I wanted to make a small "game" with a little bit of story, but I did some code and I think i did some major mistakes, here's the code
int main() {
char o, z, q, r;
string f, w = "yes", e = "no";
cout << "Hello, summoner!" << endl;
cin >> o;
cout << "You know why you are here, right?" << endl;
cin >> q;
switch ( z ) {
case 'w':
cout << "Ok";
break;
case 'e':
cout << "You are here to fight for your life!";
break;
}
return 0;
}
The second cin, the cin >> q; gets skipped every time I run the code and I don't know what to do.
It may not get skipped. You may be entering more than one characters in 'o', which is not allowed in your case and the second character automatically is stored in variable q.
Try changing the data type of the variables mentioned if you want to enter longer strings in the variables.
The second cin does not get skipped! When you enter "Hi" then std::cin >> o; will read 'H' while 'i' is still left in the stream. Then std::cin >> q; will read that 'i'. You can see the effect here:
#include <iostream>
int main() {
char first,second;
std::cin >> first;
std::cin >> second;
std::cout << first << " " << second;
}
When you type Hi then press enter the output will be
H i
If you want to store more than a single character then do not use char. For elaborated input checking you can use std::getline to read the whole line of user input and then check if it was a single character, a whole word, or something else.
PS: Don't use single letter variable names only. This is confusing and makes your code very hard to read and understand.
PPS: No offense, but don't make assumptions on what your code does. Instead use a debugger to see what actually happens. And try to be more precise on what you provide as input and what happens then. Your interpretation of "cin gets skipped" is off, and you could have seen this by inspecting the values of o and q after both cins.
Related
The code in the cont function asks the user if they want to play my game again.
The code works when receiving proper character inputs such as 'y' or 'n' as well as their respective capital letter variants, and the else block works properly to loop the function if an invalid input such as 'a' or 'c' is entered.
However during a test run, an input of 'yy' breaks the code causing the program to infinitely loop, running not only this cont function but my game function as well.
choice is stored as a char variable. I am wondering why the code even continues to run upon inputting multi-character inputs such as 'yy' or 'yes'. What's interesting is 'nn', 'ny' and other variations of multi-character inputs that begin with 'n' causes no issues and properly results in the else if block running as intended. Which prints "Thanks for playing." then ends the program.
Can variables declared as char accept inputs greater than 1 character? Does it only take the first value? And if so why does 'yy' cause a loop rather than the program running as intended by accepting a value of 'y' or 'Y'? How can I change my program so that an input of 'yy' no longer causes issues, without specific lines targeting inputs such as 'yy' or 'yes'.
#include <iostream>
#include <string> // needed to use strings
#include <cstdlib> // needed to use random numbers
#include <ctime>
using namespace std;
// declaring functions
void cont();
void game();
void diceRoll();
// variable declaration
string playerName;
int balance; // stores player's balance
int bettingAmount; // amount being bet, input by player
int guess; // users input for guess
int dice; // stores the random number
char choice;
// main functions
int main()
{
srand(time(0)); // seeds the random number, generates random number
cout << "\n\t\t-=-=-= Dice Roll Game =-=-=-\n";
cout << "\n\nWhat's your name?\n";
getline(cin, playerName);
cout << "\nEnter your starting balance to play with : $";
cin >> balance;
game();
cont();
}
// function declaration
void cont()
{
cin >> choice;
if(choice == 'Y' || choice == 'y')
{
cout << "\n\n";
game();
}
else if (choice == 'N' || choice == 'n')
{
cout << "\n\nThanks for playing.";
}
else
{
cout << "\n\nInvalid input, please type 'y' or 'n'";
cont(); // calls itself (recursive function!!!)
}
}
void game()
{
do
{
cout << "\nYour current balance is $ " << balance << "\n";
cout << "Hey, " << playerName << ", enter amount to bet : $";
cin >> bettingAmount;
if(bettingAmount > balance)
cout << "\nBetting balance can't be more than current balance!\n" << "\nRe-enter bet\n";
} while(bettingAmount > balance);
// Get player's numbers
do
{
cout << "\nA dice will be rolled, guess the side facing up, any number between 1 and 6 : \n";
cin >> guess;
if(guess <= 0 || guess > 6 )
{
cout << "\nYour guess should be between 1 and 6\n" << "Re-enter guess:\n";
}
} while(guess <= 0 || guess > 6);
dice = rand() % 6+1;
diceRoll();
if (dice == guess)
{
cout << "\n\nYou guessed correctly! You won $" << (bettingAmount * 6);
balance = balance + (bettingAmount * 6);
}
else
{
cout << "\n\nYou guessed wrong. You lost $" << bettingAmount << "\n";
balance = balance - bettingAmount;
}
cout << "\n" << playerName << ", you now have a balance of $" << balance << "\n";
if (balance == 0)
{
cout << "You're out of money, game over";
}
cout << "\nDo you want to play again? type y or n : \n";
cont();
}
void diceRoll()
{
cout << "The winning number is " << dice << "\n";
}
Does it only take the first value?
