I get an error when I try to run this. It lets me input the employee's name, and then once I press Enter it skips to the end, flying by hours and pay rate without letting me input them.
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
string EmployeeName; //Employee Name
float EmployeeHours;
float EmployeePayRate; //Emplyee Hours, Employee PayRate
cout << setw(50) << "Employee Name: ";
cin >> EmployeeName;
cout << setw(50) << "EmployeeHours: ";
cin >> EmployeeHours;
cout << setw(50) << "EmployeePayRate: ";
cin >> EmployeePayRate;
system("PAUSE");
return 0;
}
I think the comment by #AleksandarStojadinovic provides the clue to your problem. Use
std::getline(std::cin, EmployeeName);
to read the name. Rest can be as they are.
I am having an error when I try to run this. It lets me input the
employees name, and then once you press enter it skips to the end.
flies by hours and pay rate without letting me input that.
Your name must be comprised of two/three names, maybe first middle and last or something like that. cin deals space separated string and enter separated string in the same way i.e it will take each space separated string as different input. And for that if your name has two names, the first one goes right but the second one goes to a variable which is declared as float and the program terminates abnormally. That's why using cin in this case isn't right at all.
You need to use another way to avoid that, like using getline(cin,EmployeeName) .
Related
Google Code University's C++ tutorial used to have this code:
// Description: Illustrate the use of cin to get input
// and how to recover from errors.
#include <iostream>
using namespace std;
int main()
{
int input_var = 0;
// Enter the do while loop and stay there until either
// a non-numeric is entered, or -1 is entered. Note that
// cin will accept any integer, 4, 40, 400, etc.
do {
cout << "Enter a number (-1 = quit): ";
// The following line accepts input from the keyboard into
// variable input_var.
// cin returns false if an input operation fails, that is, if
// something other than an int (the type of input_var) is entered.
if (!(cin >> input_var)) {
cout << "Please enter numbers only." << endl;
cin.clear();
cin.ignore(10000,'\n');
}
if (input_var != -1) {
cout << "You entered " << input_var << endl;
}
}
while (input_var != -1);
cout << "All done." << endl;
return 0;
}
What is the significance of cin.clear() and cin.ignore()? Why are the 10000 and \n parameters necessary?
The cin.clear() clears the error flag on cin (so that future I/O operations will work correctly), and then cin.ignore(10000, '\n') skips to the next newline (to ignore anything else on the same line as the non-number so that it does not cause another parse failure). It will only skip up to 10000 characters, so the code is assuming the user will not put in a very long, invalid line.
You enter the
if (!(cin >> input_var))
statement if an error occurs when taking the input from cin. If an error occurs then an error flag is set and future attempts to get input will fail. That's why you need
cin.clear();
to get rid of the error flag. Also, the input which failed will be sitting in what I assume is some sort of buffer. When you try to get input again, it will read the same input in the buffer and it will fail again. That's why you need
cin.ignore(10000,'\n');
It takes out 10000 characters from the buffer but stops if it encounters a newline (\n). The 10000 is just a generic large value.
Why do we use:
1) cin.ignore
2) cin.clear
?
Simply:
1) To ignore (extract and discard) values that we don't want on the stream
2) To clear the internal state of stream. After using cin.clear internal state is set again back to goodbit, which means that there are no 'errors'.
Long version:
If something is put on 'stream' (cin) then it must be taken from there. By 'taken' we mean 'used', 'removed', 'extracted' from stream. Stream has a flow. The data is flowing on cin like water on stream. You simply cannot stop the flow of water ;)
Look at the example:
string name; //line 1
cout << "Give me your name and surname:"<<endl;//line 2
cin >> name;//line 3
int age;//line 4
cout << "Give me your age:" <<endl;//line 5
cin >> age;//line 6
What happens if the user answers: "Arkadiusz Wlodarczyk" for first question?
Run the program to see for yourself.
You will see on console "Arkadiusz" but program won't ask you for 'age'. It will just finish immediately right after printing "Arkadiusz".
And "Wlodarczyk" is not shown. It seems like if it was gone (?)*
What happened? ;-)
Because there is a space between "Arkadiusz" and "Wlodarczyk".
"space" character between the name and surname is a sign for computer that there are two variables waiting to be extracted on 'input' stream.
The computer thinks that you are tying to send to input more than one variable. That "space" sign is a sign for him to interpret it that way.
