Can somebody explain this to me:
(((((*(parent))->leaves))[7])->zone_id)
pointers to pointers are really confusing to me. This is the expression I got from the watch in debug mode. I am trying to access to the string zone_id inside the last element of tree (with 10 elements, each element for different number 0-9).
EDIT: this is the whole search function, hope it is enough to understand:
string Tree::search(string str, node** parent, int num) {
int value;
if (num < str.length()) {
value = boost::lexical_cast<int> (str.substr(num, 1));
if ((*parent)->leaves[value] != NULL && num != str.length() -1) {
search (str, &((*parent)->leaves[value]), num+1);
} else if (num == str.length() -1) {
if ( (*(parent)->leaves)[value]->zone_id.empty() )
cout<<"Yep.";
else
return (string) "No_results.";
}
}
}
and structure:
struct node {
string zone_id;
node* leaves [10];
};
Well let's get rid of some brackets to simplify it a bit:
(*parent)->leaves[7]->zone_id
Now it's much easier to understand. We are dereferencing parent (*parent) which gives us a pointer to some object that has an array member called leaves. So we access the element of that array with index 7, which gives us another pointer, this time pointing to an object that has a member called zone_id. We then access that zone_id member.
This is all assuming there's no operator overloading involved.
Diagrammatically (an arrow is "points to"):
________ _________ ___________ ___________
| parent |-->| *parent |-->| struct: | ,-->| struct: |
|________| |_________| | leaves[0] | | | zone_id |
| leaves[1] | | | ... |
| leaves[2] | |
| leaves[3] | |
| leaves[4] | |
| leaves[5] | |
| leaves[6] | |
| leaves[7] | --'
| leaves[8] |
| ... |
Removing the parenthesis makes it actually easier to read, in my mind:
(*parent)->leaves[7]->zone_id
So, we have a pointer to a pointer of leaves. (*parent) makes a dereference to that pointer (so fetches what it the pointer points at). So now we have a pointer to leaves, which is an array of 10, so we use element 7, and the pointer here is used to fetch the zone_id.
It does get a bit complicated, but this is far from the most complicated structure I have seen. If it helps you, you could break it down:
Parent *this_parent = *parent;
Leave *leaf = this_parent->leaves[7];
... use leaf->zone_id;
(
(
(
(
(
*(parent)
)
->leaves
)
)
[7]
)
->zone_id
)
dereference parent
access the leaves member
index the 7th element
access the zone_id member.
Related
I'm struggling with the correct mental model and understanding of std::vector.
What I thought I knew
When you create a vector of type T and then reserve N elements for the vector, the compiler basically finds and reserves a contiguous block of memory that is N * sizeof(T) bytes. For example,
// Initialize a vector of int
std::vector<int> intvec;
// Reserve contigious block of 4 4-byte chunks of memory
intvec.reserve(4); // [ | | | ]
// Filling in the memory chunks has obvious behavior:
intvec.push_back(1); // [1| | | ]
intvec.push_back(2); // [1|2| | ]
Then we can access any element in random access time because, if we ask for the kth element of the vector, we simply start at the memory address of the start of the vector and then "jump" k * sizeof(T) bytes to get to the kth element.
Custom Objects
My mental model breaks down for custom objects of unknown/varying size. For example,
class Foo {
public:
Foo() = default;
Foo(std::vector<int> vec): _vec{vec} {}
private:
std::vector<int> _vec;
};
int main() {
// Initialize a vector Foo
std::vector<Foo> foovec;
// Reserve contigious block of 4 ?-byte chunks of memory
foovec.reserve(4); // [ | | | ]
// How does memory allocation work since object sizes are unkown?
foovec.emplace_back(std::vector<int> {1,2}); // [{1,2}| | | ]
foovec.emplace_back(std::vector<int> {1,2,3,4,5}); // [{1,2}|{1,2,3,4,5}| | ]
return 0;
}
Since we don't know the size of each instance of Foo, how does foovec.reserve() allocate memory? Furthermore, how could you achieve random access time we don't know how far to "jump" to get to the kth element?
