I'm not sure how the example could make the assertion fail; I think that the point is that if 3.3 is treated as a long double by the compiler, then the result of x*3.3 will be a long double, which is more precise than the double returned by foo; but, regardless of how bar1 and bar2 are compiled, they both make a call to foo() which returns a double. Is it the case then that the behavior depends on how foo is compiled? I don't see how else the 80-bit result could propagate to the test for equality.