When using the following code to extract arguments, it works correctly for most cases, but not for IPv6 format. After a first look, seems related with the [ and ] character that I guess have a special meaning for the parsing:
This works (szServer contains 1.2.3.4):
./myapp -address "1.2.3.4"
This does not works (szServer is nil):
./myapp -address "[2001::abc]"
I am unfortunately not fluent in objective-C, so it is probably a trivial problem, but looking at NSUserDefaults documentation or searching on internet "NSUserDefaults with square bracket" does not bring any light (to me) on the topic.
int main(int argc, char *argv[])
{
#autoreleasepool {
NSUserDefaults *args = [NSUserDefaults standardUserDefaults];
NSString * szServer = [args stringForKey:#"address"];
}
}
I would also appreciate some clarification of what is happening.
It works if you quote the quotes in your argument:
./myapp -address '"[2001::abc]"'
For the command-line -<key> <value> options, the value is interpreted as an old-style (NextStep-era) property list, not just as a verbatim string.
This means that one can pass arrays and dictionaries that way, for example. If you change your code to use objectForKey: instead of stringForKey:, it will receive an array from this:
./myapp -address '("[2001::abc]")'
Or a dictionary from this:
./myapp -address '{address="[2001::abc]";}'
Because various characters can be "special" when interpreting this format, strings have to be quoted if they have any characters outside of a limited set in them. Apparently, the left square bracket is such a character.
If you need to figure out the syntax used for a given argument, you can write a simple program which constructs an array with the desired property list object in it, and log that array. Obviously, the result will include the parentheses for formatting an array, but the object in the array will also be formatted as necessary for this old-style format. (If you didn't wrap the object in a collection like an array, then it might just write itself verbatim, as when logging an NSString.)
Related
Something that is bothering me is why the term 'literal' is used to refer to instances of classes like NSString and NSArray. I had only seen the term used in reference to NSString and being naive I thought it had something to do with it 'literally' being a string, that is between quotation markers. Sorry if that sounds pathetic, but that was how I had been thinking about it.
Then today I learned that certain instances of NSArray can also be referred to as literal instances, i.e. an instance of the class created using a 'literal syntax'.
As #Linuxios notes, literal syntaxes are built into the language. They're broader than you think, though. A literal just means that an actual value is encoded in the source. So there are quite a few literal syntaxes in ObjC. For example:
1 - int
1.0 - double
1.0f - float
"a" - C-string
#"a" - NSString
#[] - NSArray
^{} - function
Yeah, blocks are just function literals. They are an anonymous value that is assignable to a symbol name (such as a variable or constant).
Generally speaking, literals can be stored in the text segment and be computed at compile time (rather than at run time). If I remember correctly, array literals are currently expanded into the equivalent code and evaluated at runtime, but #"..." string literals are encoded into the binary as static data (at least now they are; non-Apple versions of gcc used to encode an actual function call to construct static strings as I remember).
A literal syntax or a literal is just an object that was created using a dedicated syntax built into the language instead of using the normal syntax for object creation (whatever that is).
Here I create a literal array:
NSArray* a = #[#"Hello", #"World"];
Which is, for all intents and purposes equivalent to this:
NSArray* a = [NSArray arrayWithObjects:#"Hello", #"World", nil];
The first is called a literal because the #[] syntax is built into the language for creating arrays, in the same way that the #"..." syntax is built in for creating NSStrings.
the term 'literal' is used to refer to instances of classes
It's not referring to the instance really; after the object is created, the way it was created doesn't matter:
NSArray * thisWasCreatedWithALiteral = #[#1, #2];
NSArray * butWhoCares = thisWasCreatedWithALiteral;
The "literal" part is just the special syntax #[#1, #2], and
it ha[s] something to do with it 'literally' being a string, that is between quotation markers.
is exactly right: this is a written-out representation of the array, as opposed to one created with a constructor method like arrayWithObjects:
I have an instance of NSMutableString called MyMutableStr and I want access its character at index 7.
For example:
unsigned char cMy = [(NSString*) MyMutableStr characterAtIndex:7];
I think this is an ugly way; it's too much code.
My question is: Are there more simple ways in Objective-C to access the character in NSMutableString?
