问题
If I have
(float)value = 10.50
and do
int new_value = (int)value
what rules will round number?
回答1:
When a finite value of floating type is converted to an integer type, the fractional part is discarded (i.e., the value is truncated toward zero).
So in the case of -10.5, it's converted to -10.
C++11 4.9 Floating-integral conversions [conv.fpint]
An rvalue of a floating point type can be converted to an rvalue of an integer type. The conversion truncates; that is, the fractional part is discarded. The behavior is undefined if the truncated value cannot be represented in the destination type. [ Note: If the destination type is bool, see 4.12. —end note ]
回答2:
The rule is quite simple: the number simply gets truncated to its integral part, in this case, to 10. The fractional part gets dropped entirely. The same applies to negative numbers: -10.5 would be converted to -10.
回答3:
When converted to integers, the fractional part of the float is dropped, meaning the float 10.5 will be converted to the integer 10, and the float -10.5 will be converted to the integer -10.
来源：https://stackoverflow.com/questions/19079290/how-does-float-round-when-converting-it-into-integer