Checking the length of an array of characters in Objective-C - ios

Question

I'm translating a small Java library for using in an Objective-C application I'm writing.
char[] chars = sentence.toCharArray();
int i = 0;
while (i < chars.length) { ... }
Where sentence is an NSString.
I'd like to translate the above Java code to Objective-C. Here's what I have so far:
// trims sentence off white space
sentence = [sentence stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]];
const char *chars = [sentence UTF8String];
How do I the above while condition? I'm not sure of how I'm supposed to check the length of the the string after it was converted to a character array.

Answer 1

Your Objective-C string already holds a measure of its length, it's just a matter of retrieving it:
// trims sentence off white space
sentence = [sentence stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]];
NSUInteger length = sentence.length;
const char *chars = [sentence UTF8String];
But I would like to remember that even if you didn't know the length, you could use the C strlen function:
// trims sentence off white space
sentence = [sentence stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]];
const char *chars = [sentence UTF8String];
size_t length = strlen(chars);

Answer 2

Even there is already an accepted answer I want to warn of using strlen(), even in this case it might be without any problem. There are a differences between NSString and C-Strings.
A. -length (NSString) and strlen() has different semantics:
NSString is not(!) \0-terminated, but length based. It can store \0 characters. It is very easy to get different length, if there is a \0 character in the string instance:
NSString *sentence = #"Amin\0Negm";
NSLog( #"length %ld", [sentence length]); // 9
const char *chars = [sentence cStringUsingEncoding:NSUTF8StringEncoding];
size_t length= strlen(chars);
NSLog(#"strlen %ld", (long)length); // 4
length 9
strlen 4
But -UTF8String and even the used -cStringUsingEnocding: (both NSString) copy out the whole string stored in the string instance. (I think in case of -cStringUsingEncoding it is misleading, because standard string functions like strlen() always uses the first \0 as the termination of strings.)
B. In UTF8 a character can have multibytes. A char in C is one byte. (With byte not in the meaning of 8 bits, but smallest addressable unit.)
NSString *sentence = #"Αmin Negm";
NSLog( #"length %ld", [sentence length]);
const char *chars = [sentence UTF8String];
size_t length= strlen(chars);
NSLog(#"strlen %ld", (long)length);
length 9
strlen 10
WTF happened here? The "A" of Amin is no latin capital letter A but a greek capital letter Alpha. In UTF8 this takes two bytes and for pure C's strlen there are two characters!
NSLog(#"%x-%x %x-%x", 'A', 'm', (unsigned char)*chars, (unsigned char)*(chars+1) );
41-6d ce-91
The first two numbers are the codes for 'A', 'm', the second two numbers are the UTF8 code for greek capital letter Alpha (CE 91).
I do not think, that it is a good idea to simply change from NSString to char * without good reason and a complete understanding of the problems. If you do not expect such characters, use NSASCIIStringEncoding. If you expect such characters check your code again and again … or read C.
C. C supports wide characters. This is similiar to Mac OS' unichar, but typed wchar_t. There are string functions for wchar_t in wchar.h.
NSString *sentence = #"Αmin Negm";
NSLog( #"length %ld", [sentence length]);
wchar_t wchars[128]; // take care of the size
wchar_t *wchar = wchars;
for (NSUInteger index = 0; index < [sentence length]; index++)
{
*wchar++ = [sentence characterAtIndex:index];
}
*wchar = '\0';
NSLog(#"widestrlen %ld", wcslen(wchars));
length 9
widestrlen 9
D. Obviously you want to iterate through the string. The common pattern in pure C is not to use an index and to compare it to the length and definitly not to to strlen() in every loop, because it produces high costs. (C strings are not length based so the whole string has to be scanned over and over.) You simply increment the pointer to the next char:
char letter;
while ( (letter = *chars++) ) {…}
or
do
{
// *chars points to the actual char
} while (*char++);

Answer 3

int lenght = sizeof(chars) / sizeof(char)
might work, but it will (inte the best case) return same thing as
sentence.lenght
in worst case 0 because the whole pointer / sizeof thing i don't remember now