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Apr 22nd, 2008

cin.fail() - why it fails after cin.get(...)

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I'm self studying C++ and I'm actually back-tracking a bit in order to clarify things. My question revolves around why in the following code is the cin.fail() == true .

cpp Syntax (Toggle Plain Text)
#include <iostream>
using namespace std;
int main()
{
    char getdata;
    cout << "Enter one character: ";
    cin >> getdata;
    cin.get();
 
    char line[20];
    cout << "\nEnter a line of text 20 characters max: ";
    cin.get(line,20);
    cout << "\nLine of text entered : " << line;
    //cin.get();
 
    char line2[20];
    cout << "\nEnter another line of text 20 characters max: ";
    cin.get(line2,20);
    cout << "\nSecond line of text entered : " << line2;
 
    cout << "\n\nNow at the end of the program";
    cout << "\nFirst character of line[20] " << (int)line[0] << endl;
    cout << "First character of line2[20] : " << (int)line2[0] << endl;
 
    if (cin.eof())
        cout << "\ncin.eof() = true";
    else
        cout << "\ncin.eof() = false";
    if (cin.fail())
        cout << "\ncin.fail() = true";
    else
        cout << "\ncin.fail() = false";
 
    // exit routine
    cin.clear();
    while (cin.get() != '\n')
        continue;
    cin.get();
    return 0;
}
#include <iostream>
using namespace std;
int main()
{
    char getdata;
    cout << "Enter one character: ";
    cin >> getdata;
    cin.get();
    
    char line[20];
    cout << "\nEnter a line of text 20 characters max: ";
    cin.get(line,20);
    cout << "\nLine of text entered : " << line;
    //cin.get();
    
    char line2[20];
    cout << "\nEnter another line of text 20 characters max: ";
    cin.get(line2,20);
    cout << "\nSecond line of text entered : " << line2;

    cout << "\n\nNow at the end of the program";
    cout << "\nFirst character of line[20] " << (int)line[0] << endl;
    cout << "First character of line2[20] : " << (int)line2[0] << endl;
    
    if (cin.eof())
        cout << "\ncin.eof() = true";
    else
        cout << "\ncin.eof() = false";
    if (cin.fail())
        cout << "\ncin.fail() = true";
    else
        cout << "\ncin.fail() = false";
    
    // exit routine
    cin.clear();
    while (cin.get() != '\n')
        continue;
    cin.get();
    return 0;
}

The crux of the problem, intended for learning purposes, is that after the line cin.get(line,20); the newline character remains in the input queue. The very next cin.get(line2,20) , from a users point of view, is ignored or bypassed. The array line2 is in fact assigned the null character as indicated by the cout << "....line2[20]...." statement.

I'm curious or confused as to why the cin.fail() error has occurred since the array line2 has been assigned the null character.

As I've just written this, it's made me think a little. Is it the case that cin.get(array, size) leaves the newline character in the input queue and since the first character is a newline character, it's ignored and therefore nothing to assign to the array variable, hence an error state created. I'll post this question anyway.

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Apr 23rd, 2008

Re: cin.fail() - why it fails after cin.get(...)

try this slightly modified version of your program. In both cases you can only enter 19 characters, not 20, because the 20th byte is reserved for the string's null teriminator -- 0.

C++ Syntax (Toggle Plain Text)
int main()
{
    char getdata;
    cout << "Enter one character: ";
    cin >> getdata;
    cin.get();
 
    char line[20] = {0};
    cout << "\nEnter a line of text 20 characters max: ";
    cin.get(line,20);
    cout << "\nLine of text entered : " << line << "\n";
    cin.ignore();
 
    char line2[20] = {0};
    cout << "\nEnter another line of text 20 characters max: ";
    cin.get(line2,20);
    cin.ignore();
    cout << "\nSecond line of text entered : " << line2 << "\n";
 
    cout << "\n\nNow at the end of the program";
    cout << "\nFirst character of line[20] " << (int)line[0] << endl;
    cout << "First character of line2[20] : " << (int)line2[0] << endl;
 
    if (cin.eof())
        cout << "\ncin.eof() = true";
    else
        cout << "\ncin.eof() = false";
    if (cin.fail())
        cout << "\ncin.fail() = true";
    else
        cout << "\ncin.fail() = false";
 
