Embarrassing though it may be I know I am not the only one with this problem.
I have been using C/C++ on and off for many years. I never had a problem grasping the concepts of addresses, pointers, pointers to pointers, and references.
I do constantly find myself tripping over expressing them in C syntax, however. Not the basics like declarations or dereferencing, but more often things like getting the address of a pointer-to-pointer, or pointer to reference, etc. Essentially anything that goes a level or two of indirection beyond the norm. Typically I fumble with various semi-logical combinations of operators until I trip upon the correct one.
Clearly somewhere along the line I missed a rule or two that simplifies and makes it all fall into place. So I guess my question is: do you know of a site or reference that covers this matter with clarity and in some depth?
I don't know of any website but I'll try to explain it in very simple terms. There are only three things you need to understand:
variable will contain the contents of the variable. This means that if the variable is a pointer it will contain the memory address it points to.
*variable (only valid for pointers) will contain the contents of the variable pointed to. If the variable it points to is another pointer, ptr2, then *variable and ptr2 will be the same thing; **variable and *ptr2 are the same thing as well.
&variable will contain the memory address of the variable. If it's a pointer, it will be the memory address of the pointer itself and NOT the variable pointed to or the memory address of the variable pointed to.
Now, let's see a complex example:
void **list = (void **)*(void **)info.List;
list is a pointer to a pointer. Now let's examine the right part of the assignment starting from the end: (void **)info.List. This is also a pointer to a pointer.
Then, you see the *: *(void **)info.List. This means that this is the value the pointer info.List points to.
Now, the whole thing: (void **)*(void **)info.List. This is the value the pointer info.List points to casted to (void **).
I found the right-left-right rule to be useful. It tells you how to read a declaration so that you get all the pointers and references in order. For example:
int *foo();
Using the right-left-right rule, you can translate this to English as "foo is a function that returns a pointer to an integer".
int *(*foo)(); // "foo is a pointer to a function returning a pointer to an int"
int (*foo[])(); // "foo is an array of pointers to functions returning ints"
Most explanations of the right-left-right rule are written for C rather than C++, so they tend to leave out references. They work just like pointers in this context.
int &foo; // "foo is a reference to an integer"
Typedefs can be your friend when things get confusing. Here's an example:
typedef const char * literal_string_pointer;
typedef literal_string_pointer * pointer_to_literal_string_pointer;
void GetPointerToString(pointer_to_literal_string_pointer out_param)
{
*out_param = "hi there";
}
All you need to know is that getting the address of an object returns a pointer to that object, and dereferencing an object takes a pointer and turns it into to object that it's pointing to.
T x;
A a = &x; // A is T*
B b = *a; // B is T
C c = &a; // C is T**
D d = *c; // D is T*
Essentially, the & operator takes a T and gives you a T* and the * operator takes a T* and gives you a T, and that applies to higher levels of abstraction equally e.g.
using & on a T* will give you a T**.
Another way of thinking about it is that the & operator adds a * to the type and the * takes one away, which leads to things like &&*&**i == i.
I'm not sure exactly what you're looking for, but I find it helpful to remember the operator precedence and associativity rules. That said, if you're ever confused, you might as well throw in some more parens to disambiguate, even if it's just for your benefit and not the compiler's.
edit: I think I might understand your question a little better now. I like to think of a chain of pointers like a stack with the value at the bottom. The dereferencing operator (*) pops you down the stack, where you find the value itself at the end. The reference operator (&) lets you push another level of indirection onto the stack. Note that it's always legal to move another step away, but attempting to dereference the value is analogous to popping an empty stack.
Related
I have searched and searched for an answer to this but can't find anything I actually "get".
I am very very new to c++ and can't get my head around the use of double, triple pointers etc.. What is the point of them?
Can anyone enlighten me
Honestly, in well-written C++ you should very rarely see a T** outside of library code. In fact, the more stars you have, the closer you are to winning an award of a certain nature.
That's not to say that a pointer-to-pointer is never called for; you may need to construct a pointer to a pointer for the same reason that you ever need to construct a pointer to any other type of object.
In particular, I might expect to see such a thing inside a data structure or algorithm implementation, when you're shuffling around dynamically allocated nodes, perhaps?
