How to use isnan as a predicate function to std::find_if (c++11)

Question

I have a piece of code that is operating on a std::vector<double> cats and is doing something like this:

std::find_if(cats.begin(), cats.end(), std::isnan<double>);

this compiles under gcc before 4.3 but no longer compiles with gcc 4.7.2. I get an error like this:

error: no matching function for call to 'find_if(std::vector<double>::iterator, std::vector<double::iterator, <unresolved overloaded function type>)'

How can I get this to compile under newer gcc? And preferably also under the older gcc? Without getting rid of stl and manually writing the loop of course. If it's not possible to do both, the new gcc would suffice and I would leave both implementations in there with a #define.

`std::isnan`?? In the C++11 standard `isnan` is a non-template function. In the C99 it is a macro. — David Rodríguez - dribeas, Jul 10 '13 at 15:14
It used to be a template in earlier versions of GCC. In GCC 4.7 there are three overloads _and_ a template for integral types — Jonathan Wakely, Jul 10 '13 at 15:23

Jonathan Wakely · Accepted Answer · 2013-07-10T15:17:30.063

8

The overload you actually want is not a template in GCC 4.7, it is just a normal function with this signature:

bool isnan(double);

(There is a template, but SFINAE means it only applies to integral types.)

But in previous versions it was a template, so there isn't an easy portable way to take its address.

Since you've said C++11, you can use a lambda and let the body of the lambda do overload resolution to find the right overload (or template specialization):

std::find_if( cats.begin(), cats.end(), [](double d) { return std::isnan(d); } );

Otherwise for C++03 you could provide your own forwarding wrapper:

struct IsNan {
  double operator()(double d) const { return std::isnan(d); }
};
std::find_if( cats.begin(), cats.end(), IsNan() );

edited Jul 10 '13 at 15:17

answered Jul 10 '13 at 15:09

Jonathan Wakely

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very nice! I wonder if there's a line of code that would compile for both though? – Palace Chan Jul 10 '13 at 15:10
Interesting, under gcc 4.7.2 that gives "error: overloaded function with no contextual type information" – Palace Chan Jul 10 '13 at 15:13
1

The cast would also work for a template (assuming that the signature was correct: `bool (*)(double)`). – David Rodríguez - dribeas Jul 10 '13 at 15:17
@DavidRodríguez-dribeas, doh, no wonder my test using a cast didn't work! Unfortunately in GCC 4.1 it returns `int` so the cast is still not really portable – Jonathan Wakely Jul 10 '13 at 15:20
Yea the cast doesn't seem portable because of the int bool thing. So i'll require two different lines of code depending on the compiler if I don't want to litter the code with a random forwarding wrapper. – Palace Chan Jul 10 '13 at 15:26
The `IsNan` functor seems to be the only portable way of doing this +1 – David Rodríguez - dribeas Jul 10 '13 at 15:28

score 5 · Answer 2 · answered Jul 10 '13 at 15:16

5

In C++11, isnan is a set of three overloads for the different floating point types float, double, long double. You can resolve the overload you want by using the cast syntax:

std::find_if(cats.begin(), cats.end(),
             static_cast<bool (*)(double)>(std::isnan));

Of course, the use of a lambda as suggested by Jonathan might be clearer for maintainers, if your target platforms all support lambdas.

answered Jul 10 '13 at 15:16

David Rodríguez - dribeas

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This cast works for gcc 4.7.2 but it gives "error: invalid static_cast from type '' to type 'bool (*)(double)'" under 4.1.2 – Palace Chan Jul 10 '13 at 15:22
That was my first attempt, but I stupidly got the return type wrong, and in GCC 4.1's `` it returned `int` – Jonathan Wakely Jul 10 '13 at 15:22
@JonathanWakely: Ouch... so it has the C99 signature. So not really portable either. – David Rodríguez - dribeas Jul 10 '13 at 15:27

How to use isnan as a predicate function to std::find_if (c++11)

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