CODEWITHSUNDEEP
c++
Jose Miranda
Jose Miranda

Reputation: 97

String getline(cin,variable) function is skipping some lines of code

My program skips some code when I use the getline(cin,variablehere) function. I don't know whats wrong with the code. See the output below

#include <iostream>
#include <string>
using namespace std;

int main()
{
    string getfirstname;
    string lastname;
    string address;
    int contactnumber;
    cout << "Enter First name : ";
    getline(cin, getfirstname);
    cin.ignore();
    cout << "Enter Last name : ";
    getline(cin, lastname);
    cin.ignore();
    cout << "Enter Address : ";
    getline(cin, address);
    cin.ignore();
    cout << "Enter Contact number : ";
    cin >> contactnumber;
    cin.ignore();

    CurrentNumberOfContacts += 1;
    cout << "Successfully added to contact list!" << endl << endl;

    cout << "Would you like to add another contact ? [Y/N] ";
    cin >> response;

    //more lines of codes below
    return 0;
}

I have inputed 'int' as data type because it will contain numbers only

enter image description here

Upvotes: 0

Views: 3937

Answers (4)

user5946447
user5946447

Reputation: 11

The answer provided by @Galic is quite good but If you want to read a line of characters without discarding the leading spaces you need another solution.

You could do:

char a='\n';
while (a=='\n')
{
    cin.get(a);
}
cin.unget();

before doing your first getline. This assumes no trailing space resulting from a previous cin and that your first input line is not empty.

Upvotes: 1

Galik
Galik

Reputation: 48635

I recommend removing all the cin.ignore() commands.

One of the problems with user input is that the >> operator does not take the RETURN character out of the stream so if you follow it with a getline() the getline() will read the RETURN character instead of what you want to type in.

So I would change all your getline() to this:

// cin >> ws will skip any RETURN characters
// that may be left in the stream
getline(cin >> ws, lastname);

Also remove all of your cin.ignore() commands. They are not doing anything useful when used after a getline() command and if you change your getline() commands as I showed they should not be necessary at all.

So this should work:

int main()
{
    string getfirstname;
    string lastname;
    string address;
    char response;
    int contactnumber;
    int CurrentNumberOfContacts = 0;

    cout << "Enter First name : ";
    getline(cin >> ws, getfirstname);

    cout << "Enter Last name : ";
    getline(cin >> ws, lastname);

    cout << "Enter Address : ";
    getline(cin >> ws, address);

    cout << "Enter Contact number : ";
    cin >> contactnumber;

    CurrentNumberOfContacts += 1;
    cout << "Successfully added to contact list!" << endl << endl;

    cout << "Would you like to add another contact ? [Y/N] ";
    cin >> response;

    //more lines of codes below
    return 0;
}

Strictly speaking not all of your getline() functions need to employ the cin >> ws trick. I suppose the (incomplete) rules are as follows:

If you use a std::getline() after a >> then use:

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

Otherwise just use:

std::getline(cin, line);

Upvotes: 4

DavidN.
DavidN.

Reputation: 125

2 things. First, you don't really need cin.ignore() in this case as your using 

getline().

before 

cin >> variable

Second, I don't know why your program doesn't run, but I would suggest using a 

getline()

call and see if that works. But I see no reason why your code is not working.

Upvotes: 1

Dwayne Towell
Dwayne Towell

Reputation: 8603

cin >> and getline do not cooperate very well. They have different strategies for how to deal with whitespace. getline removes the newline character, but cin >> leaves it. This means that after you use cin >> to read something, there will be a newline character left waiting in the input stream for the next getline to "use". Which means it will read an empty line into the string.

Upvotes: 1

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