I have a static function:
void TextManager::printDialogue(vector<Button*>* options, int optionsCount, vector<string> outputDisplayText)
{
active = true;
buttons = *options;
buttonsCount = optionsCount;
outputText = outputDisplayText;
}
The "buttons" variable is static:
static vector<Button*> buttons;
I make a call to printDialogue in an execute function:
void WorldDoor::execute()
{
vector<Button*> buttons;
buttons.push_back(new CancelButton());
buttons.push_back(new ChangeRoomButton(room));
TextManager::printDialogue(&buttons, 2, messages); //<----
std::vector<Button*>::iterator i = buttons.begin();
for ( i = buttons.begin() ; i < buttons.end(); i++ )
{
delete * i;
}
}
For whatever reason, when I debug and have a break point inside of the printDialogue function, the values in "buttons" are perfectly fine. However, after I leave printDialogue, the strings contained in my buttons go from being readable to giving me an error message saying:
I tried passing a pointer to an array instead of using
vector<Button*>
but it was only reading the first variable. Now it is not reading anything. Could anyone please help?
There is a static member variable called buttons, and also a local variable inside execute() called buttons. You should rename to avoid confusion, otherwise, the local variable will be used instead of the static member variable inside execute().
Edit: Completely misunderstood the question.
When you do this:
vector<int*> vector1;
vector<int*> vector2;
vector1.push_back(new int(5));
vector2 = vector1;
It copies the pointers, not the value of the pointers.
So when you later iterate over the first vector and delete the dynamic memory, both vectors have pointers pointing to the same memory that you deleted, so your second vector is pointing to invalid memory.
If you are using C++11, you can use a vector of unique pointers, and std::move() one vector into another.
Otherwise, you can just call 'clear()' on the vector, without deleting the memory.
Here's how the function could be written:
void TextManager::printDialogue(vector<Button*>* options, int optionsCount, vector<string> outputDisplayText)
{
active = true;
buttons = *options;
options->clear(); //<--- Instead of crawling over it and delete-ing it.
buttonsCount = optionsCount;
outputText = outputDisplayText;
}
Everything below this was my misunderstanding the question: (contains other information that might be important)
When you do:
vector<Button*> buttons;
Inside the function, you are creating a new vector called 'buttons', which gets destroyed at the end of the function call.
If you want to access the global one, don't create a new one inside the function, or name them something different.
Example:
int myVar = 100; //Declares a variable called 'myVar' at global scope.
void func()
{
int myVar = 200; //Declares a *different* variable called 'myVar' at function scope.
std::cout << myVar << std::endl; //Prints the one inside the function, not the one outside it.
}
By the way, the variable 'static' shouldn't be used at global scope, unless the variable belongs to a class.
To make a variable global, you just put it outside of any function. To make a class member shared between all instances of that class, you declare it static so that class knows to have all instances share the one variable. It's a different thing. =)
If you have your code split into multiple files, to make a global really-truly global, you have to declare it 'extern' in your header, and not extern in one source file, and have other source files #include the header that externs it. Slightly clunky, but that's how it's done. They are working on a better system for it, but it'll be several years before it becomes standardized.
Related
The problem statement is:
Given a Binary Tree, convert this binary tree to a Doubly Linked List.
A Binary Tree (BT) is a data structure in which each node has at most two children.
A Doubly Linked List contains a previous pointer, along with the next pointer and data.
The order of nodes in Doubly Linked List must be the same as Inorder of the given Binary Tree.
The doubly linked list should be returned by taking the next pointer as right and the previous pointer as left.
You need to return the head of the Doubly Linked List.
