Finding if a string is 'numeric only' using tr1 regex

Question

tIs it possible for me to detect if a string is 'all numeric' or not using tr1 regex? If yes, please help me with a snipped as well since I am new to regex.

Why I am looking towards tr1 regex for something like this, because I don't want to create a separate function for detecting if the string is numeric. I want to do it inline in rest of the client code but do not want it to look ugly as well. I feel maybe tr1 regex might help. Not sure, any advises on this?

What does "numeric only" mean? If you mean "all of the characters are digits," why do you think a regular expression would be more effective than `std::find_if_not(begin(s), end(s), (int(*)(int))std::isdigit) == end(s)` — James McNellis, Dec 12 '11 at 18:57
@JamesMcNellis I don't think regex would be better. I just didn't think it through to find an elegant enough solution. Your looks very elegant. Thanks. — bits, Dec 12 '11 at 19:07
@bits when you say "*numeric only*" do you mean a string containing only digits (0-9) or should it be a valid (decimal) number? — Filip Roséen - refp, Dec 12 '11 at 19:09
Please see: [What’s a Number?](http://stackoverflow.com/a/4247184/433790) — ridgerunner, Dec 12 '11 at 22:46

score 1 · Accepted Answer · answered Dec 12 '11 at 19:12

1

If you just want to test whether the string has all numeric characters, you can use std::find_if_not and std::isdigit:

std::find_if_not(s.begin(), s.end(), (int(*)(int))std::isdigit) == s.end()

If you do not have a Standard Library implementation with std::find_if_not, you can easily write it:

template <typename ForwardIt, typename Predicate>
ForwardIt find_if_not(ForwardIt first, ForwardIt last, Predicate pred)
{
    for (; first != last; ++first)
        if (!pred(first))
            return first;

    return first;
}

answered Dec 12 '11 at 19:12

James McNellis

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`std::find_if_not` is not mentioned in [c++ algorithm reference](http://www.cplusplus.com/reference/algorithm/). I am not sure why? I guess I will have to implement the template as you suggested. Thanks. – bits Dec 12 '11 at 19:25
@bits: It was added in C++11, and cplusplus.com is out of date. – James McNellis Dec 12 '11 at 19:27

score 1 · Answer 2 · answered Dec 12 '11 at 19:50

1

You can use the string::find_first_not_of member function to test for numeric characters.

if (mystring.find_first_not_of("0123456789") == std::string::npos)
{
    std::cout << "numeric only!";
}

answered Dec 12 '11 at 19:50

Blastfurnace

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score 0 · Answer 3 · answered Dec 12 '11 at 18:57

The regular expression for this is rather trivial. Just try to match "\\D". This will match on any character that's not a digit. If you'd like it to include a decimal separator too, you could use "[^\\d\\.]", which translates to "not a digit or dot".

However, how about simply using strtol() to read the number? You'll be able to retrieve a pointer to the first non-number character. So, if this points to the end of the string, it's been fine. Plus side here is, you won't even need TR1 for this.

Good point. But easy to do when reversing: The string is valid, if you get a match on `"^-?\\d+$"` or `"^-?\\d+(?:\\.\\d+)?$"` — Mario, Dec 12 '11 at 19:18

Finding if a string is 'numeric only' using tr1 regex

3 Answers3