What does the & symbol mean in Objective-C? I am currently looking at data constucts and am getting really confused by it.
I have looked around the web for it but have not found an answer at all. I know this is possibly a basic Objective-C concept, but I just can't get my head around it.
For example:
int *pIntData = (int *)&incomingPacket[0];
What is the code doing with incoming packet here?
& is the C address-of unary operator. It returns the memory address of its operand.
In your example, it will return the address of the first element of the incomingPacket array, which is then cast to an int* (pointer to int)
Same thing it means in C.
int *pIntData = (int *)&incomingPacket[0];
Basically this says that the address of the beginning of incomingPacket (&incomingPacket[0]) is a pointer to an int (int *). The local variable pIntData is defined as a pointer to an int, and is set to that value.
Thus:
*pIntData will equal to the first int at the beginning of incomingPacket.
pIntData[0] is the same thing.
pIntData[5] will be the 6th int into the incomingPacket.
Why do this? If you know the data you are being streamed is an array of ints, then this makes it easier to iterate through the ints.
This statement, If I am not mistaken, could also have been written as:
int *pIntData = (int *) incomingPacket;
Related
I have trouble understanding what happens when calling &*pointer
int j=8;
int* p = &j;
When I print in my compiler I get the following
j = 8 , &j = 00EBFEAC p = 00EBFEAC , *p = 8 , &p = 00EBFEA0
&*p= 00EBFEAC
cout << &*p gives &*p = 00EBFEAC which is p itself
& and * have same operator precedence.I thought &*p would translate to &(*p)--> &(8) and expected compiler error.
How does compiler deduce this result?
You are stumbling over something interesting: Variables, strictly spoken, are not values, but refer to values. 8 is an integer value. After int i=8, i refers to an integer value. The difference is that it could refer to a different value.
In order to obtain the value, i must be dereferenced, i.e. the value stored in the memory location which i stands for must be obtained. This dereferencing is performed implicitly in C whenever a value of the type which the variable references is requested: i=8; printf("%d", i) results in the same output as printf("%d", 8). That is funny because variables are essentially aliases for addresses, while numeric literals are aliases for immediate values. In C these very different things are syntactically treated identically. A variable can stand in for a literal in an expression and will be automatically dereferenced. The resulting machine code makes that very clear. Consider the two functions below. Both have the same return type, int. But f has a variable in the return statement which must be dereferenced so that its value can be returned (in this case, it is returned in a register):
int i = 1;
int g(){ return 1; } // literal
int f(){ return i; } // variable
If we ignore the housekeeping code, the functions each translate into a sigle machine instruction. The corresponding assembler (from icc) is for g:
movl $1, %eax #5.17
That's pretty starightforward: Put 1 in the register eax.
By contrast, f translates to
movl i(%rip), %eax #4.17
This puts the value at the address in register rip plus offset i in the register eax. It's refreshing to see how a variable name is just an address (offset) alias to the compiler.
The necessary dereferencing should now be obvious. It would be more logical to write return *i in order to return 1, and write return i only for functions which return references — or pointers.
In your example it is indeed illogical to a degree that
int j=8;
int* p = &j;
printf("%d\n", *p);
prints 8 (i.e, p is actually dereferenced twice); but that &(*p) yields the address of the object pointed to by p (which is the address value stored in p), and is not interpreted as &(8). The reason is that in the context of the address operator a variable (or, in this case, the L-value obtained by dereferencing p) is not implicitly dereferenced the way it is in other contexts.
When the attempt was made to create a logical, orthogonal language — Algol68 —, int i=8 indeed declared an alias for 8. In order to declare a variable the long form would have been refint m = loc int := 3. Consequently what we call a pointer or reference would have had the type ref ref int because actually two dereferences are needed to obtain an integer value.
j is an int with value 8 and is stored in memory at address 00EBFEAC.