For example:
4
/ \
2 5
/ \
1 3
The doubly linked list would be: 1 2 3 4 5
My code is:
class BinaryTreeNode
{
public :
T data;
BinaryTreeNode<T> *left;
BinaryTreeNode<T> *right;
BinaryTreeNode(T data) {
this -> data = data;
left = NULL;
right = NULL;
}
};
void inorder(BinaryTreeNode<int>* root,BinaryTreeNode<int>* &prev,BinaryTreeNode<int>* &nroot){
if(!root) return;
inorder(root->left,prev,nroot);
if(prev == NULL) nroot=root;
else{
root->left = prev;
prev->right=root;
}
prev=root;
inorder(root->right,prev,nroot);
}
BinaryTreeNode<int>* BTtoDLL(BinaryTreeNode<int>* root) {
BinaryTreeNode<int>* prev=NULL;
BinaryTreeNode<int>* nroot=NULL;
inorder(root,prev,nroot);
return nroot;
}
I have doubt regarding root.
The root pointer works with passing by value and does not works when it is passed by reference.
When root is passed by reference, it does not work.
void inorder(BinaryTreeNode<int>*& root,BinaryTreeNode<int>*& prev,BinaryTreeNode<int>* &nroot){
if(!root) return;
inorder(root->left,prev,nroot);
if(prev == NULL) nroot=root;
else{
root->left = prev;
prev->right=root;
}
prev=root;
inorder(root->right,prev,nroot);
}
How can I know which variable should be passed by reference and which variable should by passed by value with regard to pointers?
At first glance, one problem is the API of inorder itself. You are assigning nroot = root; to see why this is a problem, consider this simpler example:
void set_x(int *x, int *y)
{
x = y;
}
Remember that pointers are really just memory addresses. So, the assignment of x = y says, "replace the memory address that x contains, and set it to whatever y contains". At no stage do you actually change the value at the memory address.
Now, considering nroot = root again, here you are just giving the local variable you have a new memory address, but not updating anything. So at your call side BTtoDLL, you provide it nullptr for the address, to which it never uses it, and never sets it. So, BTtoDLL will never see any change to nroot.
I want to find out why
x:= odsMap[segRef]
x.GetValue("#OriginDestinationKey")
works, but this does not:
odsMap[segRef].GetValue("#OriginDestinationKey")
?
The last snippet prints the following errors:
cannot call pointer method on odsMap[segRef]go
cannot take the address of odsMap[segRef]
These errors happen during compilation time (not runtime). So, my main question is why I need an intermediate variable x to access the function?
Regarding the type of the variables odsMap is a map[string] XMLElement and segRef is a string.
Thanks.
Map index expressions are not addressable, because the internals of a map may change when a new entry is added to it, so the spec intentionally does not allow taking its address (this gives greater freedom for map implementations).
This means if you store non-pointers in the map, and you want to call a method of a stored value that has a pointer receiver, that would require to take the address of the non-pointer value (to be used as the receiver), but since map index expressions are not addressable, that results in a compile-time error.
A workaround is to store pointer values in the map, so there is no need to take the address of an index expression, because it's already a pointer. An example of this can be seen in this answer: Why should constructor of Go return address? If we have this type:
type My int
func (m *My) Str() string { return strconv.Itoa(int(*m)) }
This gives the compile-time error in question:
m := map[int]My{0: My(12)}
m[0].Str() // Error!
But this works:
m := map[int]*My{}
my := My(12)
m[0] = &my // Store a pointer in the map
m[0].Str() // You can call it, no need to take the address of m[0]
// as it is already a pointer
Another option is to assign it to a local variable whose address can be taken, and call the pointer method on that. Care must be taken though, as if the method has pointer receiver, it might modify pointed object or its components (e.g. fields of a struct), which would not be reflected in the value stored in the map. If you go down this path, you might have to reassign the value to the key in the map to have the updated value.
All-in-all, if you have a value whose type has methods with pointer receiver, you're better off using it (store, pass) as a pointer and not as a non-pointer value.
See related questions:
Pointer methods on non pointer types
How can I store reference to the result of an operation in Go?
#icza's answer is the correct one.
