NSRange output of location and length

Question

An issue I have (maybe it is really easy to solve, but I am new to Obj-C. I want to print out the location and length of NSRange I created, but I get an error message. Here the code

NSRange tmpRange = [newData rangeOfData:segmentToFind options:kNilOptions range:NSMakeRange(0u, [newData length])];

NSLog(@"%@ - range location : %lu", [tmpRange location];

The error message I get is:

Bad receiver type 'NSRange' (aka 'struct _NSRange')

You have 2 placeholders, and only one argument in the `NSLog`. — Larme, Sep 10 '14 at 15:11

score 4 · Answer 1 · answered Sep 10 '14 at 15:15

4

To solutions:

A. You have to log each component separately:

NSLog( @"%lu - %lu", (long)range.location, (long)range.length );

B. You can convert it to an instance of NSString and then log the string:

NSLog( @"%@", NSStringFromRange(range) );

Output

{1, 2}

NSStringFrom…() is similar to -description and exists for many types including NSSize, NSPoint.

answered Sep 10 '14 at 15:15

Amin Negm-Awad

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+1 for mentioning the NSStringFrom... functions. – gnasher729 Sep 10 '14 at 15:35

score 1 · Accepted Answer · answered Sep 10 '14 at 15:36

NSRange starts with the letters NS, but it isn't an NSObject. It is an ordinary struct, like a struct in C or in C++. So to access location and length, you just use range.location and range.length. There are other structs like that, like NSRect, NSPoint, NSSize.

You noticed already what the compiler said: NSRange is a typedef for the type "struct _NSRange { ... }".

And then there are types like NSInteger and NSUInteger which also start with NS but which are really just primitive types (int or long, unsigned int or unsigned long).

NSRange output of location and length

2 Answers2