Like, in C language we can access a character of a string using [ ] operator:
unsigned char cMy = MyMutableStr[7];
The way of doing it is to use characterAtIndex:, but you don't need to cast it to a NSString pointer, since NSMutableString is a subclass of NSString. So it isn't that long, but if you still don't find it comfortable, I suggest to use UTF8String to obtain a C string over which you can iterate using the brackets operator:
const char* cString= [MyMutableStr UTF8String];
char first= cString[0];
But remember this (taken from NSString class reference):
The returned C string is automatically freed just as a returned object would be released; you should copy the C string if it needs to store it outside of the autorelease context in which the C string is created.
As others said characterAtIndex: but a few things you might want to consider carefully.
First you're dealing with an mutable string. You want to be careful to avoid it changing out from under you. One way is to an immutable copy and use that for the op.
Second, you're dealing with Unicode so you may want to consider normalizing your string to get a precomposed form as some visual representations may be more than one actual unichar. That's often a stumbling block for folks.
I was following a tut and found a line of code like #"%# button pressed.". I'm pretty sure the relevant part is the %#, but is the first # an escape sequence or what?
Anyways, searching symbols doesn't go well in any search engine so I thought I'd ask. I think the %# is like {0} in C#?
%# is a format specifier. Functions such as NSLog and methods such as +stringWithFormat: will replace %# with the description of the provided Objective-C or Core Foundation object argument.
For example:
NSString *myName = #"dreamlax";
NSLog (#"My name is: %#", myName);
This will log the output "My name is: dreamlax". See here for more information format specifiers.
The initial # symbol at the beginning of the string tells the compiler to create a static instance of an NSString object. Without that initial # symbol, the compiler will create a simpler C-style string. Since C-style strings are not Objective-C objects you cannot add them to NSArray or NSDictionary objects, etc.
#"some string" means this is an NSString literal.
The string as show in #"CupOverflowException", is a constant
NSString object. The # sign is used
often in Objective-C to denote
extentions to the language. A C string
is just like C and C++, "String
constant", and is of type char *
I found this page which might help - http://www.yetanotherchris.me/home/2009/6/22/objective-c-by-example-for-a-c-developer.html
It seems that you are on the right track.
I'm still fairly new to the language, but it looks like the # specifies that the variable being passed/created is an NSObject, or a compiler directive.
As mentioned above, if you use it like this:
#"someText"
you're instantiating an NSString object, and setting the text of that object to someText. If you look at a good ol' C-style format specifier such as:
..."This is some text, and this is a float: %f", myFloat);
You're creating some text and telling the compiler to put the floating point string representation of myFloat into the string. %# is a format specifier, just like %f, %d, %c, %s and any other format specifier you're used to. However, if you use %# as follows:
... "This is some text, and this is an object:%#", myObject];
What you're doing is (I believe) telling the compiler that myObject is an object, and that you want it to include the output of the description method (ie. [myObject description]) in the string that you're creating.
I am coming to Objective-C from C# without any intermediate knowledge of C. (Yes, yes, I will need to learn C at some point and I fully intend to.) In Apple's Certificate, Key, and Trust Services Programming Guide, there is the following code:
static const UInt8 publicKeyIdentifier[] = "com.apple.sample.publickey\0";
static const UInt8 privateKeyIdentifier[] = "com.apple.sample.privatekey\0";
I have an NSString that I would like to use as an identifier here and for the life of me I can't figure out how to get that into this data structure. Searching through Google has been fruitless also. I looked at the NSString Class Reference and looked at the UTF8String and getCharacters methods but I couldn't get the product into the structure.
What's the simple, easy trick I'm missing?
Those are C strings: Arrays (not NSArrays, but C arrays) of characters. The last character is a NUL, with the numeric value 0.
“UInt8” is the CoreServices name for an unsigned octet, which (on Mac OS X) is the same as an unsigned char.
static means that the array is specific to this file (if it's in file scope) or persists across function calls (if it's inside a method or function body).
const means just what you'd guess: You cannot change the characters in these arrays.
\0 is a NUL, but including it explicitly in a "" literal as shown in those examples is redundant. A "" literal (without the #) is NUL-terminated anyway.
C doesn't specify an encoding. On Mac OS X, it's generally something ASCII-compatible, usually UTF-8.
To convert an NSString to a C-string, use UTF8String or cStringUsingEncoding:. To have the NSString extract the C string into a buffer, use getCString:maxLength:encoding:.