So computer assigns "Arkadiusz" to 'name' (2) and because you put more than one string on stream (input) computer will try to assign value "Wlodarczyk" to variable 'age' (!). The user won't have a chance to put anything on the 'cin' in line 6 because that instruction was already executed(!). Why? Because there was still something left on stream. And as I said earlier stream is in a flow so everything must be removed from it as soon as possible. And the possibility came when computer saw instruction cin >> age;
Computer doesn't know that you created a variable that stores age of somebody (line 4). 'age' is merely a label. For computer 'age' could be as well called: 'afsfasgfsagasggas' and it would be the same. For him it's just a variable that he will try to assign "Wlodarczyk" to because you ordered/instructed computer to do so in line (6).
It's wrong to do so, but hey it's you who did it! It's your fault! Well, maybe user, but still...
All right all right. But how to fix it?!
Let's try to play with that example a bit before we fix it properly to learn a few more interesting things :-)
I prefer to make an approach where we understand things. Fixing something without knowledge how we did it doesn't give satisfaction, don't you think? :)
string name;
cout << "Give me your name and surname:"<<endl;
cin >> name;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << cin.rdstate(); //new line is here :-)
After invoking above code you will notice that the state of your stream (cin) is equal to 4 (line 7). Which means its internal state is no longer equal to goodbit. Something is messed up. It's pretty obvious, isn't it? You tried to assign string type value ("Wlodarczyk") to int type variable 'age'. Types doesn't match. It's time to inform that something is wrong. And computer does it by changing internal state of stream. It's like: "You f**** up man, fix me please. I inform you 'kindly' ;-)"
You simply cannot use 'cin' (stream) anymore. It's stuck. Like if you had put big wood logs on water stream. You must fix it before you can use it. Data (water) cannot be obtained from that stream(cin) anymore because log of wood (internal state) doesn't allow you to do so.
Oh so if there is an obstacle (wood logs) we can just remove it using tools that is made to do so?
Yes!
internal state of cin set to 4 is like an alarm that is howling and making noise.
cin.clear clears the state back to normal (goodbit). It's like if you had come and silenced the alarm. You just put it off. You know something happened so you say: "It's OK to stop making noise, I know something is wrong already, shut up (clear)".
All right let's do so! Let's use cin.clear().
Invoke below code using "Arkadiusz Wlodarczyk" as first input:
string name;
cout << "Give me your name and surname:"<<endl;
cin >> name;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << cin.rdstate() << endl;
cin.clear(); //new line is here :-)
cout << cin.rdstate()<< endl; //new line is here :-)
We can surely see after executing above code that the state is equal to goodbit.
Great so the problem is solved?
Invoke below code using "Arkadiusz Wlodarczyk" as first input:
string name;
cout << "Give me your name and surname:"<<endl;
cin >> name;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << cin.rdstate() << endl;;
cin.clear();
cout << cin.rdstate() << endl;
cin >> age;//new line is here :-)
Even tho the state is set to goodbit after line 9 the user is not asked for "age". The program stops.
WHY?!
Oh man... You've just put off alarm, what about the wood log inside a water?* Go back to text where we talked about "Wlodarczyk" how it supposedly was gone.
You need to remove "Wlodarczyk" that piece of wood from stream. Turning off alarms doesn't solve the problem at all. You've just silenced it and you think the problem is gone? ;)
So it's time for another tool:
cin.ignore can be compared to a special truck with ropes that comes and removes the wood logs that got the stream stuck. It clears the problem the user of your program created.
So could we use it even before making the alarm goes off?
Yes:
string name;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
int age;
cout << "Give me your age:" << endl;
cin >> age;
The "Wlodarczyk" is gonna be removed before making the noise in line 7.
What is 10000 and '\n'?
It says remove 10000 characters (just in case) until '\n' is met (ENTER). BTW It can be done better using numeric_limits but it's not the topic of this answer.
So the main cause of problem is gone before noise was made...
Why do we need 'clear' then?
What if someone had asked for 'give me your age' question in line 6 for example: "twenty years old" instead of writing 20?
Types doesn't match again. Computer tries to assign string to int. And alarm starts. You don't have a chance to even react on situation like that. cin.ignore won't help you in case like that.
So we must use clear in case like that:
string name;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
int age;
cout << "Give me your age:" << endl;
cin >> age;
cin.clear();
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
But should you clear the state 'just in case'?