Generally, though, outside of this context, if you need to pass around a reference to a pointer, you'd do just that (i.e. T*&) rather than doubling up on pointers, and even that ought to be fairly rare.
On Stack Overflow you're going to see people doing ghastly things with pointers to arrays of dynamically allocated pointers to data, trying to implement the least efficient "2D vector" they can think of. Please don't be inspired by them.
In summary, your intuition is not without merit.
An important reason why you should/must know about pointer-to-pointer-... is that you sometimes have to interface with other languages (like C for instance) through some API (for instance the Windows API).
Those APIs often have functions that have an output-parameter that returns a pointer. However those other languages often don't have references or compatible (with C++) references. That's a situation when pointer-to-pointer is needed.
It's less used in c++. However, in C, it can be very useful. Say that you have a function that will malloc some random amount of memory and fill the memory with some stuff. It would be a pain to have to call a function to get the size you need to allocate and then call another function that will fill the memory. Instead you can use a double pointer. The double pointer allows the function to set the pointer to the memory location. There are some other things it can be used for but that's the best thing I can think of.
int func(char** mem){
*mem = malloc(50);
return 50;
}
int main(){
char* mem = NULL;
int size = func(&mem);
free(mem);
}
I am very very new to c++ and can't get my head around the use of double, triple pointers etc.. What is the point of them?
The trick to understanding pointers in C is simply to go back to the basics, which you were probably never taught. They are:
Variables store values of a particular type.
Pointers are a kind of value.
If x is a variable of type T then &x is a value of type T*.
If x evaluates to a value of type T* then *x is a variable of type T. More specifically...
... if x evaluates to a value of type T* that is equal to &a for some variable a of type T, then *x is an alias for a.
Now everything follows:
int x = 123;
x is a variable of type int. Its value is 123.
int* y = &x;
y is a variable of type int*. x is a variable of type int. So &x is a value of type int*. Therefore we can store &x in y.
*y = 456;
y evaluates to the contents of variable y. That's a value of type int*. Applying * to a value of type int* gives a variable of type int. Therefore we can assign 456 to it. What is *y? It is an alias for x. Therefore we have just assigned 456 to x.
int** z = &y;
What is z? It's a variable of type int**. What is &y? Since y is a variable of type int*, &y must be a value of type int**. Therefore we can assign it to z.
**z = 789;
What is **z? Work from the inside out. z evaluates to an int**. Therefore *z is a variable of type int*. It is an alias for y. Therefore this is the same as *y, and we already know what that is; it's an alias for x.
No really, what's the point?
Here, I have a piece of paper. It says 1600 Pennsylvania Avenue Washington DC. Is that a house? No, it's a piece of paper with the address of a house written on it. But we can use that piece of paper to find the house.
Here, I have ten million pieces of paper, all numbered. Paper number 123456 says 1600 Pennsylvania Avenue. Is 123456 a house? No. Is it a piece of paper? No. But it is still enough information for me to find the house.
That's the point: often we need to refer to entities through multiple levels of indirection for convenience.
That said, double pointers are confusing and a sign that your algorithm is insufficiently abstract. Try to avoid them by using good design techniques.
A double-pointer, is simply a pointer to a pointer. A common usage is for arrays of character strings. Imagine the first function in just about every C/C++ program:
int main(int argc, char *argv[])
{
...
}
Which can also be written
int main(int argc, char **argv)
{
...
}
The variable argv is a pointer to an array of pointers to char. This is a standard way of passing around arrays of C "strings". Why do that? I've seen it used for multi-language support, blocks of error strings, etc.
Don't forget that a pointer is just a number - the index of the memory "slot" inside a computer. That's it, nothing more. So a double-pointer is index of a piece of memory that just happens to hold another index to somewhere else. A mathematical join-the-dots if you like.
This is how I explained pointers to my kids:
Imagine the computer memory is a series of boxes. Each box has a number written on it, starting at zero, going up by 1, to however many bytes of memory there is. Say you have a pointer to some place in memory. This pointer is just the box number. My pointer is, say 4. I look into box #4. Inside is another number, this time it's 6. So now we look into box #6, and get the final thing we wanted. My original pointer (that said "4") was a double-pointer, because the content of its box was the index of another box, rather than being a final result.
