当我使用扩展代码时,我得到了以下错误,我不确定他们是要求使用不同的运算符,还是根据互联网搜索修改表达式中的值。
错误:%不可用:请改用truncatingRemainder
扩展代码:
extension CMTime {
var durationText:String {
let totalSeconds = CMTimeGetSeconds(self)
let hours:Int = Int(totalSeconds / 3600)
let minutes:Int = Int(totalSeconds % 3600 / 60)
let seconds:Int = Int(totalSeconds % 60)
if hours > 0 {
return String(format: "%i:%02i:%02i", hours, minutes, seconds)
} else {
return String(format: "%02i:%02i", minutes, seconds)
}
}
}
设置分钟和秒变量时出现错误。
发布于 2016-11-09 04:00:14
CMTimeGetSeconds()
返回一个浮点数(Float64
,也称为Double
)。在Swift 2中,可以将浮点除法的余数计算为
let rem = 2.5 % 1.1
print(rem) // 0.3
在Swift 3中,这是通过
let rem = 2.5.truncatingRemainder(dividingBy: 1.1)
print(rem) // 0.3
应用于您的代码:
let totalSeconds = CMTimeGetSeconds(self)
let hours = Int(totalSeconds / 3600)
let minutes = Int((totalSeconds.truncatingRemainder(dividingBy: 3600)) / 60)
let seconds = Int(totalSeconds.truncatingRemainder(dividingBy: 60))
但是,在这种特殊情况下,首先将持续时间转换为整数会更容易:
let totalSeconds = Int(CMTimeGetSeconds(self)) // Truncate to integer
// Or:
let totalSeconds = lrint(CMTimeGetSeconds(self)) // Round to nearest integer
然后,下面的行将简化为
let hours = totalSeconds / 3600
let minutes = (totalSeconds % 3600) / 60
let seconds = totalSeconds % 60
发布于 2016-11-09 04:11:57
%
模运算符仅为整数类型定义。对于浮点类型,您需要更具体地说明您想要的IEEE754除法/余数行为的类型,因此您必须调用一个方法:remainder
或truncatingRemainder
。(如果您正在进行浮点数学运算,那么您实际上需要关注这一点和lots of other stuff,否则您可能会得到意想不到的/糟糕的结果。)
如果您真的想做整数取模,在使用%
之前,您需要将CMTimeGetSeconds
的返回值转换为整数。(请注意,如果您这样做,您将删除小数秒...这可能很重要,这取决于您在哪里使用CMTime
。例如,您想要分钟:秒:帧吗?)
根据您希望在UI中显示秒值的方式,提取CMTime
值并将其传递给NSDateFormatter
或NSDateComponentsFormatter
可能更好,这样您就可以获得适当的区域设置支持。
发布于 2017-01-10 23:49:19
在swift 3中恢复了简单的模语法:
这种语法实际上是在苹果官方的快速邮件列表here上建议的,但出于某种原因,他们选择了一种不那么优雅的语法。
infix operator %%/*<--infix operator is required for custom infix char combos*/
/**
* Brings back simple modulo syntax (was removed in swift 3)
* Calculates the remainder of expression1 divided by expression2
* The sign of the modulo result matches the sign of the dividend (the first number). For example, -4 % 3 and -4 % -3 both evaluate to -1
* EXAMPLE:
* print(12 %% 5) // 2
* print(4.3 %% 2.1) // 0.0999999999999996
* print(4 %% 4) // 0
* NOTE: The first print returns 2, rather than 12/5 or 2.4, because the modulo (%) operator returns only the remainder. The second trace returns 0.0999999999999996 instead of the expected 0.1 because of the limitations of floating-point accuracy in binary computing.
* NOTE: Int's can still use single %
* NOTE: there is also .remainder which supports returning negatives as oppose to truncatingRemainder (aka the old %) which returns only positive.
*/
public func %% (left:CGFloat, right:CGFloat) -> CGFloat {
return left.truncatingRemainder(dividingBy: right)
}
这篇简单的swift 3迁移技巧是更全面的swift 3迁移指南的一部分,其中包含许多见解(35k loc /8天迁移) http://eon.codes/blog/2017/01/12/swift-3-migration/
https://stackoverflow.com/questions/40495301
复制相似问题