Here is an example to illustrate how "value receiver" vs "pointer receiver" interact with "pointer map" vs "values map" :
https://play.golang.org/p/JVp6DirgPkU
package main
import (
"fmt"
)
// a simple type, with two methods : one with a value receiver, one with a pointer receiver
type Item struct {
name string
}
func (i Item) GetNameByValue() string {
return i.name
}
func (i *Item) GetNameByRef() string {
return i.name
}
func main() {
{
// in this map, we store *pointers* to Item values
mapByRef := make(map[int]*Item)
mapByRef[0] = &Item{"I am stored as a pointer"}
// GetNameByRef will work on a *Item : "mapByRef[0]" is already a pointer
fmt.Println("GetByRef :", mapByRef[0].GetNameByRef())
// GetNameByValue will work on a *Item : go automatically turns this into '(*mapByRef[0]).GetNameByValue()', and this is valid
fmt.Println("GetByValue :", mapByRef[0].GetNameByValue())
}
{
// in this map, we store Item values (no pointers)
mapByValue := make(map[int]Item)
mapByValue[0] = Item{"I am stored as a value"}
// GetNameByValue will work on a Item : "mapByValue[0]" has the right type
fmt.Println("GetByValue :", mapByValue[0].GetNameByValue())
// GetNameByRef will not work : go tries to turn this into : (&mapByValue[0]).GetNameByRef(),
// and go refuses to let you take the address of a value inside a map
// fmt.Println("GetByRef :", mapByValue[0].GetNameByRef())
// compiler error :
// ./prog.go:47:46: cannot call pointer method on mapByValue[0]
// ./prog.go:47:46: cannot take the address of mapByValue[0]
// you will need some way to copy the value before taking its address :
item := mapByValue[0]
fmt.Println("item.GetByRef :", item.GetNameByRef())
// same as :
fmt.Println("(&item).GetByRef :", (&item).GetNameByRef())
}
}
// Output :
//
// GetByRef : I am stored as a pointer
// GetByValue : I am stored as a pointer
// GetByValue : I am stored as a value
// item.GetByRef : I am stored as a value
// (&item).GetByRef : I am stored as a value
When the formal parameter is map, assigning a value directly to a formal parameter cannot change the actual argument, but if you add a new key and value to the formal parameter, the actual argument outside the function can also be seen. Why is that?
I don't understand the output value of the following code, and the formal parameters are different from the actual parameters.
unc main() {
t := map[int]int{
1: 1,
}
fmt.Println(unsafe.Pointer(&t))
copysss(t)
fmt.Println(t)
}
func copysss(m map[int]int) {
//pointer := unsafe.Pointer(&m)
//fmt.Println(pointer)
m = map[int]int{
1: 2,
}
}
stdout :0xc000086010
map[1:1]
func main() {
t := map[int]int{
1: 1,
}
fmt.Println(unsafe.Pointer(&t))
copysss(t)
fmt.Println(t)
}
func copysss(m map[int]int) {
//pointer := unsafe.Pointer(&m)
//fmt.Println(pointer)
m[1] = 2
}
stdout :0xc00007a010
map[1:2]
func main() {
t := map[int]int{
1: 1,
}
fmt.Println(unsafe.Pointer(&t))
copysss(t)
fmt.Println(t)
}
func copysss(m map[int]int) {
pointer := unsafe.Pointer(&m)
fmt.Println(pointer)
m[1] = 2
}
stdout:0xc00008a008
0xc00008a018
map[1:2]
I want to know if the parameter is a value or a pointer.
The parameter is both a value and a pointer.
Wait.. whut?
Yes, a map (and slices, for that matter) are types, pretty similar to what you would implement. Think of a map like this:
type map struct {
// meta information on the map
meta struct{
keyT type
valueT type
len int
}
value *hashTable // pointer to the underlying data structure
}
So in your first function, where you reassign m, you're passing a copy of the struct above (pass by value), and you're assigning a new map to it, creating a new hashtable pointer in the process. The variable in the function scope is updated, but the one you passed still holds a reference to the original map, and with it, the pointer to the original map is preserved.
