145

I have the following piece of code that prompts the user for their cat's age and name:

#include <iostream>
#include <string>

int main()
{
    int age;
    std::string name;

    std::cin >> age;
    std::getline(std::cin, name);
    
    if (std::cin)
    {
        std::cout << "My cat is " << age << " years old and their name is " << name << std::endl;
    }
}

What I find is that the age has been successfully read, but not the name. Here is the input and output:

Input:

"10"
"Mr. Whiskers"

Output:

"My cat is 10 years old and their name is "

Why has the name been omitted from the output? I've given the proper input, but the code somehow ignores it. Why does this happen?

Remy Lebeau
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David G
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    I believe `std::cin >> name && std::cin >> std::skipws && std::getline(std::cin, state)` should also work as expected. (In addition to the answers below). – jww Nov 11 '18 at 17:30

5 Answers5

176

Why does this happen?

This has little to do with the input you provided yourself but rather with the default behavior std::getline() has. When you provided your input for the age (std::cin >> age), you not only submitted the following characters, but also an implicit newline was appended to the stream when you typed Enter:

"10\n"

A newline is always appended to your input when you select Enter or Return when submitting from a terminal. It is also used in files for moving toward the next line. The newline is left in the buffer after the extraction into age until the next I/O operation where it is either discarded or read. When the flow of control reaches std::getline(), it will see "\nMr. Whiskers" and the newline at the beginning will be discarded, but the input operation will stop immediately. The reason this happens is because the job of std::getline() is to attempt to read characters and stop when it finds a newline. So the rest of your input is left in the buffer unread.

Solution

cin.ignore()

To fix this, one option is to skip over the newline before doing std::getline(). You can do this by calling std::cin.ignore() after the first input operation. It will discard the next character (the newline character) so that it is no longer in the way.

std::cin >> age;
std::cin.ignore();
std::getline(std::cin, name);

assert(std::cin); 
// Success!

std::ws

Another way to discard the whitespace is to use the std::ws function which is a manipulator designed to extract and discard leading whitespace from the beginning of an input stream:

std::cin >> age;
std::getline(std::cin >> std::ws, name);

assert(std::cin);
// Success!

The std::cin >> std::ws expression is executed before the std::getline() call (and after the std::cin >> age call) so that the newline character is removed.

The difference is that ignore() discards only 1 character (or N characters when given a parameter), and std::ws continues to ignore whitespace until it finds a non-whitespace character. So if you don't know how much whitespace will precede the next token you should consider using this.

Match the operations

When you run into an issue like this it's usually because you're combining formatted input operations with unformatted input operations. A formatted input operation is when you take input and format it for a certain type. That's what operator>>() is for. Unformatted input operations are anything other than that, like std::getline(), std::cin.read(), std::cin.get(), etc. Those functions don't care about the format of the input and only process raw text.

If you stick to using a single type of formatting then you can avoid this annoying issue:

// Unformatted I/O
std::string age, name;
std::getline(std::cin, age);
std::getline(std::cin, name);

or

// Formatted I/O
int age;
std::string firstName, lastName;
std::cin >> age >> firstName >> lastName;

If you choose to read everything as strings using the unformatted operations you can convert them into the appropriate types afterwards.

David G
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    Why not simply `if (getline(std::cin, name) && getline(std::cin, state))`? – Fred Larson Aug 19 '16 at 19:41
  • @FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string. – David G Aug 19 '16 at 20:30
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    Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then use `std::stoi()`, but then it's not so clear there's an advantage. But I tend to prefer to just use `std::getline()` for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone. – Fred Larson Aug 19 '16 at 20:35
  • @FredLarson Agreed. Maybe I'll add that in if I have the time. – David G Aug 19 '16 at 20:39
  • So what if I had a program that first inputs a number X and then runs a loop that inputs strings X number of times - how would I go about it? Should I write the first iteration code outside the loop with a cin.ignore() and then run the loop X-1 times? – Albin Apr 03 '20 at 17:23
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    @Albin The reason you might want to use `std::getline()` is if you want to capture all characters up to a given delimiter and input it into a string, by default that is the newline. If those `X` number of strings are just single words/tokens then this job can be easily accomplished with `>>`. Otherwise you would input the first number into an integer with `>>`, call `cin.ignore()` on the next line, and then run a loop where you use `getline()`. – David G Apr 03 '20 at 18:54
  • Unbelievable. So many upvotes. Nobody heard of ````std::ws````? Why not simply writing ````std::getline(std::cin>>std::ws, name);````? What is going on here? – A M Nov 19 '21 at 12:27
  • @armin That used to be in my answer before I made a major edit for simplicity. I'll put it back in. – David G Nov 19 '21 at 15:18
  • thnks you so much. – kNIG132103 Jan 12 '22 at 13:33
14