Your concept of size is flawed. A std::vector<type> has a compile time known size of space it is going to take up. It also has a run time size that it may use (this is allocated at run time and the vector holds a pointer to it). You can picture it laid out like
+--------+
| |
| Vector |
| |
| |
+--------+
|
|
v
+-------------------------------------------------+
| | | | | |
| Element | Element | Element | Element | Element |
| | | | | |
+-------------------------------------------------+
So when you have a vector of things that have a vector in them, each Element becomes the vector and then those point of to their own storage somewhere else like
+--------+
| |
| Vector |
| |
| |
+----+---+
|
|
v
+----+----+---------+---------+
| Object | Object | Object |
| with | with | with |
| Vector | Vector | Vector |
+----+----+----+----+----+----+
| | | +---------+---------+---------+---------+---------+
| | | | | | | | |
| | +--->+ Element | Element | Element | Element | Element |
| | | | | | | |
| | +-------------------------------------------------+
| | +-------------------------------------------------+
| | | | | | | |
| +--->+ Element | Element | Element | Element | Element |
| | | | | | |
| +-------------------------------------------------+
| +-------------------------------------------------+
| | | | | | |
+--->+ Element | Element | Element | Element | Element |
| | | | | |
+---------+---------+---------+---------+---------+
This way all of the vectors are next to each other, but the elements the vectors have can be anywhere else in memory. It is for this reason you don't want to use a std:vector<std::vector<int>> for a matrix. All of the sub vectors get memory to wherever so there is no locality between the rows.
Do note that this applies to all of the allocator aware containers as they do not store the elements inside the container directly. This is not true for std::array as, like a raw array, the elements are part of the container. If you have an std::array<int, 20> then it is at least sizeof(int) * 20 bytes in size.
the size of
class Foo {
public:
Foo() = default;
Foo(std::vector<int> vec): _vec{vec} {}
private:
std::vector<int> _vec;
};
is known and constant, the internal std::vector does the allocation in the heap, so there is no problem to do foovec.reserve(4);
else how a std::vector can be in the stack ? ;-)
The size of your class Foo is known at compile time, the std::vector class has a constant size, as the elements that it hold are allocated on the heap.
std::vector<int> empty{};
std::vector<int> full{};
full.resize(1000000);
assert(sizeof(empty) == sizeof(full));
Both instances of std::vector<int>, empty and full will always have the same size despite holding a different number of elements.
If you want an array which you can not resize, and it's size must be known at compile time, use std::array.
When you create a vector of type T and then reserve N elements for the vector, the compiler basically finds and reserves a contiguous block of memory
The compiler does no such thing. It generates code to request storage from the vector's allocator at runtime. By default this is std::allocator, which delegates to operator new, which will fetch uninitialized storage from the runtime system.
My mental model breaks down for custom objects of unknown/varying size
The only way a user-defined type can actually have unknown size is if it is incomplete - and you can't declare a vector to an incomplete type.
At any point in your code where the type is complete, its size is also fixed, and you can declare a vector storing that type as usual.
Your Foo is complete, and its size is fixed at compile time. You can check this with sizeof(Foo), and sizeof(foovec[0]) etc.
The vector owns a variable amount of storage, but doesn't contain it in the object. It just stores a pointer and the reserved & used sizes (or something equivalent). For example, an instance of:
class toyvec {
int *begin_;
int *end_;
size_t capacity_;
public:
// push_back, begin, end, and all other methods
};
always has fixed size sizeof(toyvec) = 2 * sizeof(int*) + sizeof(size_t) + maybe_some_padding. Allocating a huge block of memory, and setting begin to the start of it, has no effect on the size of the pointer itself.
tl;dr C++ does not have dynamically-resizing objects. The size of an object is fixed permanently by the class definition. C++ does have objects which own - and may resize - dynamic storage, but that isn't part of the object itself.
This may not be elegant. Chiefly because I am relatively new to C++, but this little program I am putting together is stumbling here.