&j gives the memory address of variable j (00EBFEAC).
int* p = &j Here you define a variable p which you define being of type int *, namely a value of an address in memory where it can find an int. You assign it &j, namely an address of an int -> which makes sense.
*p gives you the value associated with the address stored in p.
The address stored in p points to an int, so *p gives you the value of that int, namely 8.
& p is the address of where the variable p itself is stored
&*p gives you the address of the value the memory address stored in p points to, which is indeed p again. &(*p) -> &(j) -> 00EBFEAC
Think about &j itself (or even &(j)). According to your logic, shouldn't j evaluate to 8 and result in &8, as well? Dereferencing a pointer or evaluating a variable results in an lvalue, which is a value that you can assign to or take the address of.
The L in "lvalue" refers to the left in "left hand side of the assignment", such as j = 10 or *p = 12. There are also rvalues, such as j + 10, or 8, which obviously cannot be assigned to.
That's just a basic explanation. In C++ there's a lot more to it, with various classes of values (but that thread might be too advanced for your current needs).
The following is very slow for long strings:
std::string s = "long string";
K klist = DBVec::CreateList(KG , s.length());
for (int i=0; i<s.length(); i++)
{
kG(klist)[i]=s.c_str()[i];
}
It works acceptably fast (<100ms) for strings up to 100k, but slows to a crawl (tens of minutes, possibly hours) for strings of a few million characters. I don't see anything other than kG that can create nonlinearity. I don't see any reason for accessor function kG to be non-constant time, but there is just nothing else in this loop. Unfortunately I don't know how kG works due to lack of documentation.
Question: given a blob of binary data as std::string, what's the efficient way to construct a byte list?
kG is a macro defined in k.h which expands to ((x)->G0), i.e. follow the G0 pointer of the K object
http://kx.com/q/d/a/c.htm#Strings documents kp, which creates a K string object directly from a string, so presumably you could do K klist = kp(s.c_str()), which is probably faster
This works:
memcpy(kG(klist), s.c_str(), s.length());
Still wonder why that loop is not O(N).
int k[4] = {1,2,3,4};
int kk[4];
kk=k;
I get incompatible types in the assignment??
Do I have to loop and assign each value in the array, or is there an easier way?
Thank you
You have to loop, or use a library call.
One option would be memcpy(kk, k, sizeof(k));. For this you must #include <string.h>.
Yes. Arrays don't behave like primitive data types in C. You have to loop over the array and assign each value. You can't just assign one array to another (especially since kk is really an int pointer). Not too hard though:
int i;
for (i = 0; i < 4; i++)
kk[i] = k[i];
There is no copy assignment of good old C arrays. So yes, you have to loop through and copy each element.
I've been studying objective-c for a few days now. I keep coming across these two structures:
NSString * somestring
(NSString *) somestring
I understand the first simply sets a pointer to an NSString object, but what does the second construct do, and when should I use it?
What's the deal with the asterix marks?
Sorry if this question doesn't make any sense, I am completely new to this language, and haven't even reached the level of asking proper questions.
Main purpose -- I'm trying to decipher this method:
-(NSString *)pickerView:(UIPickerView *)pickerView titleForRow:(NSInteger) row forComponent: (NSInteger)component
**Also, the classes I chose don't matter. **
What you want to understand is the method declaration syntax. In Objective-C, a method has a name which looks like this doSomethingWith:andAlso:inMode:. Each colon precedes an argument. When you declare or define a method, you also specify the types of the variables and their names; the type is given in parentheses. You also prepend things with a - for instance methods and a + for static methods, as well as the (parenthesized) return type. Now, you have
- (NSString*) pickerView:(UIPickerView*)pickerView
titleForRow:(NSInteger)row
forComponent:(NSInteger)component
{
...
}
When we decipher this, we find that:
-: It's an instance method.
(NSString*): The return type, NSString*.
pickerView:: The first part of the method name.