It seems in recent times pointers themselves have become a pariah of programming. Back in the not-too-distant past, it was completely normal to pass around pointers to pointers. But with the proliferation of Java, and increasing use of pass-by-reference in C++, the fundamental understanding of pointers declined - particularly around when Java became established as a first-year computer science beginners language, over say Pascal and C.
I think a lot of the venom about pointers is because people just don't ever understand them properly. Things people don't understand get derided. So they became "too hard" and "too dangerous". I guess with even supposedly learned people advocating Smart Pointers, etc. these ideas are to be expected. But in reality there a very powerful programming tool. Honestly, pointers are the magic of programming, and after-all, they're just a number.
In many situations, a Foo*& is a replacement for a Foo**. In both cases, you have a pointer whose address can be modified.
Suppose you have an abstract non-value type and you need to return it, but the return value is taken up by the error code:
error_code get_foo( Foo** ppfoo )
or
error_code get_foo( Foo*& pfoo_out )
Now a function argument being mutable is rarely useful, so the ability to change where the outermost pointer ppFoo points at is rarely useful. However, a pointer is nullable -- so if get_foo's argument is optional, a pointer acts like an optional reference.
In this case, the return value is a raw pointer. If it returns an owned resource, it should usually be instead a std::unique_ptr<Foo>* -- a smart pointer at that level of indirection.
If instead, it is returning a pointer to something it does not share ownership of, then a raw pointer makes more sense.
There are other uses for Foo** besides these "crude out parameters". If you have a polymorphic non-value type, non-owning handles are Foo*, and the same reason why you'd want to have an int* you would want to have a Foo**.
Which then leads you to ask "why do you want an int*?" In modern C++ int* is a non-owning nullable mutable reference to an int. It behaves better when stored in a struct than a reference does (references in structs generate confusing semantics around assignment and copy, especially if mixed with non-references).
You could sometimes replace int* with std::reference_wrapper<int>, well std::optional<std::reference_wrapper<int>>, but note that is going to be 2x as large as a simple int*.
So there are legitimate reasons to use int*. Once you have that, you can legitimately use Foo** when you want a pointer to a non-value type. You can even get to int** by having a contiguous array of int*s you want to operate on.
Legitimately getting to three-star programmer gets harder. Now you need a legitimate reason to (say) want to pass a Foo** by indirection. Usually long before you reach that point, you should have considered abstracting and/or simplifying your code structure.
All of this ignores the most common reason; interacting with C APIs. C doesn't have unique_ptr, it doesn't have span. It tends to use primitive types instead of structs because structs require awkward function based access (no operator overloading).
So when C++ interacts with C, you sometimes get 0-3 more *s than the equivalent C++ code would.
The use is to have a pointer to a pointer, e.g., if you want to pass a pointer to a method by reference.
What real use does a double pointer have?
Here is practical example. Say you have a function and you want to send an array of string params to it (maybe you have a DLL you want to pass params to). This can look like this:
#include <iostream>
void printParams(const char **params, int size)
{
for (int i = 0; i < size; ++i)
{
std::cout << params[i] << std::endl;
}
}
int main()
{
const char *params[] = { "param1", "param2", "param3" };
printParams(params, 3);
return 0;
}
You will be sending an array of const char pointers, each pointer pointing to the start of a null terminated C string. The compiler will decay your array into pointer at function argument, hence what you get is const char ** a pointer to first pointer of array of const char pointers. Since the array size is lost at this point, you will want to pass it as second argument.
One case where I've used it is a function manipulating a linked list, in C.
There is
struct node { struct node *next; ... };
for the list nodes, and
struct node *first;
to point to the first element. All the manipulation functions take a struct node **, because I can guarantee that this pointer is non-NULL even if the list is empty, and I don't need any special cases for insertion and deletion:
void link(struct node *new_node, struct node **list)
{
new_node->next = *list;
*list = new_node;
}
void unlink(struct node **prev_ptr)
{
*prev_ptr = (*prev_ptr)->next;
}
To insert at the beginning of the list, just pass a pointer to the first pointer, and it will do the right thing even if the value of first is NULL.
struct node *new_node = (struct node *)malloc(sizeof *new_node);
link(new_node, &first);
Multiple indirection is largely a holdover from C (which has neither reference nor container types). You shouldn't see multiple indirection that much in well-written C++, unless you're dealing with a legacy C library or something like that.