I don't get it. Have I misunderstood arrays? The edited code is:
int diceArray [6][3][1] = {};
...
}else if (y >= xSuccess || x >= xSuccess){
// from here...
diceArray[2][1][0] = diceArray[2][1][0] + 1;
diceArray[2][1][1] = diceArray[2][1][1] + 1;
// ...to here, diceArray[2][2][0] increases by 1. I am not referencing that part of the array at all. Or am I?
}
By using comments I tracked the culprit down to the second expression. If I comment out the first one diceArray[2][2][0] does not change.
Why is diceArray[2][1][1] = diceArray[2][1][1] + 1 causing diceArray[2][2][0] to increment?
I tried..
c = diceArray[2][1][1] + 1;
diceArray[2][1][1] = c;
..as a workaround but it was just the same. It increased diceArray[2][2][0] by one.
You are indexing out of bounds. If I declare such an array
int data [3];
Then the valid indices are
data[0]
data[1]
data[2]
The analog to this is that you declare
int diceArray [6][3][1]
^
But then try to assign to
diceArray[2][1][0]
^
diceArray[2][1][1] // This is out of range
^
Since you are assigning out of range, due to pointer arithmetic you are actually assigning to the next dimension due to striding, etc.
The variable is declared as:
int diceArray [6][3][1] = {};
This is how it looks like in memory:
+---+ -.
| | <- diceArray[0][0] \
+---+ \
| | <- diceArray[0][1] > diceArray[0]
+---+ /
| | <- diceArray[0][2] /
+---+ -'
| | <- diceArray[1][0] \
+---+ \
| | <- diceArray[1][1] > diceArray[1]
+---+ /
| | <- diceArray[1][2] /
+---+ -'
. . .
. . .
. . .
+---+ -.
| | <- diceArray[5][0] \
+---+ \
| | <- diceArray[5][1] > diceArray[5]
+---+ /
| | <- diceArray[5][2] /
+---+ -'
The innermost component of diceArray is an array of size 1.
C/C++ arrays are always indexed starting from 0 and that means the only valid index in and array of size 1 is 0.
During the compilation, a reference to diceArray[x][y][z] is converted using pointer arithmetic to offset x*3*1+y*1+z (int values) using the memory address of diceArray as base.
The code:
diceArray[2][1][1] = diceArray[2][1][1] + 1;
operates on offset 8 (=2*3*1+1*1+1) inside diceArray. The same offset is computed using diceArray[2][2][0], which is a legal access inside the array.
The modern compilers are usually able to detect this kind of errors and warn you on the compilation.
This question already has answers here:
C++ Swapping Pointers
(7 answers)
Closed 6 years ago.
Here I have two swap functions
void kswap(int* a, int* b)
{
int* temp = a;
a = b;
b = temp;
}
void kswap(int* a, int* b)
{
int temp = *a;
*a = *b;
*b = temp;
}
The value only changed inside of the first function,
and the second function change the value permanently..
Can anyone tell me the different between two functions?
I thought as both functions take pointer type through parameter, the value would be changed through both functions..
In function swap, a and b are int *, aka integer pointers, that means
they contain address of an integer in memory. As seen in diagram below:
Memory
==================
+----------------+
| |
+------> | num1 = 5 |
| | |
| +----> | num2 = 6 |
| | | |
| | | |
| | |================|
| | | Function swap |
| | | |
+-(------------ a |
| | |
+------------ b |
| |
+----------------+
Here,
`*a` : should be read as : `value at address contined in a`
`*b` : should be read as : `value at address contined in b`
In first example
In first kswap, after executing below statements,
int* temp = a; /* A pointer which points to same place as 'a' */
a = b; /* 'a' will now point to where 'b' is pointing */
b = temp; /* 'b' will now point to where 'temp' is pointing
* that means where 'a' was previously pointing */
the result is:
Memory
==================
+----------------+
| |
+------> | num1 = 5 | <------+
| | | |
| +----> | num2 = 6 | |
| | | | |
| | | | |
| | |================| |
| | | Function swap | |
| | | | |
+ +------------ a | |
| | | |
+-------------- b | |
| | |
| temp -----------------+
+----------------+
Note that, neither *a or *b is assigned any value, so neither of:
`*a` : that is : `value at address contined in a`
`*b` : that is : `value at address contined in b`
are changed.
