objective-c 如何创建 NSIndexPath:indexPathWithIndexes:length 所需的“索引”：
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How to create the "indexes" required for NSIndexPath:indexPathWithIndexes:length:
提问by Ronnie Liew
The class method to create an index path with one or more nodes is:
创建具有一个或多个节点的索引路径的类方法是：
+ (id)indexPathWithIndexes:(NSUInteger *)indexes length:(NSUInteger)length
How do we create the "indexes" required in the first parameter?
我们如何创建第一个参数中所需的“索引”？
The documentation listed it as Array of indexes to make up the index pathbut it is expecting a (NSUinteger *).
该文档将其列为索引数组以构成索引路径，但它需要一个 (NSUinteger *)。
To create an index path of 1.2.3.4, is it simply an array of [1,2,3,4] ?
要创建 1.2.3.4 的索引路径，它是否只是一个 [1,2,3,4] 数组？
回答by Barry Wark
You are correct. You might use it like this:
你是对的。你可以这样使用它：
NSUInteger indexArr[] = {1,2,3,4};
NSIndexPath *indexPath = [NSIndexPath indexPathWithIndexes:indexArr length:4];
回答by Barry Wark
On iOS, you can also use this method from NSIndexPath UIKit Additions(declared in UITableView.h):
在 iOS 上，你也可以使用NSIndexPath UIKit Additions 中的这个方法（在 UITableView.h 中声明）：
+ (NSIndexPath*) indexPathForRow:(NSUInteger)row inSection:(NSUInteger)section
回答by Giao
You assumption is correct. It's as simple as a C array of NSUInteger. The length parameter is the number of elements in the indexes array.
你的假设是正确的。它就像 NSUInteger 的 C 数组一样简单。长度参数是索引数组中的元素数。
Arrays in C are often identified as a pointer (in this case NSUInteger *) with a length parameter or a known terminator such as \0 for C strings (which is just a char array).
C 中的数组通常被标识为带有长度参数或已知终止符的指针（在本例中为 NSUInteger *），例如 C 字符串中的 \0（它只是一个字符数组）。
回答by jungledev
I did this in 2 lines of code
我用两行代码做到了这一点
NSMutableArray *indexPaths = [[NSMutableArray alloc] init];
for (int i = firstIndexYouWant; i < totalIndexPathsYouWant; i++) [indexPaths addObject:[NSIndexPath indexPathForRow:i inSection:0]];
Short, clean, and readable. Free code, don't knock it.
简短，干净，可读。免费代码，不要敲它。