In the second snippet, you're accessing the underlying hash table (a copy of the pointer, but the pointer points to the same memory). You're directly manipulating the original map, because you're just changing the contents of the memory.
So TL;DR
A map is a value, containing meta information of what the map looks like, and a pointer to the actual data stored inside. The pointer is passed by value, like anything else (same way pointers are passed by value in C/C++), but of course, dereferencing a pointer means you're changing the values in memory directly.
Careful...
Like I said, slices work pretty much in the same way:
type slice struct {
meta struct {
type T
len, cap int
}
value *array // yes, it's a pointer to an underlying array
}
The underlying array is of say, a slice of ints will be [10]int if the cap of the slice is 10, regardless of the length. A slice is managed by the go runtime, so if you exceed the capacity, a new array is allocated (twice the cap of the previous one), the existing data is copied over, and the slice value field is set to point to the new array. That's the reason why append returns the slice that you're appending to, the underlying pointer may have changed etc.. you can find more in-depth information on this.
The thing you have to be careful with is that a function like this:
func update(s []int) {
for i, v := range s {
s[i] = v*2
}
}
will behave much in the same way as the function you have were you're assigning m[1] = 2, but once you start appending, the runtime is free to move the underlying array around, and point to a new memory address. So bottom line: maps and slices have an internal pointer, which can produce side-effects, but you're better off avoiding bugs/ambiguities. Go supports multiple return values, so just return a slice if you set about changing it.
Notes:
In your attempt to figure out what a map is (reference, value, pointer...), I noticed you tried this:
pointer := unsafe.Pointer(&m)
fmt.Println(pointer)
What you're doing there, is actually printing the address of the argument variable, not any address that actually corresponds to the map itself. the argument passed to unsafe.Pointer isn't of the type map[int]int, but rather it's of type *map[int]int.
Personally, I think there's too much confusion around passing by value vs passing by . Go works exactly like C in this regard, just like C, absolutely everything is passed by value. It just so happens that this value can sometimes be a memory address (pointer).
More details (references)
Slices: usage & internals
Maps Note: there's some confusion caused by this one, as pointers, slices, and maps are referred to as *reference types*, but as explained by others, and elsewhere, this is not to be confused with C++ references
In Go, map is a reference type. This means that the map actually resides in the heap and variable is just a pointer to that.
The map is passed by copy. You can change the local copy in your function, but this will not be reflected in caller's scope.
But, since the map variable is a pointer to the unique map residing in the heap, every change can be seen by any variable that points to the same map.
This article can clarify the concept: https://www.ardanlabs.com/blog/2014/12/using-pointers-in-go.html.
I've written a basic Node struct in D, designed to be used as a part of a tree-like structure. The code is as follows:
import std.algorithm: min;
alias Number = size_t;
struct Node {
private {
Node* left, right, parent;
Number val;
}
this(Number n) {val = n;}
this(ref Node u, ref Node v) {
this.left = &u;
this.right = &v;
val = min(u.val, v.val);
u.parent = &this;
v.parent = &this;
}
}
Now, I wrote a simple function which is supposed to give me a Node (meaning a whole tree) with the argument array providing the leaves, as follows.
alias Number = size_t;
Node make_tree (Number[] nums) {
if (nums.length == 1) {
return Node(nums[0]);
} else {
Number half = nums.length/2;
return Node(make_tree(nums[0..half]), make_tree(nums[half..$]));
}
}
Now, when I try to run it through dmd, I get the following error message:
Error: constructor Node.this (ulong n) is not callable using argument types (Node, Node)
This makes no sense to me - why is it trying to call a one-argument constructor when given two arguments?