Of course not.
If something goes wrong (cin >> age;) instruction is gonna inform you about it by returning false.
So we can use conditional statement to check if the user put wrong type on the stream
int age;
if (cin >> age) //it's gonna return false if types doesn't match
cout << "You put integer";
else
cout << "You bad boy! it was supposed to be int";
All right so we can fix our initial problem like for example that:
string name;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
int age;
cout << "Give me your age:" << endl;
if (cin >> age)
cout << "Your age is equal to:" << endl;
else
{
cin.clear();
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
cout << "Give me your age name as string I dare you";
cin >> age;
}
Of course this can be improved by for example doing what you did in question using loop while.
BONUS:
You might be wondering. What about if I wanted to get name and surname in the same line from the user? Is it even possible using cin if cin interprets each value separated by "space" as different variable?
Sure, you can do it two ways:
1)
string name, surname;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin >> surname;
cout << "Hello, " << name << " " << surname << endl;
2) or by using getline function.
getline(cin, nameOfStringVariable);
and that's how to do it:
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
The second option might backfire you in case you use it after you use 'cin' before the getline.
Let's check it out:
a)
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << "Your age is" << age << endl;
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
If you put "20" as age you won't be asked for nameAndSurname.
But if you do it that way:
b)
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << "Your age is" << age << endll
everything is fine.
WHAT?!
Every time you put something on input (stream) you leave at the end white character which is ENTER ('\n') You have to somehow enter values to console. So it must happen if the data comes from user.
b) cin characteristics is that it ignores whitespace, so when you are reading in information from cin, the newline character '\n' doesn't matter. It gets ignored.
a) getline function gets the entire line up to the newline character ('\n'), and when the newline char is the first thing the getline function gets '\n', and that's all to get. You extract newline character that was left on stream by user who put "20" on stream in line 3.
So in order to fix it is to always invoke cin.ignore(); each time you use cin to get any value if you are ever going to use getline() inside your program.
So the proper code would be:
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cin.ignore(); // it ignores just enter without arguments being sent. it's same as cin.ignore(1, '\n')
cout << "Your age is" << age << endl;
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
I hope streams are more clear to you know.
Hah silence me please! :-)
use cin.ignore(1000,'\n') to clear all of chars of the previous cin.get() in the buffer and it will choose to stop when it meet '\n' or 1000 chars first.
I have to prompt the user to enter their name and their height using two separate inputs, one for feet and one for inches. Then I have to display what they entered. Ex: Name, you are x feet y inches tall. The program I wrote runs but it's not working as intended. After I write an input for the first question which is the name, the program skips the other questions and ends the program.
#include <iostream>
#include <string>
using namespace std;
int main ()
{
int name;
int f;
int i;
// Start: Enter Name
cout << "What is your name?" << endl;
cin >> name;
// Enter height
cout << "How many feet tall are you tall?" << endl;
cin >> f;
// Enter inches
cout << "How many inches tall are you after feet?" << endl;
cin >> i;
// End: All info entered
cout << name << " you are " << (f) << " feet " << (i) << " inches tall." << endl;
system("pause");
return 0;
}
First and most importantly, you should change the type of name from int to string. Your current program tries to read your input to name as a number and not text.
Now, if your program is still not working as intended, make sure when you are typing in input for name there is no white-space (spaces, tabs, etc.) in between the name you are trying to input. For example, the input of John would be fine, but John Smith would not. This is because the >> operator when used with cin values all white-space the same. This means that pressing the space bar is the same as pressing the enter key to your program. If you are trying to input a full name, look into using getline to read input up until the enter key is pressed.
Getline Example:
cout << "What is your name?";
getline (std::cin,name);
Summary:
Change int name; to string name;. If you need to read multiple words for the name use getline.
when I enter (Im qwerty) as (y), the program shows "Your account has been deactivated" instead of "Your password is incorrect". I've searched for same problems but const. char and using strcmp is too complicated for me and my instructor does not use that kinds of codes.I'm very eager to know what must I do to make my program right. (Tnx in advance)
#include <iostream>
using namespace std;
int main () {
string y;
cout << "Enter Icode: ";
cin >> y;
if (y == "Im qwerty")
cout << "Your password is incorrect.";
else
cout << "Your account has been deactivated.";
cin.get();
return 0;
}
The problem is that cin >> y; reads one word, whereas there are two words in "Im qwerty". In other words, this program always outputs "Your account has been deactivated." because one word never matches two words.
