在C / C ++中从double隐式转换为int64_t是什么

Question

Could someone shed a light on the way this code behaves:

double x = 9223371036854;
int64_t y1 = /* trunc */ (x * 1000000);
int64_t y2 = round(x * 1000000);
cout << y1 << " <= " << y2 << endl;
assert( y1 <= y2 ); // fail

This code fails due to y1 actually equals to 9223371036854000000 while y2 is 9223371036853999616 .
After uncommenting trunc everything is ok (assertion verified).

Compiled by gcc-4.6.3-1ubuntu5 with g++ --std=c++0x x.cpp .

Why int64_t(round(x * 1000000)) is less than int64_t(x * 1000000) where x is double ?

And why results of int64_t(trunc(x * 1000000)) is different from int64_t(x * 1000000) ?

Answer 1

I guess I found why it works like this ( y1 == 92233710368540000000 and assert fails).

GCC optimized out run-time calculation of y1 without loosing precision which clearly happens in y2 . More intersting is that expression of literals have no such property

double x = 9223371036854;
int64_t y1 = /* trunc */ (x * 1000000);
int64_t y2 = round(x * 1000000);
cout << y1 << " <= " << y2 << endl;
assert( int64_t(9223371036854.0 * 1000000) <= int64_t(round(9223371036854e6)) ); // ok
assert( int64_t(x * 1000000) <= int64_t(round(x * 1000000)) ); // fails
assert( y1 <= y2 ); // fails

And if I'll move out 1e6 outside of expressions everything works as expected:

double x = 9223371036854.0 * 1000000;
int64_t y1 = /* trunc */ (x);
int64_t y2 = round(x);
cout << y1 << " <= " << y2 << endl;
assert( int64_t(x) <= int64_t(round(x)) ); // ok
assert( y1 <= y2 ); // ok

But probably my assumption about optimization is incorrect.