if I have a function like that:
void doSomething(int& aVar)
{
// do something
}
and I have this:
int *aVar = new int;
*aVar = 10;
doSomething(*aVar);
Why should I call *aVar? isn't aVar already an address?
No, a reference is not a pointer. References are guaranteed to not be null; you cannot say the same for a pointer. When you pass an int to a function that expects an int& the reference will be taken automatically.
P.S. Don't think of it as an address or a fancy pointer. It is a reference, or an alias to an existing object.
doSomething(int&)
wants a reference not a pointer. The way to set up that reference as a parameter is to pass in an Int. Which is why
doSomething(*aVar)
works. If you want to use a pointer in the function say
doSomething(int*)
references and pointers are not the same thing (although they have a lot in common)
The asterisk, besides multiplication has two meanings:
a) When declaring a variable: *x means "X is a pointer"
b) When using a variable: *x (when x is of pointer type) means "take whatever is pointed by x" - the opposite of &x, which means "take the address of x".
The code:
doSomething(*aVar)
just wants to dereference the pointer "aVar" (take the value of type int pointed by it) and pass this value of type int as a parameter to the function.
The variable "aVar" stores an address of some integer value, not the value itself, so you have to use the "*" operator to dereference it every time you want to access the integer, not the memory address itself.
References in C++ are quite counter-intuitive ("disguised pointers"), so if doSomething takes a reference to int, you have to call it as if you were passing an actual int value, not a pointer. Hence you need the dereference operator.
A pointer is pointing to a specific memory address and a pointer has it's own data (ie the memory address it's pointing to). When you pass aVar to the function without dereferencing (the * operator) the pointer you would be passing the memory location of the pointer, not the memory location the pointer is pointing to.
Related
Is passing pointer argument, pass by value in C++? Since i see that any change to the pointer as such is not reflected outside the method. The changes i do by dereferencing the pointer is reflected though.
In that case, is it acceptable/standard procedure to use pointer to pointer as argument to a function to modify the pointer value as such within a function?
Yes to both.
Pointers are passed by value as anything else. That means the contents of the pointer variable (the address of the object pointed to) is copied. That means that if you change the value of the pointer in the function body, that change will not be reflected in the external pointer that will still point to the old object. But you can change the value of the object pointed to.
If you want to reflect changes made to the pointer to the external pointer (make it point to something else), you need two levels of indirection (pointer to pointer). When calling functions it's done by putting a & before the name of the pointer. It is the standard C way of doing things.
When using C++, using references is preferred to pointer (henceforth also to pointer to pointer).
For the why references should be preferred to pointers, there is several reasons:
references introduce less syntaxic noise than pointers in function body
references keep more informations than pointers, than can be useful for compiler
Drawbacks of references are mostly:
they break the simple pass-by-value rule of C, what makes understanding the behavior of a function regarding of parameters (will they be changed ?) less obvious. You also need function prototype to be sure. But that is not really worse than the multiple pointer levels necessary when using C.
they are not supported by C, that can be a problem when you write code that should work with both C and C++ programs (but that's not the most usual case).
In the specific case of pointer to pointer, the difference is mostly simplicity, but using reference it may also be easy to remove both levels of pointers and pass only one reference instead of a pointer to pointer.
I understand the confusion here. The concepts of "pass by value" and "pass by reference" are not so clear even if they seem to be so.
Bear in mind that the computer does not know these concepts and does not behave according to it.
The computer does not know about the types. Hence it does not make a distinction of pointers and values.
Let me try to explain by and example:
void func1(int x) //copy some value to local variable x (of type int)
{
x = 5; //modify local variable. lost after function call
}
void func2(int *x) //copy some value to local variable x (of type int*)
{
int a;
x = &a; //modify local variable. lost after function call.
}
void func3(int *x) //copy some value to local variable x(of type int*)
{
*x = 10; //x is local but *x is not! change is saved after function call!
}
func1 and func2 are identical. Both modify a local variable. Modification is lost after function is popped off the stack.
func3 has ability to change another memory location (a variable which is not local to the function).
basically, every function call is "call by value". But in the case of a pointer type, we have a way to change the content of a remote address in memory.
