Can't use #define for convenience [closed] - ios

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Closed 9 years ago.
I may have made some syntax mistakes, but I can't understand why my code isn't working:
In top of .m file I wrote:
#define kCountry "Country";
In this case I got red warning from xCode - expected ]
Then in function body:
floatCountries = 74,2;
[[NSUserDefaults standardUserDefaults]setFloat:floatCountries forKey:kCountry];
float test= [[NSUserDefaults standardUserDefaults]floatForKey:kCountry];
NSLog(#"%f", test);
In that one - expected expression.
That may sound a bit stupid, but I want to simplify my code and use this, please help me! :)

Answer 1

Remove the semi-colon from the end of the define and add the '#':
#define kCountry #"Country"

Answer 2

Use:
#define kCountry #"Country"
You added a semicolon at the end.
And as this will be replaced for string value, you need to put #

Answer 3

To explain why this didn't work:
The semicolon
#define kCountry "Country";
In this case i got red warning from xCode - expected ]
There's no such thing as a “red warning”. Warnings are yellow; errors are red.
(Unless you turn on “Treat Warnings as Errors”; then, in a sense, all the warnings are red—because then they're errors.)
Anyway.
Preprocessor directives such as #define do not require a semicolon. If you include one, it becomes part of the directive.
The preprocessor replaces any macro with whatever you defined it as. In this case, you defined a macro named “kCountry” with the value “"Country;”. Note that the semicolon is part of the value—the directive ends at the end of the line, not at a semicolon.
Thus, when you go to use the macro:
[[NSUserDefaults standardUserDefaults]setFloat:floatCountries forKey:kCountry];
float test= [[NSUserDefaults standardUserDefaults]floatForKey:kCountry];
The preprocessor replaces the macro as follows:
[[NSUserDefaults standardUserDefaults]setFloat:floatCountries forKey:"Country";];
float test= [[NSUserDefaults standardUserDefaults]floatForKey:"Country";];
A semicolon can't go inside a statement; it must come after. Thus, the above statements are invalid.
The fix is to remove the semicolon from the line where you #defined the macro, so that the semicolon does not appear in the output:
[[NSUserDefaults standardUserDefaults]setFloat:floatCountries forKey:"Country"];
float test= [[NSUserDefaults standardUserDefaults]floatForKey:"Country"];
By the way, Xcode has an option in its Jump Bar to show you the complete preprocessed version of the code, as the compiler will see it. That can be handy when investigating macro problems.
So, that's one of the problems. The other one was…
The kind of string you used
"Country" is a C string. It's only usable with the C string and stdio libraries (including the functions printf, strcat, etc.) and various other APIs that require C strings (such as +[NSString stringWithUTF8String:] and +[NSString stringWithCString:encoding:]).
NSUserDefaults, like everything else in Cocoa that requires a string, requires a Cocoa string object—an NSString. The syntax for an NSString literal is the same, but with an # in front of it: #"Country".
Hence the version that works:
#define kCountry #"Country"
Which produces the preprocessed output:
[[NSUserDefaults standardUserDefaults]setFloat:floatCountries forKey:#"Country"];
float test= [[NSUserDefaults standardUserDefaults]floatForKey:#"Country"];
With no syntax errors and the right kind of string in both places, this is the version that will work.
Oh, and, as Anoop Vaidya already pointed out:
The number syntax
You tried to assign a number to a variable, but I think you'll find a different number there than you were expecting.
floatCountries = 74,2;
If you print the value of this variable with a statement such as:
NSLog(#"%f", floatCountries);
You'll find that the output is 2.0.
C has an operator called the comma operator, and it is simply x, y, where x and y can be any expression (ideally of types that are compatible with each other—e.g., both numbers).
The comma operator evaluates first the left-side expression, then the right-side expression, and itself evaluates to the right-side expression.
74,2 evaluates first the expression 74, and then the expression 2, and then evaluates to 2. Thus, you assign 2 (an int, which is converted automatically as needed) to the variable.
It may seem kind of silly to use this with literal numbers, and it is. The comma operator exists to be used with expressions that have side effects, such as expressions involving the ++ and -- operators.
Use of the comma operator is generally discouraged, because the resulting code is unclear: as a rule, each line should do one thing, but a line such as x = ++y, --z; does three things.
Nonetheless, it is valid, as you found. You should get a warning, though, if you have the “unused value” warning turned on (as you should), because half of the expression is, in fact, unused—you drop the 74 on the floor. Harmless, but a symptom that this isn't what you meant to do.
What you want is:
floatCountries = 74.2;