Is it possible to read infinity or NaN values using input streams?

Question

I have some input to be read by a input filestream (for example):

-365.269511 -0.356123 -Inf 0.000000

When I use std::ifstream mystream;

Answer 1

Edit: To avoid the use of a wrapper structure around a double, I enclose an istream within a wrapper class instead.

Unfortunately, I am unable to figure out how to avoid the ambiguity created by adding another input method for double. For the implementation below, I created a wrapper structure around an istream, and the wrapper class implements the input method. The input method determines negativity, then tries to extract a double. If that fails, it starts a parse.

Edit: Thanks to sehe for getting me to check for error conditions better.

struct double_istream {
    std::istream ∈

    double_istream (std::istream &i) : in(i) {}

    double_istream & parse_on_fail (double &x, bool neg);

    double_istream & operator >> (double &x) {
        bool neg = false;
        char c;
        if (!in.good()) return *this;
        while (isspace(c = in.peek())) in.get();
        if (c == '-') { neg = true; }
        in >> x;
        if (! in.fail()) return *this;
        return parse_on_fail(x, neg);
    }
};

The parsing routine was a little trickier to implement than I first thought it would be, but I wanted to avoid trying to putback an entire string.

double_istream &
double_istream::parse_on_fail (double &x, bool neg) {
    const char *exp[] = { "", "inf", "NaN" };
    const char *e = exp[0];
    int l = 0;
    char inf[4];
    char *c = inf;
    if (neg) *c++ = '-';
    in.clear();
    if (!(in >> *c).good()) return *this;
    switch (*c) {
    case 'i': e = exp[l=1]; break;
    case 'N': e = exp[l=2]; break;
    }
    while (*c == *e) {
        if ((e-exp[l]) == 2) break;
        ++e; if (!(in >> *++c).good()) break;
    }
    if (in.good() && *c == *e) {
        switch (l) {
        case 1: x = std::numeric_limits::infinity(); break;
        case 2: x = std::numeric_limits::quiet_NaN(); break;
        }
        if (neg) x = -x;
        return *this;
    } else if (!in.good()) {
        if (!in.fail()) return *this;
        in.clear(); --c;
    }
    do { in.putback(*c); } while (c-- != inf);
    in.setstate(std::ios_base::failbit);
    return *this;
}

One difference in behavior this routine will have over the the default double input is that the - character is not consumed if the input was, for example "-inp". On failure, "-inp" will still be in the stream for double_istream, but for a regular istream only "inp" will be left in the the stream.

std::istringstream iss("1.0 -NaN inf -inf NaN 1.2");
double_istream in(iss);
double u, v, w, x, y, z;
in >> u >> v >> w >> x >> y >> z;
std::cout << u << " " << v << " " << w << " "
          << x << " " << y << " " << z << std::endl;

The output of the above snippet on my system is:

1 nan inf -inf nan 1.2

Edit: Adding a "iomanip" like helper class. A double_imanip object will act like a toggle when it appears more than once in the >> chain.

struct double_imanip {
    mutable std::istream *in;
    const double_imanip & operator >> (double &x) const {
        double_istream(*in) >> x;
        return *this;
    }
    std::istream & operator >> (const double_imanip &) const {
        return *in;
    }
};

const double_imanip &
operator >> (std::istream &in, const double_imanip &dm) {
    dm.in = ∈
    return dm;
}

And then the following code to try it out:

std::istringstream iss("1.0 -NaN inf -inf NaN 1.2 inf");
double u, v, w, x, y, z, fail_double;
std::string fail_string;
iss >> double_imanip()
    >> u >> v >> w >> x >> y >> z
    >> double_imanip()
    >> fail_double;
std::cout << u << " " << v << " " << w << " "
          << x << " " << y << " " << z << std::endl;
if (iss.fail()) {
    iss.clear();
    iss >> fail_string;
    std::cout << fail_string << std::endl;
} else {
    std::cout << "TEST FAILED" << std::endl;
}

The output of the above is:

1 nan inf -inf nan 1.2
inf

Edit from Drise: I made a few edits to accept variations such as Inf and nan that wasn't originally included. I also made it into a compiled demonstration, which can be viewed at http://ideone.com/qIFVo.