Pass by value using Pointers
I'll explain it by example:
void f(int *ptr)
{
cout<<*ptr;
}
int main ()
{
int a=10;
int *aptr=&a;
f(aptr);
return 0;
}
Here, in main function a is an integer variable whose content is 10 and address is 00F8FB04 (assume).
aptr is pointer to integer, that store the address of integer variable a, so aptr content is address of integer variable a that is 00F8FB04. When we pass aptr as the function argument only content of aptr (that is address) are copies to function parameter.
So, ptr will receive the copy of content of aptr (that is address 00F8FB04)
Either a pointer to a pointer, or a reference to a pointer, is what you would use if you wanted to potentially change the pointer itself. To your original question, technically, yes, all parameters are passed by value.
Yes it is, as it is in C.
In that case, is it acceptable/standard procedure to use pointer to pointer as argument to a function to modify the pointer value as such within a function?
In which case? What do you want? You can use real references with the & modifier.
void func(type &ref);
Is passing pointer argument, pass by value in C++? Since i see that any change to the pointer as such is not reflected outside the method. The changes i do by dereferencing the pointer is reflected though.
In that case, is it acceptable/standard procedure to use pointer to pointer as argument to a function to modify the pointer value as such within a function?
Yes to both.
Pointers are passed by value as anything else. That means the contents of the pointer variable (the address of the object pointed to) is copied. That means that if you change the value of the pointer in the function body, that change will not be reflected in the external pointer that will still point to the old object. But you can change the value of the object pointed to.
If you want to reflect changes made to the pointer to the external pointer (make it point to something else), you need two levels of indirection (pointer to pointer). When calling functions it's done by putting a & before the name of the pointer. It is the standard C way of doing things.
When using C++, using references is preferred to pointer (henceforth also to pointer to pointer).
For the why references should be preferred to pointers, there is several reasons:
references introduce less syntaxic noise than pointers in function body
references keep more informations than pointers, than can be useful for compiler
Drawbacks of references are mostly:
they break the simple pass-by-value rule of C, what makes understanding the behavior of a function regarding of parameters (will they be changed ?) less obvious. You also need function prototype to be sure. But that is not really worse than the multiple pointer levels necessary when using C.
they are not supported by C, that can be a problem when you write code that should work with both C and C++ programs (but that's not the most usual case).
In the specific case of pointer to pointer, the difference is mostly simplicity, but using reference it may also be easy to remove both levels of pointers and pass only one reference instead of a pointer to pointer.
I understand the confusion here. The concepts of "pass by value" and "pass by reference" are not so clear even if they seem to be so.
Bear in mind that the computer does not know these concepts and does not behave according to it.
The computer does not know about the types. Hence it does not make a distinction of pointers and values.
Let me try to explain by and example:
void func1(int x) //copy some value to local variable x (of type int)
{
x = 5; //modify local variable. lost after function call
}
void func2(int *x) //copy some value to local variable x (of type int*)
{
int a;
x = &a; //modify local variable. lost after function call.
}
void func3(int *x) //copy some value to local variable x(of type int*)
{
*x = 10; //x is local but *x is not! change is saved after function call!
}
func1 and func2 are identical. Both modify a local variable. Modification is lost after function is popped off the stack.
func3 has ability to change another memory location (a variable which is not local to the function).
basically, every function call is "call by value". But in the case of a pointer type, we have a way to change the content of a remote address in memory.
Pass by value using Pointers
I'll explain it by example:
void f(int *ptr)
{
cout<<*ptr;
}
int main ()
{
int a=10;
int *aptr=&a;
f(aptr);
return 0;
}
Here, in main function a is an integer variable whose content is 10 and address is 00F8FB04 (assume).
aptr is pointer to integer, that store the address of integer variable a, so aptr content is address of integer variable a that is 00F8FB04. When we pass aptr as the function argument only content of aptr (that is address) are copies to function parameter.
So, ptr will receive the copy of content of aptr (that is address 00F8FB04)
Either a pointer to a pointer, or a reference to a pointer, is what you would use if you wanted to potentially change the pointer itself. To your original question, technically, yes, all parameters are passed by value.
Yes it is, as it is in C.
In that case, is it acceptable/standard procedure to use pointer to pointer as argument to a function to modify the pointer value as such within a function?
