C++:如何将 double 舍入为 int?
我有一个 double(称为 x)，本来是 55，但实际上存储为 54.999999999999943157，这是我刚刚意识到的.
I have a double (call it x), meant to be 55 but in actuality stored as 54.999999999999943157 which I just realised.
所以当我这样做时
double x = 54.999999999999943157;
int y = (int) x;
y = 54 而不是 55！
y = 54 instead of 55!
这让我困惑了很久.我怎样才能让它正确地四舍五入?
This puzzled me for a long time. How do I get it to correctly round?
推荐答案
在转换前加 0.5(如果 x > 0)或减去 0.5(如果 x <0)，因为编译器总是会截断.
add 0.5 before casting (if x > 0) or subtract 0.5 (if x < 0), because the compiler will always truncate.
float x = 55; // stored as 54.999999...
x = x + 0.5 - (x<0); // x is now 55.499999...
int y = (int)x; // truncated to 55
C++11 还引入了 std::round，它可能使用类似的逻辑，将 0.5 添加到 |x|在引擎盖下(如果有兴趣，请参阅链接)，但显然更强大.
C++11 also introduces std::round, which likely uses a similar logic of adding 0.5 to |x| under the hood (see the link if interested) but is obviously more robust.
后续问题可能是为什么浮点数没有准确存储为 55.有关说明，请参阅 this stackoverflow 答案.
A follow up question might be why the float isn't stored as exactly 55. For an explanation, see this stackoverflow answer.
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