Everything will be OK if you change your initial code in the following way:

if ((cin >> name).get() && std::getline(cin, state))
Qantas 94 Heavy
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Boris
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    Thank you. This will also work because `get()` consumes the next character. There's also `(std::cin >> name).ignore()` which I suggested earlier in my answer. – David G Mar 26 '14 at 12:14
  • "..work because get()..." Yes, exactly. Sorry for giving the answer without details. – Boris Mar 26 '14 at 13:14
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    Why not simply `if (getline(std::cin, name) && getline(std::cin, state))`? – Fred Larson Aug 19 '16 at 19:41
4

This happens because an implicit line feed also known as newline character \n is appended to all user input from a terminal as it's telling the stream to start a new line. You can safely account for this by using std::getline when checking for multiple lines of user input. The default behavior of std::getline will read everything up to and including the newline character \n from the input stream object which is std::cin in this case.

#include <iostream>
#include <string>

int main()
{
    std::string name;
    std::string state;

    if (std::getline(std::cin, name) && std::getline(std::cin, state))
    {
        std::cout << "Your name is " << name << " and you live in " << state;
    }
    return 0;
}
Input:

"John"
"New Hampshire"

Output:

"Your name is John and you live in New Hampshire"
Justin Randall
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2

Since everyone above has answered the problem for input 10\nMr Whisker\n, I would like to answer a different approach. All the solution above published the code for if the buffer is like 10\nMr Whisker\n. But what if we don't know how user will behave in giving input. The user might type 10\n\nMr. Whisker\n or 10 \n\n Mr. whisker\n by mistake. In that case, codes above may not work. So, I use the function below to take string input to address the problem.

string StringInput()  //returns null-terminated string
{
    string input;
    getline(cin, input);
    while(input.length()==0)//keep taking input as long as valid string is taken
    {
        getline(cin, input);
    }
    return input.c_str();
}

So, the answer would be:

#include <iostream>
#include <string>

int main()
{
    int age;
    std::string name;

    std::cin >> age;
    name = StringInput();
    
    std::cout << "My cat is " << age << " years old and it's name is " << name << std::endl;
    
}

Extra:

If user inputs a \n10\n \nmr. whiskey; To check whether int input is valid or not, this function can be used to check int input (program will have undefined behavior if char is given as input instead of int):


//instead of "std::cin>>age;" use "get_untill_int(&age);" in main function.
void get_Untill_Int(int* pInput)//keep taking input until input is `int or float`
{
    cin>> *pInput;
    /*-----------check input validity----------------*/
    while (!cin) 
    {
        cin.clear();
        cin.ignore(100, '\n');
        cout<<"Invalid Input Type.\nEnter again: ";
        cin >>*pInput;
    }
    /*-----------checked input validity-------------*/
}
Miraz
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2

I am really wondering. C++ has a dedicated function for eating up any remaining or whatever white spaces. It is called std::ws. And then, you can simply use

std::getline(std::cin >> std::ws, name);

That should be the idomatic approach. For each transistion between formatted to unformatted input that should be used.

If we are not talking about white spaces, but entering for example letters where a number is expected, then we should follow the CPP reference and use

.ignore(std::numeric_limits<std::streamsize>::max(), '\n'); to eliminate the wrong stuff.

Please read here

A M
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