Having said that, multiple indirection falls out of some fairly common use cases.
In both C and C++, array expressions will "decay" from type "N-element array of T" to "pointer to T" under most circumstances1. So, assume an array definition like
T *a[N]; // for any type T
When you pass a to a function, like so:
foo( a );
the expression a will be converted from "N-element array of T *" to "pointer to T *", or T **, so what the function actually receives is
void foo( T **a ) { ... }
A second place they pop up is when you want a function to modify a parameter of pointer type, something like
void foo( T **ptr )
{
*ptr = new_value();
}
void bar( void )
{
T *val;
foo( &val );
}
Since C++ introduced references, you probably won't see that as often. You'll usually only see that when working with a C-based API.
You can also use multiple indirection to set up "jagged" arrays, but you can achieve the same thing with C++ containers for much less pain. But if you're feeling masochistic:
T **arr;
try
{
arr = new T *[rows];
for ( size_t i = 0; i < rows; i++ )
arr[i] = new T [size_for_row(i)];
}
catch ( std::bad_alloc& e )
{
...
}
But most of the time in C++, the only time you should see multiple indirection is when an array of pointers "decays" to a pointer expression itself.
The exceptions to this rule occur when the expression is the operand of the sizeof or unary & operator, or is a string literal used to initialize another array in a declaration.
In C++, if you want to pass a pointer as an out or in/out parameter, you pass it by reference:
int x;
void f(int *&p) { p = &x; }
But, a reference can't ("legally") be nullptr, so, if the pointer is optional, you need a pointer to a pointer:
void g(int **p) { if (p) *p = &x; }
Sure, since C++17 you have std::optional, but, the "double pointer" has been idiomatic C/C++ code for many decades, so should be OK. Also, the usage is not so nice, you either:
void h(std::optional<int*> &p) { if (p) *p = &x) }
which is kind of ugly at the call site, unless you already have a std::optional, or:
void u(std::optional<std::reference_wrapper<int*>> p) { if (p) p->get() = &x; }
which is not so nice in itself.
Also, some might argue that g is nicer to read at the call site:
f(p);
g(&p); // `&` indicates that `p` might change, to some folks
Originally being the topic of this question, it emerged that the OP just overlooked the dereference. Meanwhile, this answer got me and some others thinking - why is it allowed to cast a pointer to a reference with a C-style cast or reinterpret_cast?
int main() {
char c = 'A';
char* pc = &c;
char& c1 = (char&)pc;
char& c2 = reinterpret_cast<char&>(pc);
}
The above code compiles without any warning or error (regarding the cast) on Visual Studio while GCC will only give you a warning, as shown here.
My first thought was that the pointer somehow automagically gets dereferenced (I work with MSVC normally, so I didn't get the warning GCC shows), and tried the following:
#include <iostream>
int main() {
char c = 'A';
char* pc = &c;
char& c1 = (char&)pc;
std::cout << *pc << "\n";
c1 = 'B';
std::cout << *pc << "\n";
}
With the very interesting output shown here. So it seems that you are accessing the pointed-to variable, but at the same time, you are not.
Ideas? Explanations? Standard quotes?
Well, that's the purpose of reinterpret_cast! As the name suggests, the purpose of that cast is to reinterpret a memory region as a value of another type. For this reason, using reinterpret_cast you can always cast an lvalue of one type to a reference of another type.
This is described in 5.2.10/10 of the language specification. It also says there that reinterpret_cast<T&>(x) is the same thing as *reinterpret_cast<T*>(&x).
The fact that you are casting a pointer in this case is totally and completely unimportant. No, the pointer does not get automatically dereferenced (taking into account the *reinterpret_cast<T*>(&x) interpretation, one might even say that the opposite is true: the address of that pointer is automatically taken). The pointer in this case serves as just "some variable that occupies some region in memory". The type of that variable makes no difference whatsoever. It can be a double, a pointer, an int or any other lvalue. The variable is simply treated as memory region that you reinterpret as another type.
As for the C-style cast - it just gets interpreted as reinterpret_cast in this context, so the above immediately applies to it.
