How is floating point overflow handled in iostreams

Question

I have some very simple code:

#include <iostream>
#include <sstream>
using namespace std;

int main()
{
  stringstream is("1.0 2.0 1e-500 1e500 12.0");
  double d = {17.0, 17.0, 17.0, 17.0, 17.0};

  for (int i=0; i < 5; ++i)
  {
    if (is >> d[i])
    {
      cout<<"Conversion succeeded"<<endl;
    }
    else
    {
      cout<<"Conversion failed"<<endl;
      is.clear();
    }
  }
  for (int i=0; i < 5; ++i) cout<<d[i]<<endl;
}

When I compile this code with g++ 4.1.2 and run it on Redhat 5.10 (same compiler), I get the output:

Conversion succeeded
Conversion succeeded
Conversion failed
Conversion failed
Conversion succeeded
1
2
0
17
17
12

When I execute the same binary on Redhat Linux 6.5 (compiler 4.4.7), I get

Conversion succeeded
Conversion succeeded
Conversion succeeded
Conversion failed
Conversion succeeded
1
2
0
1.79769e+308
12

What is the expected behavior? Underflow is succeeeding on 4.4.7 but failing on 4.1.2. Overflow fails (but still changes the value) on 4.4.7 and fails without changing anything on 4.1.2.

Is the behavior undefined or simply incorrect on one or the other?

Answer 1

According to C++11 22.4.2.1.2, the conversion should fail for overflow, but not underflow. In the case of overflow, it should still give a value of the largest representable value as well as setting failbit.

So your more recent compiler has the correct modern behaviour.

However, both of your ancient compilers predate C++11 by many years. In earlier standards, the conversion was specified to give an error if scanf would; and not to give a value in case of error. Turning to the C standard, scanf defers to strtod, which in turn specifies an error on overflow; but whether there's an error on underflow is implementation-defined.

So your older compiler is consistent with historical behaviour.