How are unicode comparisons coded? I need to test exactly as below, checking for specific letters in a string. The code below chokes: warning: comparison between pointer and integer
for (charIndex = 0; charIndex < [myString length]; charIndex++)
{
unichar testChar = [myString characterAtIndex:charIndex];
if (testChar == "A")
// do something
if (testChar == "B")
// do something
if (testChar == "C")
// do something
}
For char literals, use single quotes:
if (testChar == 'A') NSLog(#"It's an A");
Or represent the character using the code point number:
if (testChar == 0x1e01) NSLog(#"It's an A with a ring below");
The compiler sees double-quotes as a string, so builds "A" as equivalent to a const char * (which gives you there error message about the pointer).
What are you really trying to do? Doing direct character comparisons is unusual. Typically -compare: or -isEqual: would be used to compare two strings. Or NSScanner would be used to analyze the components of a string.
Related
I'm using the following technique to find the number of operators in a mathematical string.
for (int index = 0; index < [self.evaluateString length]; index++) {
unichar chars = [self.evaluateString characterAtIndex:index];
if (chars == '+' || chars == '-' || chars == '*' || chars == '/' ||chars == '^') {
self.operatorCount++;
}
}
My trainer says this method is not very good. I would like to know is there any better/more elegant method to do this. Thanks.
It seems that string is to be evaluated. Part of that evaluation is parsing. During this process mathematical operators are identified and could be counted.
The advantage over simple character counting would be to tell apart a 3 - 1 (operator) from a -1 (negative number literal).
You can do something like this. It's not particularly efficient, compared to your code, but it looks fancy.
NSString *s = #"3 + 1 - 2 * 4 / 4";
NSCharacterSet *cs = [[NSCharacterSet characterSetWithCharactersInString:#"+-*/"] invertedSet];
NSArray *a = [[s componentsSeparatedByCharactersInSet:cs] filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:#"length > 0"]];
NSLog(#"%lu", a.count);
The NSLog() will print 4, as expected.
The algorithm is simple:
Create a character set that is the inverse of the operator list.
Split the expression string by the inverted character set.
Filter out elements of zero length.
The remaining elements in the array will be the operators. The count of elements in the array is the count of operators in the expression.
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.
I am trying to make a simple calculator app. Currently, the app works perfectly. One problem: It's smart enough to change results into formatted numbers (800000 = 800,000), but not full expressions (200*600/21000 = 200*600/21,000).
I would like to be able to have a method that I could feed a string and get back a string of properly formatted numbers with operations still inside the string.
Example:
I feed the method 30000/80^2. Method gives back 30,000/80^2.
EDIT: People seem to be misunderstanding the question (Or it's possible I am misunderstanding the answers!) I want to be able to separate the numbers - 60000/200000 would separate into 60000 & 200000. I can do it from there.
Well, what's the problem? You obviously can parse the whole expression (you say calculator works), you can format single numbers (you say you can format results).
The only thing you need is to parse the expression, format all the numbers and recompose the expression...
EDIT: There is a simpler solution. For formatting, you don't need to parse the expression into a tree. You just have to find the numbers.
I suggest to create character set of all operators
NSCharacterSet* operators = [NSCharacterSet characterSetWithCharactersInString:#"+*-/^()"];
NSCharacterSet* whitespaces = [NSCharacterSet whitespaceCharacterSet];
Then split the expression using this set:
NSString* expression = [...];
NSMutableString* formattedExpression = [NSMutableString string];
NSRange numberRange = NSMakeRange(0, 0);
for (NSUInteger i = 0; i < expression.length; i++) {
unichar character = [expression characterAtIndex:i];
if ([whitespaces characterIsMember:character] || [operators characterIsMember:character]) {
if (numberRange.length > 0) {
NSString* number = [expression substringWithRange:numberRange];
NSString* formattedNumber = [self formatNumber:number];
[formattedExpression appendString:number];
numberRange.length = 0;
}
}
else if (numberRange.length == 0) {
numberRange.location = i;
numberRange.length = 1;
}
else {
numberRange.length++;
}
if ([operators characterIsMember:character]) {
[formattedExpression appendFormat:#"%C", character];
}
}
if (numberRange.length > 0) {
NSString* number = [expression substringWithRange:numberRange];
NSString* formattedNumber = [self formatNumber:number];
[formattedExpression appendString:number];
}
Note that this should work even for numbers prefixed by a sign. I am ignoring all whitespaces because if you want to have a pretty expression, you probably want to handle whitespaces differently (e.g. no space after (, space before +/-, space after - only if it's not a number sign...). In general, for handling spaces, parsing the expression into a tree would simplify matters. Also note that infix expressions are not unambiguous - that means that you should sometimes add parenthesis. However, that can't be done without parsing into a tree.
Look up NSNumberFormatter. Not only will that handle formatting of numbers, it will do so based on the user's locale.
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!");
I feel embarrassed asking this, but it seems like something I need to understand:
Why does this evaluate to true? (if I cast it to int it properly evaluates to false)
NSString *var1 = #"ABC";
NSString *var2 = #"ABCD";
if (([var1 length] - [var2 length]) > 2) NSLog(#"-1 > 2");
length is unsigned.
See docs:
length
Returns the number of Unicode characters in the receiver.
- (NSUInteger)length
Return Value
The number of Unicode characters in the receiver.