(UIPickerView*)pickerView: The first argument; its name is pickerView and it has type UIPickerView*.
titleForRow:: The second part of the method name.
(NSInteger)row: The second argument; its name is row and its type is NSInteger.
forComponent:: The third part of the method name.
(NSInteger)component: The third argument; its name is component and its type is NSInteger.
Thus, putting it all together, this defines the instance method pickerView:titleForRow:forComponent:; it returns an NSString*, and the three arguments it takes are of type UIPickerView*, NSInteger, and NSInteger, respectively. This method would be called as [obj pickerView:myPV titleForRow:myRow forComponent:myComponent].
And just for further reference: in isolation, if you have NSString* str, it declares a variable of type NSString*; and if you have (NSString*)obj, it (forcefully) converts that object to have type NSString*. This has no connection to the method declaration syntax.
Edit 1: I also saw that you were asking about the asterisks. In (Objective-)C, this represents a pointer. If you have an int x, say, then when you write int y = x and then y = 3, the original value of x is unchanged. If, however, you have an int* px, then you can write px = &x. (Note that ints, declared as int, are a completely different data type than int pointers declared as int*. Writing int y = &x is garbage, as is int* py = x, and so on.) This & is the "address of" operator; it finds where x is in memory and returns it. Now, if you write int* py = px and then py = &y, this won't change px. But if you write *px, you access the value currently stored in x, and you can change it: *px = 42 sets x to 42. For various reasons, when working with objects, people tend to like to work with references to them instead of their actual values; thus, in Objective-C, you only handle objects through pointers. What this means is that you will never see NSMutableArray x, only NSMutableArray* x; and that if you have NSMutableArray* y = x, then x and y are, roughly speaking, the same, and calling [x addObject:obj] affects y as well. There are more comprehensive tutorials out there—it's worth checking them out if you don't understand pointers—but this should suffice as an overview.
Edit 2: In another comment, you say you're coming from Ruby and Python. In Ruby (and I think Python, but I've used it less), every variable is a reference. This means that the basic use of pointers for object types should be familiar; as long as you never use & or *, they'll function in pretty much the same way. The difference between pointers and references is that you can take references to objects and create pointers to pointers. For instance, many methods end in ...error:(NSDictionary**)error. This effectively provides an extra return value; in the method, if something goes wrong, they can write *error = myErrorInfo. Since function arguments are copied, error = myErrorInfo wouldn't be visible; however, the pointer's referent is still the same, and so it can be assigned to. If you then write code such as:
NSDictionary* errorDict = nil;
[obj unsafeOperation:#"http://stackoverflow.com" error:&errorDict];
You pass in a pointer to errorDict so that the unsafeOperation:error: method can write to errorDict and you can see it.
mebFace - an asterisk represents a pointer. So whenever you see (eg) "NSString *" then it means you're dealing with a pointer to a NSString object. The method you're trying to decipher returns a pointer to a string as the result. It takes three parameters: a pointer to a UIPickerView, followed by a couple of integers. The reason that you don't use an asterisk in conjunction with the integers is because they are simple integers - not objects.
You should probably start off by reading a book on the C programming language. This is what Objective-C is built on top of.
Dave
The first one is variable declaration while second is type-casting. I.e. somestring is not of type NSString* so you type-cast it into being NSString*. I believe that the latter case is mostly (if not always) seen on the right side of assignment statement.
In ObjectiveC this comes from C and is also used in many other strictly-typed languages like Java, C# and so on.
In the updated question, your method signature defines that method returns pointer to NSString - (NSString *). And it accepts 3 parameters:
Pointer to UIPickerView object;
number of row;
number of column;
Confusingly, in Objective-c, both of your examples mean different things in different contexts.
When declaring a variable (and other places, but not everywhere else) the asterisk means 'pointer to', ie…
int foo; // An int variable
int *bar; // A pointer to an int
When it comes to objective-c objects we mostly deal with pointers to them, ie.. you will not see this..
