|
|
|
|
|
|
by a1369209993
826 days ago
|
|
|
On my machine, the complier constant-folds the multiplication, producing a single-precision add for `mul_as_float` and a convert-t-to-double, double-precision-add, convert-sum-to-single for `mul_as_double`. I missed the `+=` in your original comment, but adding a float to a double does implicitly promote it like that, so you'd actually need: t += (float)((double)0.02f * (float)17);
to achieve the "and then converting the result [of the multiplication] to float" (rather than keeping it a double for the addition) from your original comment. (With the above line in mul_as_double, your test code no longer finds a counterexample, at least when I ran it.)If you ask for higher-precision intermediates, even implicitly, floating-point compliers will typically give them to you, hoped-for efficiency of single-precision be damned. |
|
|