    // exit routine
    cin.clear();
    while (cin.get() != '\n')
        continue;
    cin.get();
    return 0;
}
int main()
{
    char getdata;
    cout << "Enter one character: ";
    cin >> getdata;
    cin.get();
    
    char line[20] = {0};
    cout << "\nEnter a line of text 20 characters max: ";
    cin.get(line,20);
    cout << "\nLine of text entered : " << line << "\n";
    cin.ignore();
    
    char line2[20] = {0};
    cout << "\nEnter another line of text 20 characters max: ";
    cin.get(line2,20);
    cin.ignore();
    cout << "\nSecond line of text entered : " << line2 << "\n";

    cout << "\n\nNow at the end of the program";
    cout << "\nFirst character of line[20] " << (int)line[0] << endl;
    cout << "First character of line2[20] : " << (int)line2[0] << endl;
    
    if (cin.eof())
        cout << "\ncin.eof() = true";
    else
        cout << "\ncin.eof() = false";
    if (cin.fail())
        cout << "\ncin.fail() = true";
    else
        cout << "\ncin.fail() = false";
    
    // exit routine
    cin.clear();
    while (cin.get() != '\n')
        continue;
    cin.get();
    return 0;
}

Last edited by Ancient Dragon; Apr 23rd, 2008 at 12:42 am.

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Apr 23rd, 2008

Re: cin.fail() - why it fails after cin.get(...)

If you use getline() instead of get() you don't have to call ignore()

C++ Syntax (Toggle Plain Text)
int main()
{
    char getdata;
    cout << "Enter one character: ";
    cin >> getdata;
    cin.get();
 
    char line[20] = {0};
    cout << "\nEnter a line of text 20 characters max: ";
    cin.getline(line,20);
    cout << "\nLine of text entered : " << line << "\n";
    //cin.ignore();
 
    char line2[20] = {0};
    cout << "\nEnter another line of text 20 characters max: ";
    cin.getline(line2,20);
    //cin.ignore();
    cout << "\nSecond line of text entered : " << line2 << "\n";
 
    cout << "\n\nNow at the end of the program";
    cout << "\nFirst character of line[20] " << (int)line[0] << endl;
    cout << "First character of line2[20] : " << (int)line2[0] << endl;
 
    if (cin.eof())
        cout << "\ncin.eof() = true";
    else
        cout << "\ncin.eof() = false";
    if (cin.fail())
        cout << "\ncin.fail() = true";
    else
        cout << "\ncin.fail() = false";
 
    // exit routine
    cin.clear();
    while (cin.get() != '\n')
        continue;
    cin.get();
    return 0;
}
int main()
{
    char getdata;
    cout << "Enter one character: ";
    cin >> getdata;
    cin.get();
    
    char line[20] = {0};
    cout << "\nEnter a line of text 20 characters max: ";
    cin.getline(line,20);
    cout << "\nLine of text entered : " << line << "\n";
    //cin.ignore();
    
    char line2[20] = {0};
    cout << "\nEnter another line of text 20 characters max: ";
    cin.getline(line2,20);
    //cin.ignore();
    cout << "\nSecond line of text entered : " << line2 << "\n";

    cout << "\n\nNow at the end of the program";
    cout << "\nFirst character of line[20] " << (int)line[0] << endl;
    cout << "First character of line2[20] : " << (int)line2[0] << endl;
    
    if (cin.eof())
        cout << "\ncin.eof() = true";
    else
        cout << "\ncin.eof() = false";
    if (cin.fail())
        cout << "\ncin.fail() = true";
    else
        cout << "\ncin.fail() = false";
    
    // exit routine
    cin.clear();
    while (cin.get() != '\n')
        continue;
    cin.get();
    return 0;
}

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Apr 23rd, 2008

Re: cin.fail() - why it fails after cin.get(...)

Thanks AD. I'm in the process of coming to grips with all or most of the basic cin uses and I was more concerned with why an error state is recorded when there is a newline left in the input buffer and then an immediate cin.get(arrayname,20) is issued.

Anyway, I've never used the ignore method before, thanks for the clue there. In lieu of the ignore function I've been using cin.get() to clear out errant data.

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