In which case? What do you want? You can use real references with the & modifier.
void func(type &ref);
While experimenting with pointers, I had the idea to change the address a pointer. I tried the following:
int* pointer;
&pointer = 0x28ff0d; //To point to an empty space in memory
But that threw out an error (Details: (here) in line 2, lvalue required as left operand [...]), so I guess it's impossible? I'm not trying to make a nullpointer, but litterally to change the memory address
For obvious reasons, setting a pointer to a hardcoded address is a bad idea. But, you can do so by reinterpret_cast<>ing the address to the type you're assigning it to, like so:
int * pointer = reinterpret_cast<int *>(0xDEADBEEF);
or
int * pointer = &some_int;
. . .
pointer = reinterpret_cast<int *>(0xDEADBEEF);
Note that dereferencing pointer in this case is undefined behavior... but doing this might still be useful for something like debugging a use-after-free, where a sentinel value like 0xDEADBEEF would clearly communicate that you had already freed that memory.
Obviously the much more common, useful, and safe case for changing the address you are pointing to would be to take the address of a different (real) piece of data:
int * pointer = &some_int;
int new_val = 1234;
pointer = &new_val; // *pointer is now 1234
You can never change the address of a variable. So whatever T is (be it an intrinsic type like int or double, a struct, a pointer or an array, the following is forbidden in C++:
T var;
&var = anything; // WRONG! &var is not a lvalue
The error lvalue required as left operand means that you can only assign to something that can receive a new value, and the address of a variable is not.
A pointer is a variable. Any variable has an address where it's located. So if you've declared a varible like e.g. int x;, then &x is the address where that variable resides.
A pointer is a variable which resides somewhere, but in addition it has a value, which happens to be an address of some other variable. So if you've declared a pointer int* x;, have written something to it, then x will represent some address you've written to it, and *x will be the value pointed to by the pointer.
So your syntactic mistake was using & where you should have just used the variable name, without &. After this is clear, be sure to read the other answers which explain further problems with your code.
int* pointer;
&pointer = 0x28ff0d;
This will throw error because &pointer is not an lvalue. On the left side of the assignment operator there must be an lvalue.
Is passing pointer argument, pass by value in C++? Since i see that any change to the pointer as such is not reflected outside the method. The changes i do by dereferencing the pointer is reflected though.
In that case, is it acceptable/standard procedure to use pointer to pointer as argument to a function to modify the pointer value as such within a function?
Yes to both.
Pointers are passed by value as anything else. That means the contents of the pointer variable (the address of the object pointed to) is copied. That means that if you change the value of the pointer in the function body, that change will not be reflected in the external pointer that will still point to the old object. But you can change the value of the object pointed to.
If you want to reflect changes made to the pointer to the external pointer (make it point to something else), you need two levels of indirection (pointer to pointer). When calling functions it's done by putting a & before the name of the pointer. It is the standard C way of doing things.
When using C++, using references is preferred to pointer (henceforth also to pointer to pointer).
For the why references should be preferred to pointers, there is several reasons:
references introduce less syntaxic noise than pointers in function body
references keep more informations than pointers, than can be useful for compiler
Drawbacks of references are mostly:
they break the simple pass-by-value rule of C, what makes understanding the behavior of a function regarding of parameters (will they be changed ?) less obvious. You also need function prototype to be sure. But that is not really worse than the multiple pointer levels necessary when using C.
they are not supported by C, that can be a problem when you write code that should work with both C and C++ programs (but that's not the most usual case).
In the specific case of pointer to pointer, the difference is mostly simplicity, but using reference it may also be easy to remove both levels of pointers and pass only one reference instead of a pointer to pointer.
I understand the confusion here. The concepts of "pass by value" and "pass by reference" are not so clear even if they seem to be so.
Bear in mind that the computer does not know these concepts and does not behave according to it.
The computer does not know about the types. Hence it does not make a distinction of pointers and values.
Let me try to explain by and example:
void func1(int x) //copy some value to local variable x (of type int)
{
x = 5; //modify local variable. lost after function call
}
void func2(int *x) //copy some value to local variable x (of type int*)
{
int a;
x = &a; //modify local variable. lost after function call.
}
void func3(int *x) //copy some value to local variable x(of type int*)
{
*x = 10; //x is local but *x is not! change is saved after function call!
}
func1 and func2 are identical. Both modify a local variable. Modification is lost after function is popped off the stack.
func3 has ability to change another memory location (a variable which is not local to the function).
basically, every function call is "call by value". But in the case of a pointer type, we have a way to change the content of a remote address in memory.
