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I know the %% method, but it's not working :(.

with NSLog

 NSLog(@"%%5B hello %%5D");

prints

%5B hello %5D

but with NSString

NSString *str = [NSString stringWithFormat:@"%%5B %@ %%5D",@"hello"];
NSLog(str);

prints

5 hello 68233204

Why?

Daniel
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2 Answers2

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It's because you left out a line of code. In your stringWithFormat: version you probably have something like

NSString *str = [NSString stringWithFormat:@"%%5B5%%5D"];
NSLog(str);

The problem is you've now doubly-interpreted format characters in the string. Remember, the first arg to NSLog() and to +[NSString stringWithFormat:] is a format string, where % characters are treated specially. In both cases, you can easily preserve the original version of the string (and strip out the duplicate %'s ahead of time) by using actual format strings, as in

NSLog(@"%@", @"%5B%5D");
Lily Ballard
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    note: calls through NSLog (`NSLog(str);`) should always use a format specification: `NSLog(@"%@",str);` – justin Apr 26 '11 at 17:48
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You do need to use stringWithFormat: if you aren't going to pass in any arguments, you should be able to just use 

NSString *str = @"%%5B%%5D";

But as to "why", what you are experiencing is occurring, I am not sure... Try passing in an argument and see if it still happens. Maybe there is a bug when no arguments are passed?

According to this post you are escaping it correctly.

String format specifiers document says you are doing it right too.

Community
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Chris Wagner
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