我一直在为Swift写一些教程。我遇到了一个TicTacToe教程,该教程试图使用Xcode 6 Beta 6进行编码。当我检查字典中的值时,出现以下错误:找不到'&&'的重载,该重载接受了提供的参数。这是我的代码。
var plays = [Int:Int]()
var whoWon = ["I":0,"you":1]
for (key,value) in whoWon {
if ((plays[6] == value && plays[7] == value && plays[8] == value) ||
(plays[3] == value && plays[4] == value && plays[5] == value) ||
(plays[0] == value && plays[1] == value && plays[2] == value) ||
(plays[6] == value && plays[3] == value && plays[0] == value) ||
(plays[7] == value && plays[4] == value && plays[1] == value) ||
(plays[8] == value && plays[5] == value && plays[2] == value) ||
(plays[6] == value && plays[4] == value && plays[2] == value) || // error appears on this line
(plays[8] == value && plays[4] == value && plays[0] == value))
{
userMessage.hidden = false
userMessage.text = "Looks like \(key) won!"
}
最佳答案
如果您在报告浏览器中查看完整的编译器输出,则将看到
信息
注意:表达过于复杂,无法在合理的时间内解决;
考虑将表达式分解为不同的子表达式
告诉您如何解决问题。
关于swift - Swift错误:找不到接受提供的参数的'&&'重载,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/25654049/