Pass by value using Pointers
I'll explain it by example:
void f(int *ptr)
{
cout<<*ptr;
}
int main ()
{
int a=10;
int *aptr=&a;
f(aptr);
return 0;
}
Here, in main function a is an integer variable whose content is 10 and address is 00F8FB04 (assume).
aptr is pointer to integer, that store the address of integer variable a, so aptr content is address of integer variable a that is 00F8FB04. When we pass aptr as the function argument only content of aptr (that is address) are copies to function parameter.
So, ptr will receive the copy of content of aptr (that is address 00F8FB04)
Either a pointer to a pointer, or a reference to a pointer, is what you would use if you wanted to potentially change the pointer itself. To your original question, technically, yes, all parameters are passed by value.
Yes it is, as it is in C.
In that case, is it acceptable/standard procedure to use pointer to pointer as argument to a function to modify the pointer value as such within a function?
In which case? What do you want? You can use real references with the & modifier.
void func(type &ref);
I would look this up, but honestly I wouldn't know where to start because I don't know what it is called. I've seen variables passed to functions like this:
myFunction((void**)&variable);
Which confuses the heck out of me cause all of those look familiar to me; I've just never seen them put together like that before.
What does it mean? I am a newb so the less jargon, the better, thanks!
void* is a "pointer to anything". void ** is another level of indirection - "pointer to pointer to anything". Basically, you pass that in when you want to allow the function to return a pointer of any type.
&variable takes the address of variable. variable should already be some kind of a pointer for that to work, but it's probably not void * - it might be, say int *, so taking its address would result in a int **. If the function takes void ** then you need to cast to that type.
(Of course, it needs to actually return an object of the right type, otherwise calling code will fail down the track when it tries to use it the wrong way.)
Take it apart piece by piece...
myFunction takes a pointer to a pointer of type void (which pretty much means it could point to anything). It might be declared something like this:
myFunction(void **something);
Anything you pass in has to have that type. So you take the address of a pointer, and cast it with (void**) to make it be a void pointer. (Basically stripping it of any idea about what it points to - which the compiler might whine about otherwise.)
This means that &variable is the address (& does this) of a pointer - so variable is a pointer. To what? Who knows!
Here is a more complete snippet, to give an idea of how this fits together:
#include <stdio.h>
int myInteger = 1;
int myOtherInt = 2;
int *myPointer = &myInteger;
myFunction(void **something){
*something = &myOtherInt;
}
main(){
printf("Address:%p Value:%d\n", myPointer, *myPointer);
myFunction((void**)&myPointer);
printf("Address:%p Value:%d\n", myPointer, *myPointer);
}
If you compile and run this, it should give this sort of output:
Address:0x601020 Value:1
Address:0x601024 Value:2
You can see that myFunction changed the value of myPointer - which it could only do because it was passed the address of the pointer.
It's a cast to a pointer to a void pointer.
You see this quite often with functions like CoCreateInstance() on Windows systems.
ISomeInterface* ifaceptr = 0;
HRESULT hr = ::CoCreateInstance(CLSID_SomeImplementation, NULL, CLSCTX_ALL,
IID_ISomeInterface, (void**)&ifaceptr);
if(SUCCEEDED(hr))
{
ifaceptr->DoSomething();
}
The cast converts the pointer to an ISomeInterface pointer into a pointer to a void pointer so that CoCreateInstance() can set ifaceptr to a valid value.
Since it is a pointer to a void pointer, the function can output pointers of any type, depending on the interface ID (such as IID_ISomeInterface).
It's a pointer to a pointer to a variable with an unspecified type. All pointers are the same size, so void* just means "a pointer to something but I have no idea what it is". A void** could also be a 2D array of unspecified type.
That casts &variable to a void** (that is, a pointer to a pointer to void).
For example, if you have something along the lines of
void myFunction(void** arg);
int* variable;
This passes the address of variable (that's what the unary-& does, it takes the address) to myFunction().
The variable is a pointer to something of undefined (void) type. The & operator returns the address of that variable, so you now have a pointer to a pointer of something. The pointer is therefore passed into the function by reference. The function might have a side effect which changes the memory referenced by that pointer. In other words, calling this function might change the something that the original pointer is referencing.