If you would like to read multiple words, the easiest is to read an entire line, e.g. replace cin >> y; with getline(cin, y);.
This is my first attempt at C++, following an example to calculate a tip through a console application. The full (working code) is shown below:
// Week1.cpp : Defines the entry point for the console application.
#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
// Declare variables
double totalBill = 0.0;
double liquour = 0.0;
double tipPercentage = 0.0;
double totalNoLiquour = 0.0;
double tip = 0.0;
string hadLiquour;
// Capture inputs
cout << "Did you drink any booze? (Yes or No)\t";
getline(cin, hadLiquour, '\n');
if(hadLiquour == "Yes") {
cout << "Please enter you booze bill\t";
cin >> liquour;
}
cout << "Please enter your total bill\t";
cin >> totalBill;
cout << "Enter the tip percentage (in decimal form)\t";
cin >> tipPercentage;
// Process inputs
totalNoLiquour = totalBill - liquour;
tip = totalNoLiquour * tipPercentage;
// Output
cout << "Tip: " << (char)156 << tip << endl;
system("pause");
return 0;
}
This works fine. However, I want to move:
cout << "Please enter your total bill\t";
cin >> totalBill;
to be the first line under:
// Capture inputs
But when I do the application breaks (it compiles, but just ignores the if statement and then prints both cout's at once.
Im scratching my head becuase I cant understand what's going on - but I'm assuming I'm being an idiot!
Thanks
Try this
// Capture inputs
cout << "Please enter your total bill\t";
cin >> totalBill;
cin.clear();
cin.sync();
See c++ getline() isn't waiting for input from console when called multiple times
Or, better yet don't use getline at all:
cout << "Please enter your total bill\t";
cin >> totalBill;
cout << "Did you drink any booze? (Yes or No)\t";
cin >> hadLiquour;
totalBill is a number, i.e. the program "consumes" everything from your input that is a number. Let's say you entered:
42.2[RETURN]
The 42.2 gets copied into totalBill. The [RETURN] doesn't match, and remains in the input buffer.
Now, when you call getline(), the [RETURN] is still sitting there... I am sure you can figure out the rest from there.
Cin doesn't remove the newline character from the stream or do type-checking. So using cin>>var; and following it up with another cin >> stringtype; or getline(); will receive empty inputs. It's best practice to NOT MIX the different types of input methods from cin.
[for more informations see link]
you may change your code as below :
cout << "Please enter your total bill\t";
getline(cin, hadLiquour); // i used the hadLiquour string var as a temp var
// so don't be confused
stringstream myStream(hadLiquour);
myStream >> totalBill;
Google Code University's C++ tutorial used to have this code:
// Description: Illustrate the use of cin to get input
// and how to recover from errors.
#include <iostream>
using namespace std;
int main()
{
int input_var = 0;
// Enter the do while loop and stay there until either
// a non-numeric is entered, or -1 is entered. Note that
// cin will accept any integer, 4, 40, 400, etc.
do {
cout << "Enter a number (-1 = quit): ";
// The following line accepts input from the keyboard into
// variable input_var.
// cin returns false if an input operation fails, that is, if
// something other than an int (the type of input_var) is entered.
if (!(cin >> input_var)) {
cout << "Please enter numbers only." << endl;
cin.clear();
cin.ignore(10000,'\n');
}
if (input_var != -1) {
cout << "You entered " << input_var << endl;
}
}
while (input_var != -1);
cout << "All done." << endl;
return 0;
}
What is the significance of cin.clear() and cin.ignore()? Why are the 10000 and \n parameters necessary?
The cin.clear() clears the error flag on cin (so that future I/O operations will work correctly), and then cin.ignore(10000, '\n') skips to the next newline (to ignore anything else on the same line as the non-number so that it does not cause another parse failure). It will only skip up to 10000 characters, so the code is assuming the user will not put in a very long, invalid line.
You enter the
if (!(cin >> input_var))
statement if an error occurs when taking the input from cin. If an error occurs then an error flag is set and future attempts to get input will fail. That's why you need
cin.clear();
to get rid of the error flag. Also, the input which failed will be sitting in what I assume is some sort of buffer. When you try to get input again, it will read the same input in the buffer and it will fail again. That's why you need
cin.ignore(10000,'\n');
It takes out 10000 characters from the buffer but stops if it encounters a newline (\n). The 10000 is just a generic large value.