In your second example you attached reference c to the memory occupied by pointer variable pc. When you did c = 'B', you forcefully wrote the value 'B' into that memory, thus completely destroying the original pointer value (by overwriting one byte of that value). Now the destroyed pointer points to some unpredictable location. Later you tried to dereference that destroyed pointer. What happens in such case is a matter of pure luck. The program might crash, since the pointer is generally non-defererencable. Or you might get lucky and make your pointer to point to some unpredictable yet valid location. In that case you program will output something. No one knows what it will output and there's no meaning in it whatsoever.
One can rewrite your second program into an equivalent program without references
int main(){
char* pc = new char('A');
char* c = (char *) &pc;
std::cout << *pc << "\n";
*c = 'B';
std::cout << *pc << "\n";
}
From the practical point of view, on a little-endian platform your code would overwrite the least-significant byte of the pointer. Such a modification will not make the pointer to point too far away from its original location. So, the code is more likely to print something instead of crashing. On a big-endian platform your code would destroy the most-significant byte of the pointer, thus throwing it wildly to point to a totally different location, thus making your program more likely to crash.
It took me a while to grok it, but I think I finally got it.
The C++ standard specifies that a cast reinterpret_cast<U&>(t) is equivalent to *reinterpret_cast<U*>(&t).
In our case, U is char, and t is char*.
Expanding those, we see that the following happens:
we take the address of the argument to the cast, yielding a value of type char**.
we reinterpret_cast this value to char*
we dereference the result, yielding a char lvalue.
reinterpret_cast allows you to cast from any pointer type to any other pointer type. And so, a cast from char** to char* is well-formed.
I'll try to explain this using my ingrained intuition about references and pointers rather than relying on the language of the standard.
C didn't have reference types, it only had values and pointer types (addresses) - since, physically in memory, we only have values and addresses.
In C++ we've added references to the syntax, but you can think of them as a kind of syntactic sugar - there is no special data structure or memory layout scheme for holding references.
Well, what "is" a reference from that perspective? Or rather, how would you "implement" a reference? With a pointer, of course. So whenever you see a reference in some code you can pretend it's really just a pointer that's been used in a special way: if int x; and int& y{x}; then we really have a int* y_ptr = &x; and if we say y = 123; we merely mean *(y_ptr) = 123;. This is not dissimilar from how, when we use C array subscripts (a[1] = 2;) what actually happens is that a is "decayed" to mean pointer to its first element, and then what gets executed is *(a + 1) = 2.
(Side note: Compilers don't actually always hold pointers behind every reference; for example, the compiler might use a register for the referred-to variable, and then a pointer can't point to it. But the metaphor is still pretty safe.)
Having accepted the "reference is really just a pointer in disguise" metaphor, it should now not be surprising that we can ignore this disguise with a reinterpret_cast<>().
PS - std::ref is also really just a pointer when you drill down into it.
Its allowed because C++ allows pretty much anything when you cast.
But as for the behavior:
pc is a 4 byte pointer
(char)pc tries to interpret the pointer as a byte, in particular the last of the four bytes
(char&)pc is the same, but returns a reference to that byte
When you first print pc, nothing has happened and you see the letter you stored
c = 'B' modifies the last byte of the 4 byte pointer, so it now points to something else
When you print again, you are now pointing to a different location which explains your result.
Since the last byte of the pointer is modified the new memory address is nearby, making it unlikely to be in a piece of memory your program isn't allowed to access. That's why you don't get a seg-fault. The actual value obtained is undefined, but is highly likely to be a zero, which explains the blank output when its interpreted as a char.
when you're casting, with a C-style cast or with a reinterpret_cast, you're basically telling the compiler to look the other way ("don't you mind, I know what I'm doing").
C++ allows you to tell the compiler to do that. That doesn't mean it's a good idea...
I was studying pointers references and came across different ways to feed in parameters. Can someone explain what each one actually means?
I think the first one is simple, it's that x is a copy of the parameter fed in so another variable is created on the stack.
As for the others I'm clueless.
void doSomething1(int x){
//code
}
void doSomething2(int *x){
//code
}
void doSomething3(int &x){
//code
}
void doSomething3(int const &x){
//code
}
I also see stuff like this when variables are declared. I don't understand the differences between them. I know that the first one will put 100 into the variable y on the stack. It won't create a new address or anything.
//example 1
int y = 100;
//example 2
int *y = 100;
//Example 3: epic confusion!
int *y = &z;
Question 1: How do I use these methods? When is it most appropriate?