The problem has nothing to do with constructors. It has to do with passing by ref. The constructor that you're trying to use
this(ref Node u, ref Node v) {...}
accepts its arguments by ref. That means that they must be lvalues (i.e. something that can be on the left-hand side of an assignment). But you're passing it the result of a function call which does not return by ref (so, it's returning a temporary, which is an rvalue - something that can go on the right-hand side of an assignment but not the left). So, what you're trying to do is illegal. Now, the error message isn't great, since it's giving an error with regards to the first constructor rather than the second, but regardless, you don't have a constructor which matches what you're trying to do. At the moment, I can think of 3 options:
Get rid of the ref on the constructor's parameters. If you're only going to be passing it the result of a function call like you're doing now, having it accept ref doesn't help you anyway. The returned value will be moved into the function's parameter, so no copy will take place, and ref isn't buying you anything. Certainly, assigning the return values to local variables so that you can pass them to the constructor as it's currently written would lose you something, since then you'd be making unnecessary copies.
Overload the constructor so that it accepts either ref or non-ref. e.g.
void foo(ref Bar b) { ... }
void foo(Bar b) { foo(b); } //this calls the other foo
In general, this works reasonably well when you have one parameter, but it would be a bit annoying here, because you end up with an exponential explosion of function signatures as you add parameters. So, for your constructor, you'd end up with
this(ref Node u, ref Node v) {...}
this(ref Node u, Node v) { this(u, v); }
this(Node u, ref Node v) { this(u, v); }
this(Node u, Node v) { this(u, v); }
And if you added a 3rd parameter, you'd end up with eight overloads. So, it really doesn't scale beyond a single parameter.
Templatize the constructor and use auto ref. This essentially does what #2 does, but you only have to write the function once:
this()(auto ref Node u, auto ref Node v) {...}
This will then generate a copy of the function to match the arguments given (up to 4 different versions of it with the full function body in each rather than 3 of them just forwarding to the 4th one), but you only had to write it once. And in this particular case, it's probably reasonable to templatize the function, since you're dealing with a struct. If Node were a class though, it might not make sense, since templated functions can't be virtual.
So, if you really want to be able to pass by ref, then in this particular case, you should probably go with #3 and templatize the constructor and use auto ref. However, personally, I wouldn't bother. I'd just go with #1. Your usage pattern here wouldn't get anything from auto ref, since you're always passing it two rvalues, and your Node struct isn't exactly huge anyway, so while you obviously wouldn't want to copy it if you don't need to, copying an lvalue to pass it to the constructor probably wouldn't matter much unless you were doing it a lot. But again, you're only going to end up with a copy if you pass it an lvalue, since an rvalue can be moved rather than copied, and you're only passing it rvalues right now (at least with the code shown here). So, unless you're doing something different with that constructor which would involve passing it lvalues, there's no point in worrying about lvalues - or about the Nodes being copied when they're returned from a function and passed into the constructor (since that's a move, not a copy). As such, just removing the refs would be the best choice.
I'm having trouble wrapping my head around how pointers, slices, and interfaces interact in Go. This is what I currently have coded up:
type Loader interface {
Load(string, string)
}
type Foo struct {
a, b string
}
type FooList []Foo
func (l FooList) Load(a, b string) {
l = append(l, Foo{a, b})
// l contains 1 Foo here
}
func Load(list Loader) {
list.Load("1", "2")
// list is still nil here
}
Given this setup, I then try to do the following:
var list FooList
Load(list)
fmt.Println(list)
However, list is always nil here. My FooList.Load function does add an element to the l slice, but that's as far as it gets. The list in Load continues to be nil. I think I should be able to just pass the reference to my slice around and have things append to it. I'm obviously missing something on how to get it to work though.
(Code in http://play.golang.org/p/uuRKjtxs9D)
If you intend your method to make changes, you probably want to use a pointer receiver.
// We also define a method Load on a FooList pointer receiver.
func (l *FooList) Load(a, b string) {
*l = append(*l, Foo{a, b})
}
This has a consequence, though, that a FooList value won't itself satisfy the Loader interface.
var list FooList
Load(list) // You should see a compiler error at this point.