I think some people are missing the point here. Everyone has explained the two constant arrays that are being set up for the tags, but if you want to use an NSString, you can simply add it to the attribute dictionary as-is. You don't have to convert it to anything. For example:
NSString *publicTag = #"com.apple.sample.publickey";
NSString *privateTag = #"com.apple.sample.privatekey";
The rest of the example stays exactly the same. In this case, there is no need for the C string literals at all.
Obtaining a char* (C string) from an NSString isn't the tricky part. (BTW, I'd also suggest UTF8String, it's much simpler.) The Apple-supplied code works because it's assigning a C string literal to the static const array variables. Assigning the result of a function or method call to a const will probably not work.
I recently answered an SO question about defining a constant in Objective-C, which should help your situation. You may have to compromise by getting rid of the const modifier. If it's declared static, you at least know that nobody outside the compilation unit where it's declared can reference it, so just make sure you don't let a reference to it "escape" such that other code could modify it via a pointer, etc.
However, as #Jason points out, you may not even need to convert it to a char* at all. The sample code creates an NSData object for each of these strings. You could just do something like this within the code (replacing steps 1 and 3):
NSData* publicTag = [#"com.apple.sample.publickey" dataUsingEncoding:NSUnicodeStringEncoding];
NSData* privateTag = [#"com.apple.sample.privatekey" dataUsingEncoding:NSUnicodeStringEncoding];
That sure seems easier to me than dealing with the C arrays if you already have an NSString.
try this
NSString *newString = #"This is a test string.";
char *theString;
theString = [newString cStringWithEncoding:[NSString defaultCStringEncoding]];
Does Objective-C have raw strings like Python's?
Clarification: a raw string doesn't interpret escape sequences like \n: both the slash and the "n" are separate characters in the string. From the linked Python tutorial:
>>> print 'C:\some\name' # here \n means newline!
C:\some
ame
>>> print r'C:\some\name' # note the r before the quote
C:\some\name
Objective-C is a superset of C. So, the answer is yes. You can write
char* string="hello world";
anywhere. You can then turn it into an NSString later by
NSString* nsstring=[NSString stringWithUTF8String:string];
From your link explaining what you mean by "raw string", the answer is: there is no built in method for what you are asking.
However, you can replace occurrences of one string with another string, so you can replace #"\n" with #"\\n", for example. That should get you close to what you're seeking.
You can use stringize macro.
#define MAKE_STRING(x) ##x
NSString *expendedString = MAKE_STRING(
hello world
"even quotes will be escaped"
);
The preprocess result is
NSString *expendedString = #"hello world \"even quotes will be escaped\"";
As you can see, double quotes are escaped, however new lines are ignored.
This feature is very suitable to paste some JS code in Objective-C files. Using this feature is safe if you are using C99.
source:
https://gcc.gnu.org/onlinedocs/cpp/Stringizing.html
How, exactly, does the double-stringize trick work?
Like everyone said, raw ANSI strings are very easy. Just use simple C strings, or C++ std::string if you feel like compiling Objective C++.
However, the native string format of Cocoa is UCS-2 - fixed-width 2-byte characters. NSStrings are stored, internally, as UCS-2, i. e. as arrays of unsigned short. (Just like in Win32 and in Java, by the way.) The systemwide aliases for that datatype are unichar and UniChar. Here's where things become tricky.
GCC includes a wchar_t datatype, and lets you define a raw wide-char string constant like this:
wchar_t *ws = L"This a wide-char string.";
However, by default, this datatype is defined as 4-byte int and therefore is not the same as Cocoa's unichar! You can override that by specifying the following compiler option:
-fshort-wchar
but then you lose the wide-char C RTL functions (wcslen(), wcscpy(), etc.) - the RTL was compiled without that option and assumes 4-byte wchar_t. It's not particularly hard to reimplement these functions by hand. Your call.
Once you have a truly 2-byte wchar_t raw strings, you can trivially convert them to NSStrings and back:
wchar_t *ws = L"Hello";
NSString *s = [NSString stringWithCharacters:(const unichar*)ws length:5];
Unlike all other [stringWithXXX] methods, this one does not involve any codepage conversions.
Objective-C is a strict superset of C so you are free to use char * and char[] wherever you want (if that's what you call raw strings).
If you mean C-style strings, then yes.