
{
// use price
}
else // reading price failed. Find out why.
{
if (!cin.eof()) // didn't hit the end of the stream
{
// clear fail flag
cin.clear();
char user_input;
if (cin >> user_input && user_input == 'q') // test for q
{
break; // note: Not return. Cannot print price if the
// program returns
}
// Not a q or not readable. clean up whatever crap is still
// in the stream
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}
else
{
// someone closed the stream. Not much you can do here but exit
cerr << "Stream closed or broken. Cannot continue.";
return -1;
}
}
}
cout << price << endl;// Undefined behaviour if price was never set.
}
Another reasonable alternative is to read all input as std::string. If the string is not "q", attempt to convert it to a double with std::stod or an std::istringstream.
When you type 1.95 in the command line, variable user_input gets assigned '1', and price gets assigned .95.

EXC_BAD_ACCESS error with arrays in C++

sorry if this is a stupid question but I'm trying to write a program that compares 7 numbers that a user inputs to 7 numbers that the computer generates(a kind of lottery simulator). However, when i try to input the 7 numbers that the user inputs the program crashes after the second input. Please help, and thanks in advance!
This is the beginning of my main:
#include <iostream>
#include <iomanip>
#include "Implementation.hpp"
using namespace std;
int main()
{
string name;
cout << "What is your name?\n";
getline(cin, name);
while(1 != 0) //I know this will never be true, I'm just doing it
//because the return statement will
{ //end the program anyways if the user inputs 2
int *userNums = new int[7];
int *winningNums = new int[7];
int cont;
int matches;
cout << "LITTLETON CITY LOTTO MODEL\n";
cout << "--------------------------\n";
cout << "1) Play Lotto\n";
cout << "2) Quit Program\n";
cin >> cont;
if(cont == 2)
return 0;
getLottoPicks(&userNums);
And this is the getLottoPicks function:
void getLottoPicks(int *picks[])
{
int numsAdded = 0, choice;
while(numsAdded <= 7)
{
cout << "Please input a valid number as your lotto decision.\n";
cin >> choice;
if(noDuplicates(*picks, choice) == false)
continue;
*picks[numsAdded] = choice;
numsAdded++;
}
}
I'm fairly certain that it is a problem with the pointers that i'm trying to use, but without them I can't actually change the arrays I don't think, and I couldn't get the function to return an array.
If you're using C++, then you're probably better using a std::vector<int>, and passing in the reference to the vector in getLottoPicks.
However, your code should only be passing the int * to the getLottoPicks, and should process < 7 items - it's the classic off-by one.
call to getLottoPicks:
getLottoPicks(userNums);
and the new getLottoPicks code:
void getLottoPicks(int *picks)
{
int numsAdded = 0, choice;
while(numsAdded < 7)
{
cout << "Please input a valid number as your lotto decision.\n";
cin >> choice;
if(noDuplicates(picks, choice) == false)
continue;
picks[numsAdded] = choice;
numsAdded++;
}
}