Why do we use:
1) cin.ignore
2) cin.clear
?
Simply:
1) To ignore (extract and discard) values that we don't want on the stream
2) To clear the internal state of stream. After using cin.clear internal state is set again back to goodbit, which means that there are no 'errors'.
Long version:
If something is put on 'stream' (cin) then it must be taken from there. By 'taken' we mean 'used', 'removed', 'extracted' from stream. Stream has a flow. The data is flowing on cin like water on stream. You simply cannot stop the flow of water ;)
Look at the example:
string name; //line 1
cout << "Give me your name and surname:"<<endl;//line 2
cin >> name;//line 3
int age;//line 4
cout << "Give me your age:" <<endl;//line 5
cin >> age;//line 6
What happens if the user answers: "Arkadiusz Wlodarczyk" for first question?
Run the program to see for yourself.
You will see on console "Arkadiusz" but program won't ask you for 'age'. It will just finish immediately right after printing "Arkadiusz".
And "Wlodarczyk" is not shown. It seems like if it was gone (?)*
What happened? ;-)
Because there is a space between "Arkadiusz" and "Wlodarczyk".
"space" character between the name and surname is a sign for computer that there are two variables waiting to be extracted on 'input' stream.
The computer thinks that you are tying to send to input more than one variable. That "space" sign is a sign for him to interpret it that way.
So computer assigns "Arkadiusz" to 'name' (2) and because you put more than one string on stream (input) computer will try to assign value "Wlodarczyk" to variable 'age' (!). The user won't have a chance to put anything on the 'cin' in line 6 because that instruction was already executed(!). Why? Because there was still something left on stream. And as I said earlier stream is in a flow so everything must be removed from it as soon as possible. And the possibility came when computer saw instruction cin >> age;
Computer doesn't know that you created a variable that stores age of somebody (line 4). 'age' is merely a label. For computer 'age' could be as well called: 'afsfasgfsagasggas' and it would be the same. For him it's just a variable that he will try to assign "Wlodarczyk" to because you ordered/instructed computer to do so in line (6).
It's wrong to do so, but hey it's you who did it! It's your fault! Well, maybe user, but still...
All right all right. But how to fix it?!
Let's try to play with that example a bit before we fix it properly to learn a few more interesting things :-)
I prefer to make an approach where we understand things. Fixing something without knowledge how we did it doesn't give satisfaction, don't you think? :)
string name;
cout << "Give me your name and surname:"<<endl;
cin >> name;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << cin.rdstate(); //new line is here :-)
After invoking above code you will notice that the state of your stream (cin) is equal to 4 (line 7). Which means its internal state is no longer equal to goodbit. Something is messed up. It's pretty obvious, isn't it? You tried to assign string type value ("Wlodarczyk") to int type variable 'age'. Types doesn't match. It's time to inform that something is wrong. And computer does it by changing internal state of stream. It's like: "You f**** up man, fix me please. I inform you 'kindly' ;-)"
You simply cannot use 'cin' (stream) anymore. It's stuck. Like if you had put big wood logs on water stream. You must fix it before you can use it. Data (water) cannot be obtained from that stream(cin) anymore because log of wood (internal state) doesn't allow you to do so.
Oh so if there is an obstacle (wood logs) we can just remove it using tools that is made to do so?
Yes!
internal state of cin set to 4 is like an alarm that is howling and making noise.
cin.clear clears the state back to normal (goodbit). It's like if you had come and silenced the alarm. You just put it off. You know something happened so you say: "It's OK to stop making noise, I know something is wrong already, shut up (clear)".
All right let's do so! Let's use cin.clear().
Invoke below code using "Arkadiusz Wlodarczyk" as first input:
string name;
cout << "Give me your name and surname:"<<endl;
cin >> name;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << cin.rdstate() << endl;
cin.clear(); //new line is here :-)
cout << cin.rdstate()<< endl; //new line is here :-)
We can surely see after executing above code that the state is equal to goodbit.
Great so the problem is solved?
