Setting Integer value in Objective c - ios

I have recently started programming in iOS.. I am going through a code snippet that declares the following variables:
int rc = 0X00;
sqlite3_stmt *pStmt = 0X00;
FMStatement *stat = 0X00;
BOOL abc = 0X00;
what does this mean?? I read somewhere that setting 0X00 in a reference variable means setting it to NULL (in C). But what does setting a BOOL type variable and an int type variable to 0X00 mean??

I suggest you read up about the basics of programming languages, specifically, C programing with pointers. Objective-C is a superset of C and follows many similar rules.
But to your question:
The 0x in front of the literal values in the code (0x00) specifies that the value is interpreted as hexadecimal rather than decimal. But 0x00(hex) is the same as 0(dec).
int rc = 0x00; //same as int rc = 0;
int is a primitive type in both Obj-C and C that specifies an integer, effectively you are initializing the variable. In the C language you must initialize variables otherwise they could be pointing at a random piece of memory.
Therefore, examine this code:
int a;
int b = 0;
//a is NOT equal to b!
In C, the variable 'a' has not be initialized and therefore its not typically safe to assume that it will be initialized to 0. Always initialize your variable.
If you did a printf, or an NSLog of the variable 'a' you will see that it prints some huge number and it doesnt make sense (sometimes this is compiler dependent)
The same can be said for a BOOL. Although setting a BOOL to 0 is the same as setting it to false;
BOOL flag = 0; //The same as saying BOOL flag = false;
Now for the final part of your code:
FMStatement *stat = 0X00;
Often in Objective-C if you are dealing with pointers and objects you need to initialise the pointer to point at some memory address. The actual memory address is usually determined by the stack/heap and you don't need to worry about that. But you do need to ensure that the pointer isn't pointing to the wrong location (known as a garbage pointer).
To do this, we simply set our pointer to nil. eg:
FMStatement *stat = nil; //This pointer is now safe. Although memory still hasnt been allocated for it yet
This is usually taken care of for you though when you immediately allocate the memory for an object, therefore in this case you don't need to worry about initializing the pointer to nil:
FMStatement *stat = [[FMStatement alloc]init];
Like I said, I recommend you read about basic C programming, allocations, pointers, datatypes, initialising etc, once you have a grasp of this, then move to Objective-C which then builds ontop of it with Object-Oriented stuff.
Good luck.

0X00 is simply 0 in hexadecimal notation. So,
int rc = 0X00;
is the same as
int rc = 0;
Same for BOOL variables, where 0 is the same as NO. Using 0X00 is odd -- it'd make more sense to use 0 or NO where appropriate, and use nil for the pointers.

Related

Passing an object that undergoes default argument promotion to 'va_start'

This is my first Xcode app and objective-c so give me some slack :)
I tried googling on the issue but I cannot see any help regarding Xcode and app development. I added the error masseages after //
- (id)initWithBytes:(int8_t)byte1, ... { //Error: 1. Parameter of type 'int8_t' (aka 'signed char') is declared here
va_list args;
va_start(args, byte1); //Error: Passing an object that undergoes default argument promotion to 'va_start' has undefined behavior
unsigned int length = 0;
for (int8_t byte = byte1; byte != -1; byte = va_arg(args, int)) {
length++;
}
va_end(args);
if ((self = [self initWithLength:length]) && (length > 0)) {
va_list args;
va_start(args, byte1); // Error: Passing an object that undergoes default argument promotion to 'va_start' has undefined behavior
int i = 0;
for (int8_t byte = byte1; byte != -1; byte = va_arg(args, int)) {
_array[i++] = byte;
}
va_end(args);
}
return self;
}
Thank you in advance!!
va_start() saves the pointer to the first argument passed to the function into a va_list.
The arguments themselves are passed via a hardware stack.
The issue with int8_t comes from the way the hardware stack is implemented. (in x86 at least)
Just like the SSE and MMX does, the stack requires elements stored on it to have an alignment equal to a multiple of 16bits, so everything passed to the function WILL have at least 16 bits of size, regardless of its type.
But the problem is va_arg() doesn't know about all that. Historically, it was a macro, and all it does is returning a pointer stored in va_list, and incrementing va_list by sizeof(type).
So, when you retrieve the next argument, the pointer returned does not point to the next argument but one byte before it, or not - depending on whether the va_arg is a macro or a compiler built-in function.
And this is what a warning is about.
IMO at least. Pardon my English, It's my 2nd language.

Objective-C passing object to function is always by reference or or by value?

