I've got a dictionary initialized like so...
keyDictionary = [[NSDictionary dictionaryWithObjects:values forKeys:keys]retain];
where keys is an NSArray of the alphabet and other characters and values is an NSArray of unsigned chars, which are the USB hex keycodes for those characters.
The USB key codes are hex values that range from 0x04 to 0xE7. I'm trying to create a map between these two depending on what key is pressed on the keyboard.
The values array is created like so...
NSArray *values = [NSArray arrayWithObjects:
[NSNumber numberWithUnsignedChar:0x04]/*A*/,
[NSNumber numberWithUnsignedChar:0x05]/*B*/, /*ETC*/];
So ideally when I run this code...
where character == #"I"
- (uint8) getUSBCode:(NSString *)character
{
NSNumber *val = [keyDictionary objectForKey:character];
return (uint8)[val unsignedCharValue];
}
I would expect to get back 0x0C, but I'm getting 12 back as an int (which after I thought about it, makes sense). I need the hex value preserved. I do NOT need a string value. I need a straight conversion to the hex value or a better way to store
uint8 is just a typedef unsigned char.
EDIT I was not clear when I posted this earlier. Here's what I need.
I need the hex value of these codes because they are being sent over the internal company network. In addition, the pressed key's value is being converted from big endian (or little, it's escaping me right now which one it is) to the other, then being transmitted over an internal network. I understand that these values are stored in binary, but I need to transmit them in hex.
Also, I stated I was getting 12 back from the function. I was reading 12 from the debugger, not actually getting the value. That might be why I was getting confused.
12 (in base 10) is 0x0c.
If you want to print it out in hex, use the %x format specifier e.g.
NSLog(#"Hex value of char is 0x%02x", (unsigned int) c);
If you want to see it in hex in the debugger (assuming Xcode 3.2.x) right click on the variable and select hexadecimal as the format.
You know that an int is stored in binary (i.e. the 'hex' value is always and never preserved), so I'm interpreting your question as pertaining to printing to the screen.
You should be able to use a format specifier for that -- something like %0x.
The value that's returned from your -getUSBCode: method isn't two decimal digits, it's one eight-bit byte. Both "12" and "0x0C" are strings that represent that byte's value, so saying you want "0x0C" but don't want a string is a contradiction.
Related
I need to convert values like 1393443048683555715 to HEX. But, first of all, i cann't display it as decimal using NSLog(), for example.
Ok, it works:
NSLog(#"%qu", 1393443048683555706);
But what about converting to HEX. What type i have to use to store this big value?
NSLog([NSString stringWithFormat: #"%x", 1393443048683555706]);
// result eb854b7a. It's incorrect result!
but i forgot to say that this big number represented as string #"1393443048683555706" (not int)
You can use %qi and %qu format specifiers with NSLog to display 64-bit integers. Your constant appears to fit in 64-bit signed number, with the limits of:
[−9223372036854775808 to 9223372036854775807]
The "x" format specifier is for 32-bit numbers; you need to use either "qx" or "qX" (depending on whether you want the letter values to be uppercase or not). These are the formatters for unsigned long long values, see:
https://developer.apple.com/library/mac/#documentation/Cocoa/Conceptual/Strings/Articles/formatSpecifiers.html#//apple_ref/doc/uid/TP40004265-SW1
Next, you should not pass a string as you have done above directly to NSLog - this can cause a crash.
NSLog(string); // bad!!
NSLog(#"%#", string); // good
So if your value comes as a string, you'll want to do this:
NSString *longNumber = #"1393443048683555706";
NSLog(#"%qx", [longNumber longLongValue]);
If the string value can't be coerced to a number, longLongValue will return 0. I'll leave it to you do handle the error (and bounds) checking - see NSString for details.
If you want to save the hex value as a string, do this:
NSString *hexRepresentation = [NSString stringWithFormat:#"%qx", [longNumber longLongValue]];
Again, best to take care for error handling.
I know it is possible to use the methods compare: and isEqualToString:, and I suppose isEqualToString is the most efficient method If you know it´s an string. But my question is, is there another way to do it more efficiently? Like comparing char by char or something like that.