Invoke below code using "Arkadiusz Wlodarczyk" as first input:
string name;
cout << "Give me your name and surname:"<<endl;
cin >> name;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << cin.rdstate() << endl;;
cin.clear();
cout << cin.rdstate() << endl;
cin >> age;//new line is here :-)
Even tho the state is set to goodbit after line 9 the user is not asked for "age". The program stops.
WHY?!
Oh man... You've just put off alarm, what about the wood log inside a water?* Go back to text where we talked about "Wlodarczyk" how it supposedly was gone.
You need to remove "Wlodarczyk" that piece of wood from stream. Turning off alarms doesn't solve the problem at all. You've just silenced it and you think the problem is gone? ;)
So it's time for another tool:
cin.ignore can be compared to a special truck with ropes that comes and removes the wood logs that got the stream stuck. It clears the problem the user of your program created.
So could we use it even before making the alarm goes off?
Yes:
string name;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
int age;
cout << "Give me your age:" << endl;
cin >> age;
The "Wlodarczyk" is gonna be removed before making the noise in line 7.
What is 10000 and '\n'?
It says remove 10000 characters (just in case) until '\n' is met (ENTER). BTW It can be done better using numeric_limits but it's not the topic of this answer.
So the main cause of problem is gone before noise was made...
Why do we need 'clear' then?
What if someone had asked for 'give me your age' question in line 6 for example: "twenty years old" instead of writing 20?
Types doesn't match again. Computer tries to assign string to int. And alarm starts. You don't have a chance to even react on situation like that. cin.ignore won't help you in case like that.
So we must use clear in case like that:
string name;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
int age;
cout << "Give me your age:" << endl;
cin >> age;
cin.clear();
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
But should you clear the state 'just in case'?
Of course not.
If something goes wrong (cin >> age;) instruction is gonna inform you about it by returning false.
So we can use conditional statement to check if the user put wrong type on the stream
int age;
if (cin >> age) //it's gonna return false if types doesn't match
cout << "You put integer";
else
cout << "You bad boy! it was supposed to be int";
All right so we can fix our initial problem like for example that:
string name;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
int age;
cout << "Give me your age:" << endl;
if (cin >> age)
cout << "Your age is equal to:" << endl;
else
{
cin.clear();
cin.ignore(10000, '\n'); //time to remove "Wlodarczyk" the wood log and make the stream flow
cout << "Give me your age name as string I dare you";
cin >> age;
}
Of course this can be improved by for example doing what you did in question using loop while.
BONUS:
You might be wondering. What about if I wanted to get name and surname in the same line from the user? Is it even possible using cin if cin interprets each value separated by "space" as different variable?
Sure, you can do it two ways:
1)
string name, surname;
cout << "Give me your name and surname:"<< endl;
cin >> name;
cin >> surname;
cout << "Hello, " << name << " " << surname << endl;
2) or by using getline function.
getline(cin, nameOfStringVariable);
and that's how to do it:
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
The second option might backfire you in case you use it after you use 'cin' before the getline.
Let's check it out:
a)
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << "Your age is" << age << endl;
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
If you put "20" as age you won't be asked for nameAndSurname.
But if you do it that way:
b)
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cout << "Your age is" << age << endll
everything is fine.
WHAT?!
Every time you put something on input (stream) you leave at the end white character which is ENTER ('\n') You have to somehow enter values to console. So it must happen if the data comes from user.
b) cin characteristics is that it ignores whitespace, so when you are reading in information from cin, the newline character '\n' doesn't matter. It gets ignored.
a) getline function gets the entire line up to the newline character ('\n'), and when the newline char is the first thing the getline function gets '\n', and that's all to get. You extract newline character that was left on stream by user who put "20" on stream in line 3.
So in order to fix it is to always invoke cin.ignore(); each time you use cin to get any value if you are ever going to use getline() inside your program.
So the proper code would be:
int age;
cout << "Give me your age:" <<endl;
cin >> age;
cin.ignore(); // it ignores just enter without arguments being sent. it's same as cin.ignore(1, '\n')
cout << "Your age is" << age << endl;
string nameAndSurname;
cout << "Give me your name and surname:"<< endl;
getline(cin, nameAndSurname);
cout << "Hello, " << nameAndSurname << endl;
I hope streams are more clear to you know.
Hah silence me please! :-)
use cin.ignore(1000,'\n') to clear all of chars of the previous cin.get() in the buffer and it will choose to stop when it meet '\n' or 1000 chars first.