I have an enum and I want to pass one of its value to an Objective-C method expecting char * directly instead of creating another variable. What should I do?
e.g.
typedef enum { value1 = 0xAA, value2 = 0xBB, value3 = 0xCC } myValue;
NSMutableData *data = [[NSMutableData alloc] init];
// want to pass in the enum directly as char * but won't work
[data appendBytes:(char *){ value1 } length: 1];
// this will work, why? and is this the best way?
[data appendBytes:(char []){ value1 } length: 1];
Your example of using (char[]){value1} to get a char* that points to value1 interpreted as a char is perfectly legitimate. Equivalent code might look like
char achar = value1;
[data appendBytes:&achar length:1];
But your (char[]){value1} is obviously one fewer lines of code, so I say go with that.
For any who are confused, (char[]){value1} is a compound literal. It's a way of constructing any sort of value, including aggregates, as an expression. In this particular case, it's constructing an array of chars, with the initializer containing one element, so this constructs a char[1], which then gets implicitly converted to a pointer when passed to the method. Interestingly, you can even take the address of compound literals (although the address is only going to be valid for the full expression in which the compound literal is found), so an equivalent in this case would be &(char){value1}.
Related
i got the following char array in Objective-C (Xcode):
char *incomeMessage;
NSString *str = [[NSString alloc] initWithBytes:data.bytes length:data.length encoding:NSUTF8StringEncoding];
incomeMessage = [str UTF8String];
NSLog(#"%c", incomeMessage[0]);
NSLog(#"%c", incomeMessage[1]);
NSLog(#"%c", incomeMessage[2]);
NSLog(#"%c", incomeMessage[3]);
NSLog(#"%c", incomeMessage[4]);
NSLog(#"%c", incomeMessage[5]);
For example I get some results like this in console:
"3
2
6
1
8
4"
Now i want to replace the char in incomeMessage[2] by 4:
incomeMessage[2] = '4';
But then it gives me the error:
EXC_BAD_ACCESS
Do you have an idea, how to solve the problem?
According to the reference documentation, UTF8String returns a read-only (const char*) reference to the string data.
The reference material goes on to note:
This C string is a pointer to a structure inside the string object,
which may have a lifetime shorter than the string object and will
certainly not have a longer lifetime. Therefore, you should copy the C
string if it needs to be stored outside of the memory context in which
you use this property.
So I'd suggest following their advice and creating a copy of the array and then performing your modifications against that.
For example: http://ideone.com/mhjwZW
You might have better luck with something like:
NSString* str = [[NSString alloc] initWithBytes:data.bytes length:data.length encoding:NSUTF8StringEncoding];
char* incomeMessage = malloc([str lengthOfBytesUsingEncoding:NSUTF8StringEncoding] + 1);
strcpy(incomeMessage, [str UTF8String]);
//now you can change things
incomeMessage[2] = '4';
//do this when you're done
free(incomeMessage);
Although, is there any particular reason why you want to use a C-string/character array as opposed to an NSMutableString? I think you might find replaceCharactersInRange:withString: a better approach generally. See also: stringByReplacingCharactersInRange:withString:.
i got the following char array in Objective-C (Xcode)
You don't, you know. All you have is a pointer. You have not set aside any actual memory; there is no array there.
incomeMessage = [str UTF8String];
All you've done in that line is repoint the pointer incomeMessage at your string's UTF8String. A string's UTF8String is immutable. Note this passage in the docs:
you should copy the C string if it needs to be stored outside of the memory context in which you use this property.
So basically, if you want to write into an array of char, your first task should be to make an array of char.
I understand pointers work with addresses and not the data itself. This is why I need to use the address-of (&) operator below as I need to assign the address of num to the pointer and not the actual value of num (40).
int num = 40;
int *numPtr = #
Therefore i'm confused as to why I can do this.
NSString *str = #"hello";
I've created a pointer str but instead of giving it an address i'm able to assign it some data, a literal string.
I thought pointers could only hold memory addresses so why am I able to directly assign it some data?
For someone trying to get their head around pointers and objects this is very confusing.