By reading the documentation:
The comparison uses the canonical representation of strings, which for a particular string is the length of the string plus the Unicode characters that make up the string. When this method compares two strings, if the individual Unicodes are the same, then the strings are equal, regardless of the backing store. “Literal” when applied to string comparison means that various Unicode decomposition rules are not applied and Unicode characters are individually compared. So, for instance, “Ö” represented as the composed character sequence “O” and umlaut would not compare equal to “Ö” represented as one Unicode character.
and:
When you know both objects are strings, this method is a faster way to check equality than isEqual:.
it seems that it's the best method available, to compare strings and that it does exactly what you need, that is: first it checks for length (if 2 strings have different length, is not necessary to check each char contained), then if the length it's the same it compares each char. Simple and efficient!
isEqualToString: is faster if you know both objects are strings, as the documentation states.
You could try converting both string to C strings and then use strcmp. Doubt it'll actually be any quicker though.
char *str1 = [myNSString1 UTF8String];
char *str2 = [myNSString2 UTF8String];
BOOL isEqual = strcmp(str1,str2);
I am trying to generate a numerical string by padding the number with zeroes to the left.
0 would become 00000
1 would become 00001
10 would become 00010
I want to create five character NSString by padding the number with zeroes.
I read this Create NSString by repeating another string a given number of times but the output is an NSMutableString.
How can I implement this algorithm with the output as an NSString?
Best regards.
You can accomplish this by calling
[NSString stringWithFormat:#"%05d", [theNumber intValue]];
where theNumber is the NSString containing the number you want to format.
For further reading, you may want to look at Apple's string formatting guide or the Wikipedia entry for printf.
One quick & simple way to do it:
unsigned int num = 10; // example value
NSString *immutable = [NSString stringWithFormat:#"%.5u", num];
If you actually really want to use the long-winded approach from the example you read, you can send a “copy” message to a mutable string to get an immutable copy. This holds for all mutable types.
NSScanner has the instance method -scanHexInt: for converting a hexadecimal string representation of an int to an int. I'd like to invert this: I'd like a method that takes an int and returns a hexadecimal representation.
Is there a ready made method in the docs?
[To be more relevant, I'd like to take a string comprising a single kanji and return its unicode point.]
If you check the documentation on string format specifiers, you'll see that there are several specifiers you can use to get a hexadecimal string representation of a number. For instance, you could use:
[NSString stringWithFormat:#"%x", foo]
I'm still learning, and I'm just stuck. I want the user to enter any number and in result, my program will do this equation:
x = 5*y
(y is the number the user adds, x is outcome)
How would I do this? I'm not sure if I'm suppose to add in an int or NSString. Which should I use, and should I enter anything in the header files?
I'm not sure if I'm suppose to add in an int or NSString.
Well, one of these is a numeric type and the other is a text type. How do you multiply text? (Aside from repeating it.)
You need a numeric type.
I would caution against int, since it can only hold integers. The user wouldn't be able to enter “0.5” and get 2.5; when you converted the “0.5” to an int, the fractional part would get lopped off, leaving only the integral part, which is 0. Then you'd multiply 5 by 0, and the result you return to the user would be 0.
Use double. That's a floating-point type; as such, it can hold fractional values.
… should I enter anything in the header files?
Yes, but what you enter depends on whether you want to use Bindings or not (assuming that you really are talking about Cocoa and not Cocoa Touch).
Without Bindings, declare an outlet to the text field you're going to retrieve the multiplier from, and another to the text field you're going to put the product into. Send the input text field a doubleValue message to get the multiplier, and send the output text field a setDoubleValue: message with the product.
With Bindings, declare two instance variables holding double values—again, one for the multiplier and one for the product—along with properties exposing the instance variables, then synthesize the properties, and, finally, bind the text fields' value bindings to those properties.
If you're retrieving the NSString from a UI, then it's pretty simple to do:
NSString * answer = [NSString stringWithFormat:#"%d", [userInputString integerValue]*5];
This can be done without any objective C. That is, since Objective-C is a superset of C, the problem can be solved in pure C.
#include <stdio.h>
int main(void)
{
int i;
fscanf(stdin, "%d", &i);
printf("%d\n", i * 5);
}
In the above the fscanf takes care of converting the character(s) read on the standard input to a number and storing it in i.
However, if you had characters from some other source in a char* and needed to convert them to an int, you could create an NSString* with the – initWithCString:encoding: and then use its intValue method, but in this particular problem that simply isn't needed.