I have the code below, the credit and load successfully saves to the object but the code and title only takes the first letter of the string. I tried course[noOfCourses_].setCode(Code) at first, but it gives me an error:
cannot initialize a parameter type char with lvalue char[7].
Another problem is that the input skips the input prompt for the title and goes directly to credit prompt.
void addCourse() {
Course*course=new Course[MAX_NO_RECS];
char Code[MAX_COURSECODE_LEN];
char title[20];
int credit;
int load;
cout << "Course Code: ";
cin.clear();
cin >> Code;
cin.clear();
course[noOfCourses_].setCode(Code[0]);
cin.clear();
cout << "Course Title: ";
cin.getline(title, 20);
cin.clear();
course[noOfCourses_].setTitle(title[0]);
cin.clear();
cout << "Credits: ";
cin >> credit;
course[noOfCourses_].setCredits(credit);
cout << "Study Load: ";
cin >> load;
course[noOfCourses_].setLoad(load);
noOfCourses_++;
courseList_[noOfCourses_]=course;
}
Your first way was right. You should use course[noOfCourses_].setCode(Code) (by the way: it would be better if the variable Code would not start with a capital letter).
The problem is in your Cource::setCode function. It seems that the parameter is a char. To fix it, you should change it to a char*
void Cource::setCode(char* code) {
...
}
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 am trying to ask the user to enter the names of 3 of their friends, however, it only asks one question and writes the answer from my first one in the second and third.
#include <iostream>
using namespace std;
int main()
{
char first_name;
cout << "Please enter a name: ";
cin >> first_name;
cout << first_name << endl;
char second_name;
cout << "Please enter a name: ";
cin >> second_name;
cout << second_name << endl;
char third_name;
cout << "Please enter a name: ";
cin >> third_name;
cout << third_name << endl;
return 0;
}
You should probably be using string in your code to take the input of names. In names you are probably passing more than one character. The first one is read by first_name and any further character will be read by the following character, specifically cin>>second_name and cin>>third_name would read the 2nd and 3rd character of your input.
char a;
char b;
cin>>a; //will only read one character and stop
cin>>b; //will read the second character of the input...
//be that after pressing enter(a Enter b) or continuous input (ab)
cout<<a<<" "<<b; //will output 1st and 2nd character only
This will happen even if you don't press the Enter key explicitly and this is why your program uses the answer of first question(which is probably more than 1 character since it is a name) in your code as the answer to 2nd and 3rd questions as well.
So for your purpose, you are better of using string to take input from the users.
Hope this clears your doubt !
You tried to hold a lot of chars (one word) in one char who can hold only one char.
#include <iostream>
#include <string> // We need a string, container to hold a chars. Something like array of chars but have a few difference.
using namespace std; // You should avoid using this but in that short code this doesn't matter
int main()
{
// You don't need separate container for this code
// Then we create one container to holds all of inputs
string input;
cout << "Please enter a name: ";
cin >> input; // Put input from user in our input(string)
cout << input << endl; // Print input
// Next code is the same as above
cout << "Please enter a name: ";
cin >> input;
cout << input << endl;
cout << "Please enter a name: ";
cin >> input;
cout << input << endl;
return 0;
}
I special avoided a few elements like using function because this must be simple as possible.
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.
Something like this
cout << "Enter the number of columns: " ;
cin >> input ;
while( input != int ){
cout << endl <<"Column size must be an integer"<< endl << endl;
cout << "Enter the number of columns: " ;
cin >> input ;
}
cin will do this for you, kind of. cin will fail if it receives something that is not of the same type as input. What you can do is this:
int input;
while(!(cin >> input))
{
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << endl <<"Column size must be an integer"<< endl << endl;
cout << "Enter the number of columns: " ;
}
The cin.clear() clears the error bits, and cin.ignore() clears the input stream. I'm using number_limits to get the maximum size of the stream, that requires that you #include<limits>. Alternatively you can just use a big number or a loop.
You can't do it like that; input has to have some concrete type.
The simplest approach that will work is to read a string from cin, then convert it to an integer in a second step with strtol or one of its relatives, and issue an error message if strtol doesn't consume the whole string.
#include<iostream.h>
using namespace std;
int main()
{
int x;
int input;
while(!0){
cout<<"Enter your option :";
cout<<"1 .Enter Column size :"<<endl;
cout<<"2.Exit "<<endl;
cin>>x;
switch(x)
{
case 1: cout << "Enter the number of columns: "<<endl ;
cin>>input;
if(input>0)
cout << "The number of columns: is "<<input<<endl ;
else
cout << "Enter the number of columns as integer "<<endl ;
case 2:exit(0);
}
};
return 0;
}
Many of the answers here use the cin's built in filter. While these work to prevent a char or string from being entered, they do not prevent a float entry. When a float is entered, it is accepted and the decimal value remains in the buffer. This creates problems with later requests to cin. The following code will check the cin error flag and also prevent float inputs.
*note: The cin.ignore statement may require some tweaking to fully bullet proof the code.
void main()
{
int myint;
cout<<"Enter an integer: ";
intInput(myint);
}
void intInput(int &x)
{
bool valid = true; //flag used to exit loop
do
{
cin>>x;
//This 'if' looks for either of the following conditions:
//cin.fail() returned 'true' because a char was entered.
//or
//cin.get()!='\n' indicating a float was entered.
if(cin.fail() || cin.get()!='\n')
{
cout<<"Error. The value you entered was not an integer."<<endl;
cout<<"Please enter an integer: ";
cin.clear(); //clears cin.fail flag
cin.ignore(256,'\n'); //clears cin buffer
valid = false; //sets flag to repeat loop
}
else valid = true; //sets flag to exit loop
}while(valid == false);
}
This is a very basic solution to your problem that newer programers should find useful for people trying to break their programs. Eventually there are more advanced and efficient ways to do this.
int input;
int count = 1;
while(count == 1){ //this is just a simple looping design
cin >> input;
if(cin.fail()){ //If the input is about to crash your precious program
cin.clear(); //Removes the error message from internal 'fail safe'
cin.ignore(std::numeric_limits<int>::max(), '\n'); //Removes the bad values creating the error in the first place
count = 1; //If there is an error then it refreshes the input function
}
else{
count--; //If there is no error, then your program can continue as normal
}
}
Here is the advanced code: stackoverflow.com/questions/2256527/