Question 2: When do I declare variables in that way?
Examples would be great.
P.S. this is one the main reasons I didn't learn C++ as Java just has garbage collection. But now I have to get into C++.
//example 1
int y = 100;
//example 2
int *y = 100;
//Example 3: epic confusion!
int *y = &z;
I think the problem for most students is that in C++ both & and * have different meanings, depending on the context in which they are used.
If either of them appears after a type within an object declaration (T* or T&), they are type modifiers and change the type from plain T to a reference to a T (T&) or a pointer to a T (T*).
If they appear in front of an object (&obj or *obj), they are unary prefix operators invoked on the object. The prefix & returns the address of the object it is invoked for, * dereferences a pointer, iterator etc., yielding the value it references.
It doesn't help against confusion that the type modifiers apply to the object being declared, not the type. That is, T* a, b; defines a T* named a and a plain T named b, which is why many people prefer to write T *a, b; instead (note the placement of the type-modifying * adjacent the object being defined, instead of the type modified).
Also unhelpful is that the term "reference" is overloaded. For one thing it means a syntactic construct, as in T&. But there's also the broader meaning of a "reference" being something that refers to something else. In this sense, both a pointer T* and a reference (other meaning T&) are references, in that they reference some object. That comes into play when someone says that "a pointer references some object" or that a pointer is "dereferenced".
So in your specific cases, #1 defines a plain int, #2 defines a pointer to an int and initializes it with the address 100 (whatever lives there is probably best left untouched ), and #3 defines another pointer and initializes it with the address of an object z (necessarily an int, too).
A for how to pass objects to functions in C++, here is an old answer from me to that.
From Scott Myers - More Effective C++ -> 1
First, recognize that there is no such thing as a null reference. A reference must always refer to some object.Because a reference must refer to an object, C++ requires that references be initialized.
Pointers are subject to no such restriction. The fact that there is no such thing as a null reference implies that it can be more efficient to use references than to use pointers. That's because there's no need to test the validity of a reference before using it.
Another important difference between pointers and references is that pointers may be reassigned to refer to different objects. A reference, however, always refers to the object with which it is initialized
In general, you should use a pointer whenever you need to take into account the possibility that there's nothing to refer to (in which case you can set the pointer to null) or whenever you need to be able to refer to different things at different times (in which case you can change where the pointer points). You should use a reference whenever you know there will always be an object to refer to and you also know that once you're referring to that object, you'll never want to refer to anything else.
References, then, are the feature of choice when you know you have something to refer to, when you'll never want to refer to anything else, and when implementing operators whose syntactic requirements make the use of pointers undesirable. In all other cases, stick with pointers.
Read S.Lippmann's C++ Premier or any other good C++ book.
As for passing the parameters, generally when copying is cheap we pass by value. For mandatory out parameters we use references, for optional out parameters - pointers, for input parameters where copying is costly, we pass by const references
Thats really complicated topic. Please read here: http://www.goingware.com/tips/parameters/.
Also Scott Meiers "Effective C++" is a top book on such things.
void doSomething1(int x){
//code
}
This one pass the variable by value, whatever happens inside the function, the original variable doesn't change
void doSomething2(int *x){
//code
}
Here you pass a variable of type pointer to integer. So when accessing the number you should use *x for the value or x for the address
void doSomething3(int &x){
//code
}
Here is like the first one, but whatever happens inside the function, the original variable will be changed as well
int y = 100;
normal integer
//example 2
int *y = 100;
pointer to address 100
//Example 3: epic confusion!
int *y = &z;
pointer to the address of z
void doSomething1(int x){
//code
}
void doSomething2(int *x){
//code
}
void doSomething3(int &x){
//code
}
And i am really getting confused between them?
The first is using pass-by-value and the argument to the function will retain its original value after the call.
The later two are using pass-by-reference. Essentially they are two ways of achieving the same thing. The argument is not guarenteed to retain its original value after the call.
Most programmers prefer to pass large objects by const reference to improve the performance of their code and provide a constraint that the value will not change. This ensures the copy constructor is not called.
Your confusion might be due to the '&' operator having two meanings. The one you seem to be familiar with is the 'reference operator'. It is also used as the 'address operator'. In the example you give you are taking the address of z.
A good book to check out that covers all of this in detail is 'Accelerated C++' by Andrew Koening.