NSString myString;
you will always see this
NSString *myString;
The reason why is quite involved, but is to do with where the memory for the object is allocated (the heap) and what happens when you pass a variable to a function or method. A value passed as an argument to a method is copied. It turns out that copying our string when we pass it to a method is probably not what we intended, so we pass a pointer to the string instead, and that is copied. Imagine a string that contains the complete works of Shakesphere, roughly 5 million bytes. It is inefficient to copy it every time we pass it to a method, but a pointer to it is only 4 bytes, and a pointer gives us just as good access to the data because it, well, points to it. This approach wouldn't get us anything for the int argument tho, as the int and the pointer to the int are going to be the same size (or ballpark, at least).
Your second example is a method or a method declaration. It might help to compare it to function definition in something like javascript:
function pickerViewTitleForRowForComponent( pickerView, row, component ) {
...do something here
return result;
}
Objective-c does things slightly better, sorry, i mean differently. The above would basically look like
- pickerView: pickerView titleForRow: row forComponent: component {
...do something here
return result;
}
The arguments are the same: pickerView, row and component. However, the above isn't a valid method because it is missing type info. ie. We must explicitly say what type each argument is and also the type of the object returned from the method.
- (return type)pickerView:(argument type)pickerView titleForRow:(argument type)row forComponent:(argument type)component {
...do something here
return result;
}
As you can probably see, in your example:
return type > a pointer to an NSString object -- (NSString *)
argument1 type > a pointer to a UIPickerView object -- (UIPickerView *)
argument2 type > an NSInteger -- (NSInteger)
argument3 type > an NSInteger -- (NSInteger)
I have a HTTP connector in my iPhone project and queries must have a parameter set from the username using the Fowler–Noll–Vo (FNV) Hash.
I have a Java implementation working at this time, this is the code :
long fnv_prime = 0x811C9DC5;
long hash = 0;
for(int i = 0; i < str.length(); i++)
{
hash *= fnv_prime;
hash ^= str.charAt(i);
}
Now on the iPhone side, I did this :
int64_t fnv_prime = 0x811C9DC5;
int64_T hash = 0;
for (int i=0; i < [myString length]; i++)
{
hash *= fnv_prime;
hash ^= [myString characterAtIndex:i];
}
This script doesn't give me the same result has the Java one.
In first loop, I get this :
hash = 0
hash = 100 (first letter is "d")
hash = 1865261300 (for hash = 100 and fnv_prime = -2128831035 like in Java)
Do someone see something I'm missing ?
Thanks in advance for the help !
In Java, this line:
long fnv_prime = 0x811C9DC5;
will yield in fnv_prime the numerical value -2128831035, because the constant is interpreted as an int, which is a 32-bit signed value in Java. That value is then sign-extended when written in a long.
Conversely, in the Objective-C code:
int64_t fnv_prime = 0x811C9DC5;
the 0x811C9DC5 is interpreted as an unsigned int constant (because it does not fit in a signed 32-bit int), with numerical value 2166136261. That value is then written into fnv_prime, and there is no sign to extend since, as far as the C compiler is concerned, the value is positive.
Thus you end up with distinct values for fnv_prime, which explains your distinct results.
This can be corrected in Java by adding a "L" suffix, like this:
long fnv_prime = 0x811C9DC5L;
which forces the Java compiler to interpret the constant as a long, with the same numerical value than what you get with the Objective-C code.
Incidentally, 0x811C9DC5 is not a FNV prime (it is not even prime); it is the 32 bit FNV "offset basis". You will get incorrect hash values if you use this value (and more hash collisions). The correct value for the 32 bit FNV prime is 0x1000193. See http://www.isthe.com/chongo/tech/comp/fnv/index.html
It is a difference in sign extension assigning the 32-bit value 0x811C9DC5 to a 64-bit var.
Are the characters in Java and Objective-c the same? NSString will give you unichars.