为什么Swift中的startIndex不等于endIndex移到String中的startIndex位置? [英] Why is startIndex not equal to endIndex shifted to startIndex position in String in Swift?

问题描述

查看程序:

let s = "1"
print(s.startIndex)
print(s.index(before: s.endIndex))
print(s.index(before: s.endIndex) == s.startIndex)

它返回:

Index(_rawBits: 0)
Index(_rawBits: 256)
true

因此，用rawBits 0和256表示字符串中的相同位置.为什么?

解决方案

索引的原始位是实现细节.如您在示例中所见，两个值相等(对于==，它们返回true).

对于当前的实现方式，设置了位8，该位不属于该位置. 这是缓存到的偏移量值告诉您下一个字素有一个字节(直到您计算endIndex才知道).

Look at the program:

let s = "1"
print(s.startIndex)
print(s.index(before: s.endIndex))
print(s.index(before: s.endIndex) == s.startIndex)

It returns:

Index(_rawBits: 0)
Index(_rawBits: 256)
true

So, the same position in the string is represented with rawBits 0 and 256. Why?

解决方案

The raw bits of the index are an implementation detail. As you see in your example, the two values are equal (they return true for ==).

As to the current implementation, bit 8 is set, which is not part of the position. That's a cached value for the offset to the next grapheme cluster, which is 1 byte away. It's telling you that there's one byte to the next grapheme (which it didn't know until you calculated the endIndex).

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