Reading Different Input Types C++

Question

I am attempting to create a student object which takes the name, id, email, and three grades which are integers.

My code is quite simple and is as follows:

studentObj* newStudent = new studentObj;

cout << "Student First Name: ";
getline(cin, newStudent->name);
cout << "Student ID: ";
getline(cin, newStudent->id);
cout << "Student Email: ";
getline(cin, newStudent->email);
cout << "Grade 1: ";
cin >> newStudent->gradeOne;
cout << "Grade 2: ";
cin >> newStudent->gradeTwo;
cout << "Term Grade: ";
cin >> newStudent->termGrade;

cout << "Student Name: " + newStudent->name << endl;
cout << "Student ID: " + newStudent->id << endl;
cout << "Student Email: " + newStudent->email << endl;
cout << "Grade 1: " + newStudent->gradeOne << endl;
cout << "Grade 2: " + newStudent->gradeTwo << endl;

I assumed this would run flawlessly, but unfortunately that is not the case. It seems to be an issue mixing the getline() and cin.

The output is:

Student Name: Test Tester
Student ID: abcdef
Student Email: email@test.com
rade 1:
ade 2:
m Grade:

I've tried adding cin.ignore(numeric_limits<streamsize>::max(),'\n'); in a few places, but no luck. Any suggestions? `

Psychic debugging session. Your input was: *Grade 1: `1`*, *Grade 2: `2`*, *Term Grade: `3`*. — IInspectable, Sep 12 '13 at 22:10
Why do you *need* to allocate the `newStudent` from dynamic memory? C++ is not Java or C#. Try something like `studentObj newStudent;`. — Thomas Matthews, Sep 12 '13 at 23:49
@ThomasMatthews Because I have a vector, `vector students`, and when I do `students.push_back(newStudent)` it throws an error, so this is how I was able to get around it. — Barry Tormey, Sep 13 '13 at 02:39
Try `std::vector` **without** the '*'. The vector will push a copy of what you give it. — Thomas Matthews, Sep 13 '13 at 15:41

Nemanja Boric · Accepted Answer · 2013-09-12T22:08:55.447

You can't add string literals to the integers (well, you can, but in your case you will not get anything meaningful - you will make an offset - so the string output will be rade 1: because "Grade 1" + 1 will point to the string literal rade 1).

cout << "Student Name: " << newStudent->name << endl;
cout << "Student ID: " << newStudent->id << endl;
cout << "Student Email: " << newStudent->email << endl;
cout << "Grade 1: " << newStudent->gradeOne << endl;
cout << "Grade 2: " << newStudent->gradeTwo << endl;

score -1 · Answer 2 · answered Sep 12 '13 at 22:05

Your studentObj::gradeOne and studentObj::gradeTwo members are integer values. The string literal Grade 1: has type char*. An expression of type pointer + integer is the same as pointer[integer]. When you try to print the grade you are really producing a pointer into the string literal (if you are lucky).

Reading Different Input Types C++

2 Answers2