So as seen in above picture, num1 is still 5, and num2 is still 6.
Only thing that has happended is that a is pointing to num2, and b is
pointing to num1.
In second example
In second kswap, after executing below statements,
int temp = *a; /* An int variable which will contain the same value as the
* value at adress contained in a */
*a = *b; /* value at address contained in 'a' will be equal to value
* at address contained in 'b' */
*b = temp; /* value at address contained in 'b' will be equal to value
* contained in 'temp' */
the result is:
Memory
==================
+----------------+
| |
+------> | num1 = 6 |
| | |
| +----> | num2 = 5 |
| | | |
| | | |
| | |================|
| | | Function swap |
| | | |
+-(------------ a |
| | |
+------------ b |
| |
| temp = 5 |
+----------------+
Note that, both *a or *b are assigned new value, so both:
`*a` : that is : `value at address contained in a`
`*b` : that is : `value at address contained in b`
are changed.
And as seen in above picture, num1 is now 6, and num2 is now 5. So in the second example, values of variables num1 and num2 are permanently changed.
Assume each function is called as:
void f()
{
int x = 101, y = 999;
kswap(&x, &y);
}
Remember that in C++ arguments are passed by value, so kswap receives the values of the addresses where x, y reside. The rest of the answer is inlined in the code comments below.
The kswap that works.
void kswap(int* a, int* b)
{
int temp = *a; // `a` is the address of `int x`
// `*a` is the integer value at address `a`
// i.e. the value of `x` so temp == 101 now
*a = *b; // same as above `*b` is the value of `y` i.e. 999
// now this integer value is copied to the address where `a` points
// effectively overwriting the old `x` value `101` with `999`
*b = temp; // finally, this copies the value in `temp` i.e. 101
// to the address where `b` points and overwrites
// the old `y` value `999`, which completes the swap
}
The kswap which does not work.
void kswap(int* a, int* b)
{
int* temp = a; // this copies `a` i.e. the address of `x`
// to local variable `temp`
a = b; // this copies `b` to `a`
// since arguments `a` and `b` are pointers and passed by value
// this only modifies the value of variable `a`
// it does **not** change `x` or its address in any way
b = temp; // this copies 'temp' to 'b', same comments as above
// now 'a' holds the address of `y` and `b` holds the address
// of `x` but **neither** 'x' nor 'y' values have been modified
// and pointer variables `a`, `b` go out of scope as soon as
// the function returns, so it's all a big no-op in the end
}
The first function swaps the addresses, but not outside the scope of the function.
The second function swaps the values, and outside the function's scope.
Adding the * to the name, means you want the value, not where it's at.
i am trying to cast a void** pointer to an int** 2D array in C
here is the code that i am trying to work with (with all the extraneous bits removed):
\*assume that i have a data structure called graph with some
*element "void** graph" in it and some element "int order" */
void initialise_graph_data(graph_t *graph)
{
void **graph_data = NULL;
int (*matrix)[graph->order];
size_t size = (graph->order * graph->order) * sizeof(int);
graph_data = safe_malloc(size); /*safe malloc works fine*/
matrix = (int(*)[graph->order])graph_data;
graph->graph = graph_data;
}
when i compile that, it works fine, but gives me a warning that variable 'matrix' is set but not used. i dont really want to have to use the interim matrix variable because the function is just supposed to initialise the array, not put anything in it; but if i try to cast graph_data directly to an int** when i am assiging it to graph->graph like so:
graph->graph = (int(*)[graph->order])graph_data;
it gives me an assignment from incompatible pointer type warning.
am i just not casting it properly? does anyone have any suggestions as to how i can make it work without the interim "matrix" variable? or if not, what i can do with that variable so that it doesnt give me the warning that it is set but not used?
