When a float variable goes out of the float limits, what happens?

Question

I remarked two things:

1. std::numeric_limits<float>::max()+(a small number) gives: std::numeric_limits<float>::max().

2. std::numeric_limits<float>::max()+(a large number like: std::numeric_limits<float>::max()/3) gives inf.

Why this difference? Does 1 or 2 results in an OVERFLOW and thus to an undefined behavior?

Edit: Code for testing this:

1.

float d = std::numeric_limits<float>::max();
float q = d + 100;
cout << "q: " << q << endl;

2.

float d = std::numeric_limits<float>::max();
float q = d + (d/3);
cout << "q: " << q << endl;

Answer 1

Formally, the behavior is undefined. On a machine with IEEE floating point, however, overflow after rounding will result in Inf. The precision is limited, however, and the results after rounding of FLT_MAX + 1 are FLT_MAX.

You can see the same effect with values well under FLT_MAX. Try something like:

float f1 = 1e20;     // less than FLT_MAX
float f2 = f1 + 1.0;
if ( f1 == f2 ) ...

The if will evaluate to true, at least with IEEE arithmetic. (There do exist, or at least have existed, machines where float has enough precision for the if to evaluate to false, but they aren't very common today.)

Answer 2

It depends on what you are doing. If the float "overflow" comes in an expression which is directly returned, i.e.

return std::numeric_limits::max() + std::numeric_limits::max();

the operation might not result in an overflow. I cite from the C standard [ISO/IEC 9899:2011]:

The return statement is not an assignment. The overlap restriction of subclause 6.5.16.1 does not apply to the case of function return. The representation of floating-point values may have wider range or precision than implied by the type; a cast may be used to remove this extra range and precision.

See here for more details.