为什么在Scala中使用参数调用capitalize时会调用charAt

Question

scala> "Foo".capitalize(2)    
res18: Char = o

Answer 1

从StringLike trait doc复制:

def capitalize: String
Returns this string with first character converted to upper case

你实际在做的是

"foo" 的.capitalize.apply(2)

文档再次:

def apply(n: Int): Char
Return element at index n