unichar myChar = [myString characterAtIndex:0];
[myNSMutableArray addObject:myChar];
I am trying to insert the first char of a string into an array, to create an array of chars. the first line does not give me an error. The second line however, provides the following error: warning: passing argument 1 of 'addObject:' makes pointer from integer without a cast
This also crashes the application with a "bad address" error. I thought this error was due to a problem with memory allocation. Can someone shed some light on this.
You can only add objects to an array. unichar is a primitive data type. You have to wrap it in an NSNumber. A unichar is an unsigned short, so you can use:
[myNSMutableArray addObject:[NSNumber numberWithUnsignedShort:[myString characterAtIndex:0]]];
One option would be to add the character to your array as a string:
unichar myChar = [myString characterAtIndex:0];
NSString * charString = [NSString stringWithFormat:#"%c", myChar];
[myNSMutableArray addObject:charString];
Note that this is probably overkill.
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'm trying to loop through a NSString, character by character, but I'm getting a EXC_BAD_ACCESS error. Do you have an idea how to do this right? I've been googling for hours now but can't figure it out.
Here is my code (.m):
self.textLength = [self.text length];
for (int position=0; position < self.textLength; position++) {
NSLog(#"%#", [self.text characterAtIndex:position]);
if ([[self.text characterAtIndex:position] isEqualToString:#"."]){
NSLog(#"it's a .");
}
}
Thanks a lot!
Characters are not object. characterAtIndex returns unichar, which is actually an integer type unsigned short. You need to use %C instead of %# in NSLog. Also character is not a NSString, so you can't send it isEqualToString. You need to use ch == '.' to compare ch against '.'.
unichar ch = [self.text characterAtIndex:position];
NSLog(#"%C", ch);
if (ch == '.') {} // single quotes around dot, not double quotes
Note that, 'a' is character, "a" is C string and #"a" is NSString. They all are different types.
When you are using %# with unichar ch in NSLog, it is trying to print an object from memory location ch which is invalid. Thus you are getting a EXC_BAD_ACCESS.
characterAtIndex: returns a unichar, so you should use NSLog(#"%C", ...) instead of #"%#".
You also cannot use isEqualToString for a unichar, just use == '.' is fine.
If you want to find the position of all '.'s, you can use rangeOfString. Refer to:
String Programming Guide: Searching, Comparing, and Sorting Strings
Position of a character in a NSString or NSMutableString
characterAtIndex: returns a unichar, which is declared as typedef unsigned short unichar; The format specifier you are using in your calls to NSLog are incorrect, you could just do NSLog(#"%u",[self.text characterAtIndex:position]); or NSLog(#"%C",[self.text characterAtIndex:position]); if you want the actual character to print out.
Also, as a result of unichar being defined the way that it is, it's not a string, so you cannot compare it to other strings. Try something like:
unichar textCharacter = '.';
if ([self.text characterAtPosition:position] == testCharacter) {
// do stuff
}
If you want to find the location of a character in a string you can use this:
NSUInteger position = [text rangeOfString:#"."].location;
if the character or text is not found you will get a NSNotFound:
if(position==NSNotFound)
NSLog(#"text not found!");
What's the easiest way to get the Unicode value from an NSString? For example,
NSString *str = "A";
NSString *hex;
Now, I want to set the value of hex to the Unicode value of str (i.e. 0041)... How would I go about doing that?
The unichar type is defined to be a 16-bit unicode value (eg, as indirectly documented in the description of the %C specifier), and you can get a unichar from a given position in an NSString using characterAtIndex:, or use getCharacters:range: if you want to fill a C array of unichars from the NSString more quickly than by querying them one by one.
NSUTF32StringEncoding is also a valid string encoding, as are a couple of endian-specific variants, in case you want to be absolutely future proof. You'd get a C array of those using the much more longwinded getBytes:maxLength:usedLength:encoding:options:range:remainingRange:.
EDIT: so, e.g.
NSString *str = #"A";
NSLog(#"16-bit unicode values are:");
for(int index = 0; index < [str length]; index++)
NSLog(#"%04x", [str characterAtIndex:index]);
You can use
NSData * u = [str dataUsingEncoding:NSUnicodeStringEncoding];
NSString *hex = [u description];
You may replace NSUnicodeStringEncoding by NSUTF8StringEncoding, NSUTF16StringEncoding (the same as NSUnicodeStringEncoding) or NSUTF32StringEncoding, or many other values.
See here
for more
As per my assignment I have to take in input from a user via a console to be used with NSString.
At the moment I have
char* name[100]; // declaring char array
NSString* firstName; // declaring the NSString
printf("Please enter first name \n");
printf("=> ");
scanf("%s", &name);
firstName = [NSString stringWithCString:name encoding:NSASCIIStringEncoding];
This works, however I am getting this warning
Incompatible pointer types sending 'char [100]' to parameter of type
'const char '
I don't want to be having these errors coming up in the code, I would like to also mention I'm using Xcode 4.2.
Can anyone explain to me why I'm getting these errors, and if I can possibly overcome them?
Many thanks in advance!
Change this:
char* name[100];
to
char name[100];
The first form creates an array of 100 pointers to char. The second one creates an array of 100 char elements. What might be confusing, is that name in that last case, is in fact a pointer, pointing to the first of these 100 char elements.
As printed with NSLog is assigned to initialize a NSString.
NSLog(#"%s", arrayChar);
NSString *str = [NSString stringWithFormat:#"%s", arrayChar];
NSLog(#"Array to String: %#",str);
How do I convert a char to an NSString in Objective-C?
Not a null-terminated C string, just a simple char c = 'a'.
You can use stringWithFormat:, passing in a format of %c to represent a character, like this:
char c = 'a';
NSString *s = [NSString stringWithFormat:#"%c", c];
You can make a C-string out of one character like this:
char cs[2] = {c, 0}; //c is the character to convert
NSString *s = [[NSString alloc] initWithCString:cs encoding: SomeEncoding];
Alternatively, if the character is known to be an ASCII character (i. e. Latin letter, number, or a punctuation sign), here's another way:
unichar uc = (unichar)c; //Just extend to 16 bits
NSString *s = [NSString stringWithCharacters:&uc length:1];
The latter snippet with surely fail (not crash, but produce a wrong string) with national characters. For those, simple extension to 16 bits is not a correct conversion to Unicode. That's why the encoding parameter is needed.
Also note that the two snippets above produce a string with diferent deallocation requirements. The latter makes an autoreleased string, the former makes a string that needs a [release] call.