Yes, the >> formatted extraction operator, when called for a single char value, will read the first non-whitespace character, and stop. Everything after it remains unread.
why does 'yy' cause a loop
Because the first "y" gets read, for the reasons explained above. The second "y" remains unread.
This is a very common mistake and a misconception about what >> does. It does not read an entire line of typed input. It only reads a single value after skipping any whitespace that precedes it.
Your program stops until an entire line of input gets typed, followed by Enter, but that's not what >> reads. It only reads what it's asked to read, and everything else that gets typed in remains unread.
So the program continues to execute, until it reaches this part:
cin >> bettingAmount;
At this point the next unread character in the input is y. The >> formatted extraction operator, for an int value like this bettingAmount, requires numerical input (following optional whitespace). But the next character is not numerical. It's the character y.
This results in the formatted >> extraction operator failing. Nothing gets read into bettingAmount. It remains completely unaltered by the >> operator. Because it is declared in global scope it was zero-initialized. So it remains 0.
In addition to the >> extraction operator failing, as part of it failing it sets the input stream to a failed state. When an input stream is in a failed state all subsequent input operation automatically fail without doing anything. And that's why your program ends up in an infinite loop.
Although there is a way to clear the input stream from its failed state this is a clumsy approach. The clean solution is to fix the code that reads input.
If your intent is to stop the program and enter something followed by Enter then that's what std::getline is for. The shown program uses it to read some of its initial input.
The path of least resistance is to simply use std::getline to read all input. Instead of using >> to read a single character use std::getline to read the next line of typed in input, into a std::string, then check the the string's first character and see what it is. Problem solved.
cin >> bettingAmount;
And you want to do the same thing here. Otherwise you'll just run into the same problem: mistyped input will result in a failed input operation, and a major headache.
Why do you need this headache? Just use std::getline to read text into a std::string, construct a std::istringstream from it, then use >> on the std::istringstream, and check its return value to determine whether it failed, or not. That's a simple way to check for invalid input, and if something other than numeric input was typed in here, you have complete freedom on how to handle bad typed in input.
So I am trying to allow a user to make a selection but Im having issue. the following lines
cout << "\nWhat would you like to do? " << endl
<< "(1) Display the entire List" << endl
<< "(2) to exit" << endl;
getline(cin,answer);
answerint=stoi(answer);
switch (answerint) {
case 1:
showLIST(myLIst);
break;
case 2:
return;
default:
cout << "\nPlease choose from the list";
}
So this works fine if the user chooses a integer, but it barfs if a character is received. How can I make this less breakable?
It seems you want something like this:
if (std::cin >> answerint) {
switch (answerint) {
// ...
}
}
Converting the integer directly from the stream puts the stream into failure state in case something different than an int is entered. You may want to add a manipulator to make sure there are no non-space characters on the end of the line.
std::string answer;
int answerint;
while (std::getline(std::cin, answer))
{
std::istringstream iss(answer);
char c; // seek non-whitespace after the number?
if (!(iss >> answerint) || (iss >> c))
answerint = 0;
switch (answerint)
{
...as you had...
}
}
The code above uses getline to ensure a complete line of text is parsed into answer, after which it creates a distinct std::istringstream from that single line. That way, we can use >> knowing it won't skip over newlines are consume or wait for further input.
if (!(iss >> answerint) || (iss >> c)) checks whether there isn't a number to parse into answerint, or after having parsed a number an additional non-whitespace character appears on the line. For example, the line typed might have been 2r - we'll consider that an error and reprompt for a selection, assigning 0 to answerint to ensure we reach the default label in the switch, thereby printing cout << "\nPlease choose from the list"; which seems as appropriate for 2r as for 99 or whatever....
Using getline and a separate istringstream this way, you can reliably attempt parsing of the next line of input from a known state.
I actually wrote a function to convert string into ascii values.
However I managed to confuse my self and don't understand why my own code works.
here it is:
void convertToString()
{
char redo;
int letter;
int length;
do {
cout<< "How long is your word \n";
cin >> length;
cout << "Type in the letter values \n";
for (int x = 0; x < length; x++) {
cin >> letter;
cout << char (letter);
}
cout << "\n To enter another word hit R" << endl;
cin >> redo;
} while (redo == 'R');
}
In the terminal I can type in all the ASCII values I want with out changing line, however I though this would cause a problem, anyways my question is, is hitting the enter button the same as hitting space? if not i dont understand how my code is able to print out the chars since i write it all in one line...Does it assign the interger "letter" a new value everytime there is a space?
Please help/explain
This is to expand a bit on what Igor said in his comment and to give a little example.