In objective-c I am passing NSMutableDictionary to function and modifying it inside function it returns modified mutable dictionary :
NSMutableDictionary *obj2 = [[NSMutableDictionary alloc]initWithObjectsAndKeys:#"hello",#"fname",nil];
[self callerDictionary:obj2];
NSLog(#"%#",obj2[#"fname"]);//printing "Hi"
-(void)callerDictionary:(NSMutableDictionary*)obj
{
obj[#"fname"] = #"Hi";
}
Technically, Objective C always passes parameter by value, as does C, but practically when you pass an object you need to pass a pointer. While this pointer is passed by value, the semantics of Objective-C give the same effect as if you had passed an object reference; if you modify the objected that is pointed to by the pointer then you are modifying the same object instance that is pointed to in the calling context. The common terminology used in Objective C programming is "object reference" even though it is really a pointer value.
You can see from the * in the method signature that it is a pointer (or object reference in the common usage). If you are passing an intrinsic type, such as an int then it is passed by value unless you explicitly declare the method as requiring a reference:
For example:
-(void) someFunction:(int *)intPointer {
*intPointer = 5;
}
would be called as
int someInteger = 0;
[self someFunction: &someInteger];
// someInteger is now 5
The distinction between a pointer value and a true object reference can be seen in comparison to Swift which uses true references;
If I have
-(void)someFunction:(NSString *)someString {
int length = [someString length];
}
and then do
NSMutableArray *array = [NSMutableArray new];
[someFunction: (NSString *)array];
I will get a runtime exception since array doesn't have a length method, but the compiler can't confirm the type I am passing since it is a pointer.
If I attempted the equivalent in Swift then I will get a compile time error since it knows that the type coercion will always fail
All objects in Objective C passed by reference.
All C types such as NSUInteger, double etc. passed by value
C and Objective-C always pass parameters by value. Objective-C objects are always accessed through a reference (i.e. a pointer). There is a difference between a variable type (int, pointer, etc.) and the way variables are passed as function parameters. The use of the term reference in both scenarios can cause confusion.
by-value:
void f(int a) {
a = 14;
}
int a = 5;
NSLog(#"%d", a); // prints: 5
f(a);
NSLog(#"%d", a); // prints: 5
The value 5 is printed both times because the function f() is given a copy of the value of a, which is 5. The variable referenced within the function is not the same variable that was passed in; it is a copy.
In C++, you can have functions that take parameters by reference.
by-reference:
void f(int &a) {
a = 14;
}
int a = 5;
NSLog(#"%d", a); // prints: 5
f(a);
NSLog(#"%d", a); // prints: 14
Note the & in the function signature. In C++ (but not C, nor Objective-C), this means that the parameter is passed by reference. What this means is that a reference (pointer) to a is passed to the function. Within the function, the a variable is implicitly dereferenced (remember, it's really a pointer, but you don't treat it as one), and the original a variable declared outside the function is changed.
In C and Objective-C, passing a pointer to a function is functionally equivalent to using a reference parameter in C++. This is because a copy of the address is given to the function (remember, the parameter is still passed by value), and that address points to the same object instance that the original pointer does. The reason you don't see any explicit pointer dereferencing within the function (similar to the C++ reference) is because Objective-C syntax for object access always implicitly dereferences -- being within a function doesn't change this behavior.