A pointer to a FooList value, though, will satisfy the Loader interface.
var list FooList
Load(&list)
Complete code below:
package main
import "fmt"
/////////////////////////////
type Loader interface {
Load(string, string)
}
func Load(list Loader) {
list.Load("1", "2")
}
/////////////////////////////
type Foo struct {
a, b string
}
// We define a FooList to be a slice of Foo.
type FooList []Foo
// We also define a method Load on a FooList pointer receiver.
func (l *FooList) Load(a, b string) {
*l = append(*l, Foo{a, b})
}
// Given that we've defined the method with a pointer receiver, then a plain
// old FooList won't satisfy the Loader interface... but a FooList pointer will.
func main() {
var list FooList
Load(&list)
fmt.Println(list)
}
I'm going to simplify the problem so it's easier to understand. What is being done there is very similar to this, which also does not work (you can run it here):
type myInt int
func (a myInt) increment() { a = a + 1 }
func increment(b myInt) { b.increment() }
func main() {
var c myInt = 42
increment(c)
fmt.Println(c) // => 42
}
The reason why this does not work is because Go passes parameters by value, as the documentation describes:
In a function call, the function value and arguments are evaluated in the usual
order. After they are evaluated, the parameters of the call are passed by value
to the function and the called function begins execution.
In practice, this means that each of a, b, and c in the example above are pointing to different int variables, with a and b being copies of the initial c value.
To fix it, we must use pointers so that we can refer to the same area of memory (runnable here):
type myInt int
func (a *myInt) increment() { *a = *a + 1 }
func increment(b *myInt) { b.increment() }
func main() {
var c myInt = 42
increment(&c)
fmt.Println(c) // => 43
}
Now a and b are both pointers that contain the address of variable c, allowing their respective logic to change the original value. Note that the documented behavior still holds here: a and b are still copies of the original value, but the original value provided as a parameter to the increment function is the address of c.
The case for slices is no different than this. They are references, but the reference itself is provided as a parameter by value, so if you change the reference, the call site will not observe the change since they are different variables.
There's also a different way to make it work, though: implementing an API that resembles that of the standard append function. Again using the simpler example, we might implement increment without mutating the original value, and without using a pointer, by returning the changed value instead:
func increment(i int) int { return i+1 }
You can see that technique used in a number of places in the standard library, such as the strconv.AppendInt function.
It's worth keeping a mental model of how Go's data structures are implemented. That usually makes it easier to reason about behaviour like this.
http://research.swtch.com/godata is a good introduction to the high-level view.
Go is pass-by-value. This is true for both parameters and receivers. If you need to assign to the slice value, you need to use a pointer.
Then I read somewhere that you shouldn't pass pointers to slices since
they are already references
This is not entirely true, and is missing part of the story.
When we say something is a "reference type", including a map type, a channel type, etc., we mean that it is actually a pointer to an internal data structure. For example, you can think of a map type as basically defined as:
// pseudocode
type map *SomeInternalMapStructure
So to modify the "contents" of the associative array, you don't need to assign to a map variable; you can pass a map variable by value and that function can change the contents of the associative array pointed to by the map variable, and it will be visible to the caller. This makes sense when you realize it's a pointer to some internal data structure. You would only assign to a map variable if you want to change which internal associative array you want it to point to.
However, a slice is more complicated. It is a pointer (to an internal array), plus the length and capacity, two integers. So basically, you can think of it as:
// pseudocode
type slice struct {
underlyingArray uintptr
length int
capacity int
}
So it's not "just" a pointer. It is a pointer with respect to the underlying array. But the length and capacity are "value" parts of the slice type.
So if you just need to change an element of the slice, then yes, it acts like a reference type, in that you can pass the slice by value and have the function change an element and it's visible to the caller.
However, when you append() (which is what you're doing in the question), it's different. First, appending affects the length of the slice, and length is one of the direct parts of the slice, not behind a pointer. Second, appending may produce a different underlying array (if the capacity of the original underlying array is not enough, it allocates a new one); thus the array pointer part of the slice might also be changed. Thus it is necessary to change the slice value. (This is why append() returns something.) In this sense, it cannot be regarded as a reference type, because we are not just "changing what it points to"; we are changing the slice directly.