How to try reading as int, then if it fails read as string?

Question

I'm trying to repeatedly read integers, and for tokens that are not integers, to read and handle them as strings. I thought the following would work:

string s;
int x;

while(1) {
    if (cin >> x)                                     // try reading as int
        cout << "read int: " << x << endl;
    else {                                            // try reading as string
        cin.clear();

        if (cin >> s) 
            cout << "read string: " << s << endl;
        else
            break;                                    // eof
    }
}

However, for input (with newline at end)

1 2 a

it gives output

read int: 1
read int: 2

It seems that after clearing the state of the stream is not exactly the same as before, and in fact cin >> s fails. Why is this, and how can I achieve my desired behaviour?

To clarify, I can't parse the string as an int because I have not "learned" that yet.

Is it true that you *can't* (meaning school project) or "don't want to"? Concerning parse that is. — ChiefTwoPencils, Jul 06 '13 at 01:50
@KiraChow: Either you're approaching the assignment wrong, or the assignment is stupid. Either way, the solution will depend on a more thorough understanding of the details of your course than is realistic to convey in a StackOverflow question. — user2357112, Jul 06 '13 at 01:53
Well then maybe you've learned ways of determining if a specific char is alpha and such? Then you could still read as string and determine if it's a number or not. Meaning your traverse the input string and deduce by it's contents if it could possibly be a number. — ChiefTwoPencils, Jul 06 '13 at 01:53
wait, I'm a little confused. Is your problem that it's not printing out `read string: a` at the end or something else? A quick [test](http://ideone.com/9xpYnz) shows all 3 outputs printed as expected. — greatwolf, Jul 06 '13 at 01:55
@greatwolf Yes, I've tested it on ideone and I know it works there. That's why I specified that there is a newline at the end and that it's from the file. Because in my environment, it doesn't work. — Kira Chow, Jul 06 '13 at 01:57
Are you saying you're piping the file content from stdout to your program's stdin because `cin` is being used here, not `fstream`. — greatwolf, Jul 06 '13 at 02:03
Can you add what platform environment this is failing under + compiler toolchain version used if possible. — greatwolf, Jul 06 '13 at 02:08
@greatwolf Thanks for trying so much to help, but I may go with one of the given answers (we did cover string streams). If you're interested, it's in Linux with g++ 4.4.5. — Kira Chow, Jul 06 '13 at 02:14

score 5 · Answer 1 · edited May 23 '17 at 11:50

5

clear does not rewind the stream; it only clears the error flags, as the name suggests. To solve your problem, read the input as a string, then try to parse it as an int. The link explains the details of the string-int conversion better than I could.

edited May 23 '17 at 11:50

Community

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answered Jul 06 '13 at 01:48

user2357112

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I should have mentioned that this probably not a possible solution. – Kira Chow Jul 06 '13 at 01:50
Kira, it's the only real solution. Once cin has been read, it cannot be unread. – Chelsea Urquhart Jul 06 '13 at 01:51

scaryrawr · Accepted Answer · 2013-07-06T02:20:17.703

Why not read it as a string, then try a conversion, so if the conversion fails you can treat it as a string.

In C++11 there is std::stoi. Which will throw an exception if a conversion cannot be found.

Another way would be read it in as a string, and then parse using a stringstream.

If you cannot use an existing string parsing method, why not write your own?

// Returns false if not an int, otherwise true and outint will hold the correct value.
bool strtoint(const std::string str, int *outint) {
    int val = 0;
    for (int x = str.length() - 1; x >= 0; x--) {
        if (!(str[x] <= '9' && str[x] >= '0') || !(str[x] == '-')) {
            return false;
        } else if (str[x] == '-') {
            val *= -1;
        } else {
            val += (int)((str[x] - '0') * pow(10, (str.length() - 1) - x));
        }
    }

    *outint = val;
    return true;
}

Or something kinda like that, I'm not sure if the logic is completely correct, but I think it's on the right track.

Thanks for the effort, I was hoping it would not come to this. :) I'll accept yours because you mentioned string streams, which we did cover (although did not learn of its particular use in parsing). — Kira Chow, Jul 06 '13 at 02:23
It probably shouldn't, but sometimes it's just a fun rewrite existing things. — scaryrawr, Jul 06 '13 at 02:25

How to try reading as int, then if it fails read as string?

2 Answers2