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.
Related
I can't find an easy way to generate a one digit sized substring from an NSString. The NSString looks like "(N3)" but I need to store just the number portion "3" in an substring. Using:
NSString *subString = [dis substringFromIndex:2];
returns "3)".
Any ideas?
Well, if the string is one digit and its position is invariant, you could just use:
int Foo(NSString * dis) {
const unichar c = [dis characterAtIndex:2];
assert(isdigit(c) && "not a number");
return c - '0';
}
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.
The following is tried to print out N number of spaces (or 12 in the example):
NSLog(#"hello%#world", [NSString stringWithCharacters:" " length:12]);
const unichar arrayChars[] = {' '};
NSLog(#"hello%#world", [NSString stringWithCharacters:arrayChars length:12]);
const unichar oneChar = ' ';
NSLog(#"hello%#world", [NSString stringWithCharacters:&oneChar length:12]);
But they all print out weird things such as hello ÔÅÓñüÔÅ®Óñü®ÓüÅ®ÓñüÔ®ÓüÔÅ®world... I thought a "char array" is the same as a "string" and the same as a "pointer to a character"? The API spec says it is to be a "C array of Unicode characters" (by Unicode, is it UTF8? if it is, then it should be compatible with ASCII)... How to make it work and why those 3 ways won't work?
You can use %*s to specify the width.
NSLog(#"Hello%*sWorld", 12, "");
Reference:
A field width, or precision, or both, may be indicated by an asterisk
( '*' ). In this case an argument of type int supplies the field width
or precision. Applications shall ensure that arguments specifying
field width, or precision, or both appear in that order before the
argument, if any, to be converted.
This will get you what you want:
NSLog(#"hello%#world", [#"" stringByPaddingToLength:12 withString:#" " startingAtIndex:0]);
I think the issue you have is you are misinterpreting what +(NSString *)stringWithCharacters:length: is supposed to do. It's not supposed to repeat the characters, but instead copy them from the array into a string.
So in your case you only have a single ' ' in the array, meaning the other 11 characters will be taken from whatever follows arrayChars in memory.
If you want to print out a pattern of n spaces, the easiest way to do that would be to use -(NSString *)stringByPaddingToLength:withString:startingAtIndex:, i.e creating something like this.
NSString *formatString = #"Hello%#World";
NSString *paddingString = [[NSString string] stringByPaddingToLength: n withString: #" " startingAtIndex: 0];
NSLog(formatString, paddingString);
This is probably the fastest method:
NSString *spacesWithLength(int nSpaces)
{
char UTF8Arr[nSpaces + 1];
memset(UTF8Arr, ' ', nSpaces * sizeof(*UTF8Arr));
UTF8Arr[nSpaces] = '\0';
return [NSString stringWithUTF8String:UTF8Arr];
}
The reason your current code isn't working is because +stringWithCharacters: expects an array with a length of characters of 12, while your array is only 1 character in length {' '}. So, to fix, you must create a buffer for your array (in this case, we use a char array, not a unichar, because we can easily memset a char array, but not a unichar array).
The method I provided above is probably the fastest that is possible with a dynamic length. If you are willing to use GCC extensions, and you have a fixed size array of spaces you need, you can do this:
NSString *spacesWithLength7()
{
unichar characters[] = { [0 ... 7] = ' ' };
return [NSString stringWithCharacters:characters length:7];
}
Unfortunately, that extension doesn't work with variables, so it must be a constant.
Through the magic of GCC extensions and preprocessor macros, I give you.... THE REPEATENATOR! Simply pass in a string (or a char), and it will do the rest! Buy now, costs you only $19.95, operators are standing by! (Based on the idea suggested by #JeremyL)
// step 1: determine if char is a char or string, or NSString.
// step 2: repeat that char or string
// step 3: return that as a NSString
#define repeat(inp, cnt) __rep_func__(#encode(typeof(inp)), inp, cnt)
// arg list: (int siz, int / char *input, int n)
static inline NSString *__rep_func__(char *typ, ...)
{
const char *str = NULL;
int n;
{
va_list args;
va_start(args, typ);
if (typ[0] == 'i')
str = (const char []) { va_arg(args, int), '\0' };
else if (typ[0] == '#')
str = [va_arg(args, id) UTF8String];
else
str = va_arg(args, const char *);
n = va_arg(args, int);
va_end(args);
}
int len = strlen(str);
char outbuf[(len * n) + 1];
// now copy the content
for (int i = 0; i < n; i++) {
for (int j = 0; j < len; j++) {
outbuf[(i * len) + j] = str[j];
}
}
outbuf[(len * n)] = '\0';
return [NSString stringWithUTF8String:outbuf];
}
The stringWithCharaters:length: method makes an NSString (or an instance of a subclass of NSString) using the first length characters in the C array. It does not iterate over the given array of characters until it reaches the length.
The output you are seeing is the area of memory 12 Unicode characters long starting at the location of your passed 1 Unicode character array.
This should work.
NSLog(#"hello%#world", [NSString stringWithCharacters:" " length:12]);
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!");
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.