No you are not assigning a literal string to it, # makes a NSString object with the string value hello.
In most C languages strings are just an array of char, where char is a primitive type like int like in your example.
There is a reason you put an # before string literals (when you want an NSString and not a C string) in objective-c
#"String" is basically equivalent to [NSString stringWithCString:"string"] which returns a pointer to an NSString object containing the value "string"
It is the same way 1 is a c type integer, but #1 is a NSNumber representing the value of 1. If you see an # it means "this is shorthand for creating an object". (#[] for NSArrays, #{} for NSDictionarys, #(), #123, #YES, #NO for NSNumbers, and #"" for NSString)
C does not have strings. Usually char arrays are used to represent them.
NSString *str = #"hello";
can be thought of as short hand (literal) for:
char charArray[] = "hello";
NSString *str = [[NSString alloc] initWithBytes:charArray length:sizeof(charArray) encoding:NSUTF8StringEncoding]; // disregard character encoding for this example
or
unichar bla[] = {'h', 'e', 'l', 'l', 'o'};
str = [[NSString alloc] initWithCharacters:bla length:sizeof(bla)];
So an object is created and thus you need a pointer.
Is it possible to replace a character from a c string after converting it from NSString via the UTF8string method?
For example take the code below. It is to format a string with particular rule.
- (NSString *)formatString:(NSString *)input {
if (input.length==0) {
return #"";
}
//code to determine rule
....
....
// substitute output format with input characters
if (rule) {
input = [input substringFromIndex:prefix.length];
char *string = (char *)[rule UTF8String];
int repCount = 0;
for (int i=0; i<rule.length; i++) {
if (string[i] == '#') {
if (repCount < input.length)
string[i] = [input characterAtIndex:repCount++];//bad access
else
string[i] = ' ';
}
}
NSMutableString *output = [NSMutableString stringWithCString:string encoding:NSUTF8StringEncoding];
...
... //do something with the output
return output;
} else {
return input;
}
}
Initially string[0] has '#' and it should get replaced with the character in the input. This is not happening.
In a word, NO. That buffer doesn't belong to you so leave it alone.
A couple of issues:
You are casting UTF8String, which returns a const char *, to char *. UTF8String is, by definition, returning a read-only string and you should use it as such. (You really should use casts sparingly, if at all. Certainly never use casts to override const qualifiers for variables.)
If you want to perform this C-string manipulation, you have to copy the string to your own buffer. For example, use getCString or getCharacters methods (but only after you've created a buffer to receive them, and remember to add a character for the NULL terminator).
By the way, you're also returning characterAtIndex, which is a unichar (which can be larger than 8-bits), and using it in your char * buffer (8-bits per character). I'd be wary about mixing and matching those without being very careful. It is best to pick one and stick with it (and unichar offers a little more tolerance for those non-8-bit characters).
Perhaps you check for this earlier, but you're setting string to be those characters after the prefix, and then proceed to check the next rule.length number of characters. But, as far as I can tell, you have no assurances that string actually has that many characters left in it. You should test for that, or else that will also cause problems.
Personally, I'd retire this whole C-string algorithm and employ the appropriate NSString and/or NSMutableString methods to do whatever replacement you wanted, e.g. stringByReplacingCharactersInRange, stringByReplacingOccurrencesOfString, or the equivalent NSMutableString methods, replaceCharactersInRange or replaceOccurrencesOfString.
Is there any easy way to convert an Objective-C holding class of NSStrings into parameters for a function accepting a variable list of char *? Specifically I have a function like:
-(void)someFunction:(NSSomething *) var
that I want to forward to a C function like
void someCFunction(char * var, ...)
Is there an easy way to go about this?
No, you can only do what you want if the number of arguments you're passing is known at compile time. If you just want to convert a single string, use the -UTF8String message:
// Example with two strings
NSString *str1 = ...;
NSString *str2 = ...;
someCFunction([str1 UTF8String], [str2 UTF8String]); // etc.