thanks
The compiler is right, an array of arrays (or a pointer to an array) is not the same as a pointer to a pointer. Just think about how they would be laid out in memory:
A matrix of size MxN in the form of an array of arrays:
+--------------+--------------+-----+----------------+--------------+-----+------------------+
| matrix[0][0] | matrix[0][1] | ... | matrix[0][N-1] | matrix[1][0] | ... | matrix[M-1][N-1] |
+--------------+--------------+-----+----------------+--------------+-----+------------------+
A and the same "matrix" in the form of pointer to pointer:
+-----------+-----------+-----------+-----+
| matrix[0] | matrix[1] | matrix[2] | ... |
+-----------+-----------+-----------+-----+
| | |
| | V
| | +--------------+--------------+-----+
| | | matrix[2][0] | matrix[2][1] | ... |
| | +--------------+--------------+-----+
| |
| V
| +--------------+--------------+-----+
| | matrix[1][0] | matrix[1][1] | ... |
| +--------------+--------------+-----+
|
V
+--------------+--------------+-----+
| matrix[0][0] | matrix[0][1] | ... |
+--------------+--------------+-----+
It doesn't matter if you allocate the correct size, the two variables simply are incompatible which is what your compiler is telling you.
I am trying to understand the concept of linked lists. So far this is what I know and where I'm having problems to understand.
//create node
struct list
{
int id; //member var
list* next; //pointer to link next list item
}
//int main()
//create list head and set it to NULL
list* head = NULL;
//instantiate list node
list* newList = new list;
//insert a list
newList->id = 20;
newList->next = NULL;
This I do not really understand what's going on.
newList->next = head;
head = newList;
NOTE: The memory "locations" I will reference in this answer are there purely for example are not meant to mimic the actual location these pointers might or might not ever point to.
Draw these relationships out on paper to visualize the results. Let's break it down by lines.
list *head = NULL;
And here is our visualization:
*head (0x00)
+-----------+
| |
| NULL |
| |
+-----------+
Now, we follow these next lines:
list *newList = new list;
newList->id = 20;
newList->next = NULL;
And that visualization:
*head (0x00) *newList (0x3a)
+-----------+ +----+------+
| | | id | next |
| NULL | +----+------+
| | | 20 | NULL |
+-----------+ +----+------+
And finally we end with your last bit:
newList->next = head;
And that alters the visualization thusly (reordered for clarity):
*newList (0x3a) +->*head (0x00)
+----+------+ | +-----------+
| id | next | | | |
+----+------+ | | NULL |
| 20 | head------+ | |
+----+------+ +-----------+
This has created the "link" that gives a LinkedList it's name. You link nodes together by some form of a reference. So what you've done is created a "head" or beginning of the list, and then you've created a secondary node in the list and placed it (logically) before head. Normally you'd then reassign your reference to newList to head since it's the new beginning of the list.
The next step would likely be (and I'm sure this what you meant with the erroneous bit that I ask about at the end of this question):
head = newList;
Which now changes the visualization to this:
*head (0x3a) +---> (0x00)
+----+------+ | +-----------+
| id | next | | | |
+----+------+ | | NULL |
| 20 | 0x00----+ | |
+----+------+ +-----------+
Also, what about the following line?
head = n; // What is 'n'? Where did you get it from? It doesn't appear anywhere else in your sample
EDIT
I altered the visualization to more accurately reflect what would be realistic. I wasn't paying attention to what I was doing when I posted the original answer so many thanks to José and his comment for bringing it to my attention the visualization was inaccurate.
In addition to altering the visuals I added a bit more information and wanted to take it a step further in saying that here's how you would use this linked list to loop over it's records.
list *node = head;
while (node != NULL) {
std::cout << "The id is " << node->id << std::endl;
node = node->next;
}
newList->next = head;
head = newList;
You start out with head set to NULL.
Then, you create a new node and let its next point to head, which is NULL.
Then, you set head to be the new node. Now, head points to a list that has one item in it.