According to a KNK Modern Approach Textbook, The expression of ++(*p) is that “ Increment *p first; value of expression is *p after increment”.
When I first interpreted this, I thought it would first fetch a value of *p to which pointer points, and then increment the value. So its result is 11 if *p is 10.
Am I wrong?
Related
I noticed this interesting thing about the max() and min() functions in SV LRM (1800-2012) 7.12 (Array manipulation methods). I tried out the max() and min() functions in a dummy SV file
int a[3] = {0,5,5};
int q[$];
int b;
q = a.max(); // legal
b = a.max(); // illegal
The illegal statement error was
Incompatible complex type assignment
Type of source expression is incompatible with type of target expression.
Mismatching types cannot be used in assignments, initializations and
instantiations. The type of the target is 'int', while the type of the
source is 'int$[$]'.
So I commented out the illegal statement and tested it. It compiled and ran fine but I was hoping to get some more insight as to why the function returns a queue and not a single element - I printed out the contents of q and the size, but the size is still 1 and 5 is being printed just once. Kind of redundant then to make the max() and min() functions return a queue ?
The "SystemVerilog for Verification" book by Chris Spear and Greg Tumbush has a good explanation on this topic in Chapter 2.6.2, which I am quoting below:
"The array locator methods find data in an unpacked array. At first
you may wonder why these return a queue of values. After all, there
is only one maximum value in an array. However, SystemVerilog needs a
queue for the case when you ask for a value from an empty queue or
dynamic array."
It returns a queue to deal with empty queues and when the with () conditions have no matches. The the empty queue return is a a way to differentiate a true match from no matches.
Consider the below code. to find the minimum value of a that is greater than 5. a has data but none of its entries have above 5. b is empty, so it will return an empty. c will return 7.
int a[3] = '{0,5,5};
int b[$] = '{};
int c[4] = '{0,15,5,7};
int q[$];
q = a.min() with (item > 5); // no items >5, will return an empty queue
q = b.min(); // queue is empty, will return an empty queue
q = c.min() with (item > 5); // will return a queue of size 1 with value 7
I believe the example results as per Greg's answer is not correct.
As per System Verilog Language:
min() returns the element with the minimum value or whose expression evaluates to a minimum.
max() returns the element with the maximum value or whose expression evaluates to a maximum.
So, when with expression is evaluated, the resultant value will be:
a.min() with (item > 5); {0,0,0} -> Minimum is 0 and corresponding item is 5.
c.min() with (item > 5); {0,1,0,1}-> Minimum is 0 and corresponding item is 5.
Since, example demonstrates the usage of min, the result will be:
q = a.min() with (item > 5); // A queue of size 1 with value 5.
q = c.min() with (item > 5); //A queue of size 1 with value 5.
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).
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.
I have two NSStrings with the same value
this failed for me:
if (button.controlName == controlName) {
return button;
}
this worked:
if ([button.controlName compare: controlName] == NSOrderedSame) {
return button;
}
Is this just how strings are compared in objective c? Or should the first statement have worked as well? Why might the first statement have failed? I know it worked for other strings.
The strings it does not work for are initialized like this:
button.controlName = [NSString stringWithFormat:#"controlName%d", i]
With NSString you should use isEqualToString: instead of compare.
[button.controlName isEqualToString:controlName]
Read more the why (and also why it worked for some other strings)
Objective-C is a fairly thin layer on top of standard C. As a result obj-c, just as in normal c, doesn't have operator overloading.
NSString *controlName = #"bobDole";
The above code creates a pointer to the string #"bobDole", controlName is not the value itself, but instead is really just a long integer that says the memory address of an object.
When using pointers and comparing them using the == operator like (mutableCopy is being used to prevent the compiler from optimizing out the validity of this example.)
NSString *string1 = #"bobDole";
NSString *string2 = [string1 mutableCopy];
NSLog(#"%d", string1 == string2);
The above code will always print false (or zero in this case), even though both objects are NSStrings, and both contain the value of #"bobDole". This is because the value of string1 is actually a hex number like 0x0123456 and string2 could be something like 0x0987654. So really the above comparison looks like this to the computer:
NSLog(#"%d", 0x0123456 == 0x0987654);
So when comparing strings (or any other object), always use one of the isEqual methods, never use the == operator.
Now as to why it worked for some other strings:
As mentioned above when using the == operator you're actually doing pointer comparison. You'll also notice in my above example I used mutableCopy instead of the following:
NSString *string1 = #"bobDole";
NSString *string2 = #"bobDole";
The reason I did such was that the compiler will look at those two statements know they share the same immutable value and optimize them so they point at the same value in memory. Thus making the pointer values of the two identical.
The compiler also makes the same optimizations for these methods of string initialization.
NSString *string3 = [NSString stringWithString:#"bobDole"];
NSString *string4 = [NSString stringWithString:string1];
NSString *string5 = [string1 copy];
Because of this optimization by the compiler and runtime all 5 pointers point to the same memory location and are thus equal to each other when compared via ==.
This may be kinda long, but I tried to make it accessible and understandable. Hope it helps.
1st statement just compares pointers, but not string values, so to compare strings you should use -isEqualToString as Bryan points.
There's a method called isEqualToString to compare two NSStrings.
if([button.controlName isEqualToString:controlName])
...
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;