Question

Doesn’t C++ offer any guarantee about keeping variables intact on input failure? With older versions of gcc, a program like this one keeps the -1 value of i on failure (for instance if a letter is typed instead of a number on input). With Ubuntu 10.10 (gcc 4.4.5), i is reset to zero in case of input failure.

#include <iostream>

int main()
{
 int i = -1;
 std::cin >> i;
 std::cout << "i = " << i << '\n';
 return 0;
}

This behavior breaks a lot of my code. I suppose the gcc people know what they are doing, and it is likely to be my mistake. If anyone knows the standard, I’d like to know what it says about this situation.

Thanks.

Answer 1

Don’t rely on the variable. Rely on the state of the stream:

if (std::cin >> i) // "if (!std::cin.fail())" would also work
{
    // ok
}
else
{
    // error
}

As for why the behavior has changed, that’s because the C++ standard has evolved:

From C++03:

If an error occurs, val is unchanged;
otherwise it is set to the
resulting value.

From C++0x (well.. from the last draft I have access to):

The numeric value to be stored can be
one of:

• zero, if the conversion function fails to convert the entire field.
• the most positive (or negative) representable value, if the field to
  be converted to a signed integer type
  represents a value too large positive
  (or negative) to be represented in
  val.
• the most positive representable value, if the field to be converted to
  an unsigned integer type represents a
  value that cannot be represented in
  val.
• the converted value, otherwise.