当浮点数变量超出浮动限制时，会发生什么？

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, ie

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.