But if the number of strings will vary at runtime, you'll need to use a different API, if one is available. For example, if there's an API that took an array of strings, you could convert the Objective-C array into a C array:
// This function takes a variable number of strings. Note: in C/Objective-C
// (but not in C++/Objective-C++), it's not legal to convert 'char **' to
// 'char *const *', so you may sometimes need a cast to call this function
void someCFunction(const char *const *stringArray, int numStrings)
{
...
}
...
// Convert Objective-C array to C array
NSArray *objCArray = ...;
int numStrings = [objCArray count];
char **cStrArray = malloc(numStrings * sizeof(char*));
for (int i = 0; i < count; i++)
cStrArray[i] = [[objCArray objectAtIndex:i] UTF8String];
// Call the function; see comment above for note on cast
someCFunction((const char *const *)cStrArray, numStrings);
// Don't leak memory
free(cStrArray);
This would do the trick:
NSString *string = #"testing string"
const char * p1=[string UTF8String];
char * p2;
p2 = const_cast<char *>(p1);
Yes, this can be done, and is explained here:
How to create a NSString from a format string like #"xxx=%#, yyy=%#" and a NSArray of objects?
And here:
http://www.cocoawithlove.com/2009/05/variable-argument-lists-in-cocoa.html
With modifications for ARC here:
How to create a NSString from a format string like #"xxx=%#, yyy=%#" and a NSArray of objects?
Also, variable arguments are not statically or strongly typed, as the other poster seems to be suggesting. In fact, there is no clear indication in the callee of how many arguments you really have. Determining the number of arguments generally breaks down into having to either specify the number by an count parameter, using a null terminator, or inferring it from a format string a la (s)print* . This is frankly why the C (s)print* family of functions has been the source of many errors, now made much much safer by the XCode / Clang / GCC compiler that now warns.
As an aside, you can approach statically typed variable arguments in C++ by creating a template method that accepts an array of an unspecified size. This is generally considered bad form though as the compiler generates separate instances for each size of array seen by by the compiler (template bloat).
I have a need to map char values to int values in Objective-C. I know NSDictionary is out because it deals with reference types, and these are values. The map will be used while iterating through an NSString. Each character in the string will be converted to an integer value. All the integers will be summed together.
Using NSDictionary seems like a bad fit because of all the type coercion I'd have to do. (Converting values types, char and int, to reference types.)
I figure I'll have to drop down to C to do this, but my experience with C libraries is very limited.
Is there something most C developers use that will map char values to int values?
Edit for clarification
The C# equivalent would be a Dictionary<char,int>.
In pseudocode, I'd like to the following:
for (int i = 0; i < [string length]; i++) {
char current = [string characterAtIndex:i];
int score = map[current]; // <- I want map without boxing
// do something with score
}
Char to int?
char aChar = 'a';
int foo = (int) aChar;
Done. No need for a hash or anything else. Even if you wanted to map char -> NSNumber, an array of 256 char's (char being a signed 8 bit type) is very little overhead.
(Unless I entirely misparsed your question -- are you asking for (char*)? ... i.e. C strings? Show some code.).
If I understand correctly, you want to store chars and ints in a dictionary, as keys and values. However, NSDictionary only accepts objects. The solution? Wrap the chars and ints in the NSNumber object:
NSDictionary *dict = [NSDictionary dictionaryWithObjectsAndKeys:
[NSNumber numberWithInt:1],
[NSNumber numberWithChar:'a'],
[NSNumber numberWithInt:2],
[NSNumber numberWithChar:'b'],
nil];
Or if you don't want boxing, why not just make a function that takes chars and returns ints?
int charToScore(char character)
{
switch (character) {
case 'a':
return 1;
case 'b':
return 2;
default:
return 0;
}
}
#Pontus has the correct answer in Objective-C, but if you're willing to use C++, you can use std::map<char, int> (or the still-slightly-nonstandard unordered_map<char, int>.)
To use C++ from within Objective-C, you must rename the file from Whatever.m to Whatever.mm--this tells GCC that the file contains Objective-C++, which allows you to use the Objective-C syntax with C++ underpinnings.