The best time to use those methods is when it's more efficient to pass around references as opposed to entire objects. Sometimes, some data structure operations are also faster using references (inserting into a linked list for example). The best way to understand pointers is to read about them and then write programs to use them (and compare them to their pass-by-value counterparts).
And for the record, knowledge of pointers makes you considerably more valuable in the workplace. (all too often, C++ programmers are the "mystics" of the office, with knowledge of how those magical boxes under the desks process code /semi-sarcasm)
I am trying to learn C++ via some web tutorials. I don't have a compiler available to me, otherwise I would try this out. I'm not sure what is meant by a const pointer. Does that just mean it always points to the same memory address? Why would you ever want to do that? Would the following code be legal?
...
int * const aPointer = new int;
... //do something with aPointer
delete aPointer;
... //do something else, including possibly more 'new' statements
aPointer = new int;
...
A simple way to remember how const works with pointers is to remember that it always applies to whatever is to the left of it, unless it's the left-most keyword, in which case it applies to the right.
Examples:
Pointer to a constant char:
The pointer can be changed to point to something else, but the char it initally points to cannot change value.
const char * p;
Constant pointer to a char:
The pointer cannot be changed to point to anything else, but the char it points to can change value.
char *const p;
Constant pointer to a constant char:
The pointer cannot be changed to point to anything else, and the char it points to cannot change value.
const char *const p;
Const pointer could mean a few different things. I recommend checking out the C++ FAQ Lite on the matter.
You can have:
const int* p;
int* const p;
const int* const p;
All three mean different things. Yes, it's kind of confusing.
In your case, you have the second, which means you have a constant pointer to a non-constant object. That is, you can change the value of the integer via the pointer, but you can't change what the pointer points to. So the code you posted would not be legal.
You can't learn to drive a car just by reading books.
Get yourself a C++ compiler if you want to learn C++. g++ is free of charge, as well as Visual Studio 2008 Express Edition.
As for your question, a const pointer is a zone of memory that is ready only. Example: A class may provide read-only access to an internal buffer.
Note that you also have the const pointer that is also const, aka
const char * const p
In that case, even the value of the pointer cannot be modified.
G'day,
To remember this easily you can use the trick that Scott Meyers describes in his excellent book "Effective C++" (sanitised Amazon link)
You draw a line through the declaration where the asterisk is located.
If the keyword const appears to the left of the line, then you can't change the value of the item that you're pointing to.
If the keyword const appears to the right of the line, then you can't change the pointer to point to another location.
If const appears on both sides, then you can't change the pointer and you can't change the value of what you're pointing to.
BTW That book is excellent, and while not for a beginner, is definitely a way of taking your C++ knowledge to the next level! Highly recommended.
HTH
cheers,
A const pointer means that you can change the value of the variable which is being pointed to, but you can't change where the pointer is pointed. I don't use them often myself, but a common application for const pointers is in defining specific memory segments that you need to address. See this question for more information.
As an aside, you should try to get a compiler on your computer if you can. I've shown myself many times that human brains are poor C++ compilers.
Your code is not legal. You can't assign to aPointer (except using copy-style initialisation, which in fact is not assignment even though it looks like it) if aPointer is declared const like that.
But usually when people say "a const pointer", they mean const int * aPointer, not int * const aPointer as you have in your code. The whole internet will explain the difference at the drop of a hat. As far as I know, the term "const pointer" isn't defined in the standard, so we're free to do this even though it's potentially confusing. "Pointer-to-const" would be an unambiguous description, and a pointer-to-const is much more commonly used than a pointer-which-is-itself-const.
A pointer-which-is-itself-const is used to refer to something, where you won't want the pointer to refer to a different thing at any point in its life. For instance, this is a pointer-which-is-itself-const, because "this object" remains the same object through the execution of a member function. The C++ language has opted not to let you decide part way through that you want to assign some other value to this, to make it refer to some other object. In C++ references often serve that purpose too, since they cannot be "re-seated" to change the referand.
In your code, the pointer cannot move, but the data pointed to can change.
It's legal up to the first delete. A subsequent new would not work because it's an assignment to a constant.