How to convert switch state into integer in ios

I am using five switches for handling different types of notifications. To remember the state of the switch, I am thinking of converting state of five switches into an integer. For example, if my switches status is as follow, 01010 then the integer should be 10. Please help me how to achieve this.
At first extract each switch value and store it in a single string
Now convert the string to decimal /integer value like this:-
NSString * binarystring = #"01010";
long decimalValue = strtol([binarystring UTF8String], NULL, 2);
NSLog(#"%ld", decimalValue );
Edit
Get all switch control value in single string:-
NSString *binarystring = [[NSString alloc] initWithFormat:#"%i%i%i%i%i",self.switch1.isOn,self.switch2.isOn,self.switch3.isOn,self.switch4.isOn,self.switch5.isOn];
(Why bother encoding your 5 switch values into a single integer? Storing 5 Booleans is not hard. That said the question is how to do it...)
Important aside: BOOL values are not 0 and 1
Objective-C is a superset of C, and in the original C there was no Boolean type - instead it just used an integer type with the interpretation that 0 was false and anything else was true.
Objective-C defines BOOL as a signed char, that is an 8-bit signed integer type (as characters are just an integer type in C). So in Objective-C 0 is false, and -128..-1, 1..127 are all true. NO is defined as 0 and YES as 1, but various operations may result in other values.
To get a 0 or 1 from a BOOL b you can use the conditional operator:
b ? 1 : 0
However the built in logical operators by definition will always return 0 or 1 and never any of the other possible values. The ! operator is logical not, and two not's get you back to where you started so:
!!b
will also give you a 0 or 1.
In any code that takes a BOOL and tries to use it as a 0 or 1 you should really use one of the above (or an equivalent).
One way to solve it: using strings
Your question has been interpreted as using a string as an intermediary during the encoding. First assume the class has your five buttons stored in an instance variable as a simple array (it will allow us to loop):
const int kSWITCH_COUNT = 5; // let's not hard code it everywhere
#implemention MyClass
{
Switch *switches[kSWITCH_COUNT];
}
then the string method goes something like:
- (void) stringMethod
{
NSMutableString *binarystring = NSMutableString.new;
// build up the string one value at a time, note the !! so we only get 0 or 1 values
for (int ix = 0; ix < kSWITCH_COUNT; ix++)
[binarystring appendFormat:#"%d", !!switches[ix].isOn];
long decimalValue = strtol([binarystring UTF8String], NULL, 2);
NSLog(#"Encoded: 0x%lx", decimalValue);
}
This method works, but it is rather a circuitous way of getting to the result - you have 5 integer (Boolean) values and you want to combine them into an integer, why involve strings?
A better way to solve it: using integers
(Objective-)C provides bitwise operators to do shifts, or, and, etc. operations which treat integer types as an ordered collection of bits - which is what they are on a computer.
The << operator shifts left, e.g. 0x1 << 1 produces 0x2, i.e. << 1 is equivalent to multiplication by 2. The | operator is bitwise or, e.g. 0x1 << 1 | 1produces0x3`. The answer to your question now follows easily:
- (void) shiftMethod
{
unsigned int encoded = 0;
for (int ix = 0; ix < kSWITCH_COUNT; ix++)
encoded = (encoded << 1) | !!switches[ix].isOn;
NSLog(#"Encoded: 0x%x", encoded);
}
If you don't like shifts and ors you can use multiplication and addition:
encoded = encoded * 2 + !!switches[ix].isOn;
The above solves the problem directly, no converting to/from intermediate strings. It happens to be a lot faster as well, but in the overall scheme of an application neither approach is probably going to take a significant proportion of the execution time and you shouldn't select based on that.
A Third Way
If you are going to wish to set/get the individual bits of an integer a lot you can use struct types with bit-field widths. These let you set/get the bits of an integer directly - no shifting etc. required - and you may find them useful, but they are rather "low level". Any good book on C will show you how to use these.
HTH

How do I allocate an array at runtime in Rust?

Once I have allocated the array, how do I manually free it? Is pointer arithmetic possible in unsafe mode?
Like in C++:
double *A=new double[1000];
double *p=A;
int i;
for(i=0; i<1000; i++)
{
*p=(double)i;
p++;
}
delete[] A;
Is there any equivalent code in Rust?
Based on your question, I'd recommend reading the Rust Book if you haven't done so already. Idiomatic Rust will almost never involve manually freeing memory.
As for the equivalent to a dynamic array, you want a vector. Unless you're doing something unusual, you should avoid pointer arithmetic in Rust. You can write the above code variously as:
// Pre-allocate space, then fill it.
let mut a = Vec::with_capacity(1000);
for i in 0..1000 {
a.push(i as f64);
}
// Allocate and initialise, then overwrite
let mut a = vec![0.0f64; 1000];
for i in 0..1000 {
a[i] = i as f64;
}
// Construct directly from iterator.
let a: Vec<f64> = (0..1000).map(|n| n as f64).collect();
It is completely possible to allocate a fixed-sized array on the heap:
let a = Box::new([0.0f64; 1000]);
Because of deref coercion, you can still use this as an array:
for i in 0..1000 {
a[i] = i as f64;
}
You can manually free it by doing:
std::mem::drop(a);
drop takes ownership of the array, so this is completely safe. As mentioned in the other answer, it is almost never necessary to do this, the box will be freed automatically when it goes out of scope.

Why the need to use enif_compare?

Why does
int enif_compare(ERL_NIF_TERM lhs, ERL_NIF_TERM rhs)
need to be used instead of just
if( lhs == rhs ) return 1;
I believe it matters that I am comparing atoms.
ERL_NIF_TERM is an opaque datatype and, to the best of my knowledge, is more akin to a pointer than a value. In fact, here's the definition: typedef unsigned long ERL_NIF_TERM (technically there are a few variants, but they're all integers with the same size as a memory address on the system)
So, you must use enif_compare for the same reason you must use str_cmp when comparing C strings: the referenced values may be identical, but the values you get are not representative of that.
Consider:
char a[] = "test";
char b[] = "test";
printf("%d\n", a == b);
Logically, you and I know that the strings are identical, but the values a and b are actually pointers to the contained value. So when you do a regular compare (==), it's comparing the pointers, not the underlying value. Since they are distinct values within the function, they are allocated to different memory addresses, and as a result, a != b, but str_cmp(a, b) == true

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