As Igor said, istream::operator>>(&int) will read non-whitespace. This means for each call on the operator, it scans along the input stream (what you typed in) for non-whitespace and reads until the next whitespace again. The next call will pick up where you left off. So, entering a space or a newline is exactly the same for this situation where you're taking in an int.
You can verify this with a simple bit of code that scans until EOF:
#include <iostream>
int main()
{
int number;
while (std::cin >> number)
{
std::cout << number << std::endl;
}
return 0;
}
This will wait for user entry to be complete (pressing enter), but print a new line for each integer in your input as separated by whitespace. So "1 2 3 4" will print each of those numbers on separate lines, regardless of if you separate them with spaces, tabs, or newlines.
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.
So I'm having some problems when using cin and switch statements.
First, the program prompts for an int:
cout << "How many registers would you like to use?: ";
cin >> kassor;
Later on, the user is asked to make a choice:
cout << endl << "Press \"S\" to run the queue simulator and/or make a timestep." << endl;
cout << "Press \"Q\" to quit." << endl << endl;
cout << "Choice: ";
cin >> choice;
switch (choice)
{
case 's':
case 'S':
{
cout << "test";
nya = newCustomers();
break;
}
case 'q':
case 'Q':
{
return 0;
}
}
Here, the 'q' option works fine, but the 's' option does not. It 'hangs', as if still waiting for input. I have tried various cin.ignore() and such, but to no avail.
What puzzles me is that
switch (choice)
{
case 's':
case 'S':
{
cout << "test";
nya = newCustomers();
break;
Gives nothing, but the following:
switch (choice)
{
case 's':
case 'S':
{
cout << "test";
cin.ignore(1024, '\n');
nya = newCustomers();
break;
outputs 'test'.
My main question here is: Is the problem cin or something in the case: s ?
Thank's in advance :
It looks like the function newCustomers is getting hung up on a stray character or two after the input to choice. That would explain why the ignore call "fixes" the problem, and it's the solution suggested in the comment by #TheBuzzSaw.
To see this more clearly, change
cout << "test";
to
cout << "test\n";
or to
cout << "test" << std::endl;
On some systems the console is line-buffered, so you won't see any output until the program writes a newline. Inserting endl flushes the output buffer, so you should see the message even if subsequent code hangs. Or change it to:
cerr << "test\n";
cerr is more aggressive about flushing buffers, precisely because it's used for error output that you don't want to miss.
Short Answer:
I believe cin is failing somewhere or it is the cause of unexpected behavior
I cannot pinpoint it without more code.
Edit: I cannot post a comment on your post, so I figured I would post here. The "hanging" could be an infinite loop. If you are doing something like what I am doing and looping to get input and fail to clear the error bits cin will constantly fail. If you are not cout'ing something each time you ask for input it could just be quietly sitting there looping through that input-gathering function. Put breakpoints in all of your loops and step through with the debugger to verify none of your loops are infinite looping.
Try adding a std::cin.clear(); before the .ignore() in your second test case. See if that stops your hanging.
Explanation:
cin can fail. This can cause weird behavior. A common way for it to fail is if you are reading into an integer and you get character input. Once cin fails it sets a fail bit and from then on, cin does not behave like you would expect.
I would recommend not using a bare cin >> choice, because it can fail and you wont know it. I would abstract this out to a method that gets the proper input you want.
I personally keep a tiny utility library(.h) and .cpp around to include in projects where I am using common functionality I have already coded.
I have a readIntBetween() function which accepts 2 integers and reads from standard input an integer between those two numbers. If the user does not provide the right input (integer over or under bounds, or input containing a character) I ask them to re-enter the input.
In this function I make sure to clear the fail bits when I have detected a failure, and I ignore something like 200 characters to "flush" it out.
Here is the code to my readIntBetween function. I hope it helps you diagnose your error and fix it:
int readIntBetween(int lower, int upper, std::string prompt)
{
bool goodVal = false;
int value = -1;
do
{
std::cout << prompt ;
std::cin >> value;
if ( value >= lower && value <= upper)//check to make sure the value is in between upper and lower
goodVal = true;
else
{
if(std::cin.fail())
{
std::cout << "\tError - invalid format for integer number" << std::endl;
//clear all status bit including fail bit
std::cin.clear();
//flush input buffer to discard invalid input
std::cin.ignore(1000, '\n');
}
else
std::cout << "Value is not valid, must be between " << lower<<" and "<<upper<<"."<<std::endl;
}
}while(!goodVal);
return value;
}
If your cin is being used inside a loop, I recommend a different approach.
char buffer[64];
cin.getline(buffer, 64);
char choice = buffer[0];
switch (choice)
{
...
}
It is possible your cin is not being reset, and any extraneous characters are being fed into the subsequent requests for input. Grab more input and only process its first character.