It's relatively unusual to see this, more common is to see where the data pointed to is unchangeable but the pointer can move.
int bar;
int baz;
const int * foo = &bar;
*foo = 4; // illegal
foo = &baz; // legal
having both pointer and value being const is common with strings
const wchar_t * const myString = L"String that will never change.";
Since you're new to C++, for the answer to this question and many other questions you may have or don't know you have, check out the C++ FAQ
I don't have a compiler available to me, otherwise I would try this out
Everyone has a C++ Compiler available to them: http://www.online-compiler.com
There are many others, but this seems to work...
As it has already been pointed out the perhaps most common const pointer is
const char* p;
The variable p can change, but the data p points to is unmodifable.
However, moving the const keyword to the left of the asterisk does not alter the meaning of the declaration:
char const* p;
I prefer the later since it becomes much easier to remember where to place the const keywords when declaring const pointers to const pointers:
char const* const* p;
Again, the variable p can change and the data pointed to is unmodifiable. Furthermore, the data is declared as const pointers meaning that it points to data that cannot be modified.
The more common notation for this type is
const char* const* p;
Placing the const keyword immediately to the left of the asterisk it modifies (or ampersand for reference) makes it easy to create complex types involving the const keyword. For example, a pointer to const pointers:
char const** p;
and a const pointer to pointers:
char* const* p;
Remember to "read" pointer declarations from the right to the left, and don't declare more than one pointer in each statement to avoid a lot of confusion.
Is there some kind of subtle difference between those:
void a1(float &b) {
b=1;
};
a1(b);
and
void a1(float *b) {
(*b)=1;
};
a1(&b);
?
They both do the same (or so it seems from main() ), but the first one is obviously shorter, however most of the code I see uses second notation. Is there a difference? Maybe in case it's some object instead of float?
Both do the same, but one uses references and one uses pointers.
See my answer here for a comprehensive list of all the differences.
Yes. The * notation says that what's being pass on the stack is a pointer, ie, address of something. The & says it's a reference. The effect is similar but not identical:
Let's take two cases:
void examP(int* ip);
void examR(int& i);
int i;
If I call examP, I write
examP(&i);
which takes the address of the item and passes it on the stack. If I call examR,
examR(i);
I don't need it; now the compiler "somehow" passes a reference -- which practically means it gets and passes the address of i. On the code side, then
void examP(int* ip){
*ip += 1;
}
I have to make sure to dereference the pointer. ip += 1 does something very different.
void examR(int& i){
i += 1;
}
always updates the value of i.
For more to think about, read up on "call by reference" versus "call by value". The & notion gives C++ call by reference.
In the first example with references, you know that b can't be NULL. With the pointer example, b might be the NULL pointer.
However, note that it is possible to pass a NULL object through a reference, but it's awkward and the called procedure can assume it's an error to have done so:
a1(*(float *)NULL);
In the second example the caller has to prefix the variable name with '&' to pass the address of the variable.
This may be an advantage - the caller cannot inadvertently modify a variable by passing it as a reference when they thought they were passing by value.
Aside from syntactic sugar, the only real difference is the ability for a function parameter that is a pointer to be null. So the pointer version can be more expressive if it handles the null case properly. The null case can also have some special meaning attached to it. The reference version can only operate on values of the type specified without a null capability.
Functionally in your example, both versions do the same.
The first has the advantage that it's transparent on the call-side. Imagine how it would look for an operator:
cin >> &x;
And how it looks ugly for a swap invocation
swap(&a, &b);
You want to swap a and b. And it looks much better than when you first have to take the address. Incidentally, bjarne stroustrup writes that the major reason for references was the transparency that was added at the call side - especially for operators. Also see how it's not obvious anymore whether the following
&a + 10
Would add 10 to the content of a, calling the operator+ of it, or whether it adds 10 to a temporary pointer to a. Add that to the impossibility that you cannot overload operators for only builtin operands (like a pointer and an integer). References make this crystal clear.
Pointers are useful if you want to be able to put a "null":
a1(0);
Then in a1 the method can compare the pointer with 0 and see whether the pointer points to any object.
One big difference worth noting is what's going on outside, you either have:
a1(something);
or:
a1(&something);
I like to pass arguments by reference (always a const one :) ) when they are not modified in the function/method (and then you can also pass automatic/temporary objects inside) and pass them by pointer to signify and alert the user/reader of the code calling the method that the argument may and probably is intentionally modified inside.