I am confused about the meaning of "pass by reference" in C and C++.
In C, there are no references. So I guess pass by reference means passing a pointer. But then why not call it pass by pointer?
In C++, we have both pointers and references (and stuff like iterators that lies close). So what does pass by reference mean here?
In colloquial usage, "pass by reference" means that, if the callee modifies its arguments, it affects the caller, because the argument as seen by the callee refers to the value as seen by the caller.
The phrase is used independent of the actual programming language, and how it calls things (pointers, references, whatever).
In C++, call-by-reference can be done with references or pointers. In C, call-by-reference can only be achieved by passing a pointer.
"Call by value":
void foo( int x )
{
// x is a *copy* of whatever argument foo() was called with
x = 42;
}
int main()
{
int a = 0;
foo( a );
// at this point, a == 0
}
"Call by reference", C style:
void foo( int * x )
{
// x is still a *copy* of foo()'s argument, but that copy *refers* to
// the value as seen by the caller
*x = 42;
}
int main()
{
int a = 0;
foo( &a );
// at this point, a == 42
}
So, strictly speaking, there is no pass-by-reference in C. You either pass the variable by-value, or you pass a pointer to that variable by-value.
In C, there are no references
There are no reference variables. But you can refer to objects using pointers. Therefore pointers are "references" from an abstract point of view.
But then why not call it pass by pointer?
You can call it pass by pointer. Reference is a more general term than pointer. It is often preferable to use the more general term when you want to discuss abstractions and want to ignore implementation details. You would call it pass by reference for the same reason that you call a variable "integer" rather than "int32_t".
In C++, we have both pointers and references (and stuff like iterators that lies close). So what does pass by reference mean here?
Depends on context. Often it means that the function argument is a reference variable, but it may also refer to a pointer, iterator, a reference wrapper... anything that referes to an object.
Reference is an abstract concept that exists beyond c and c++; even beyond programming. In c++, the term is ambiguous with reference variables and the context and convention (which isn't universal) determines the meaning.
In C, there are no any reference variables, but you can pass by reference with using pointers.
In wikipedia, there is this definition.
In call-by-reference evaluation (also referred to as pass-by-reference), a function receives an implicit reference to a variable used as argument, rather than a copy of its value. So this term is for type of parameter passing as mentioned by Thomas. So yes, since C is older than C++, also this idea is older than C++.
However, in C++ both pointers and references can be used for passing to the function(Call by address and call by reference). Actually they are working the same way, they have only a few differences.
Once a reference is created, it cannot be later made to reference
another object; it cannot be reseated. This is often done with
pointers.
References cannot be NULL. Pointers are often made NULL to indicate
that they are not pointing to any valid thing.
A reference must be initialized when declared. There is no such
restriction with pointers
With these differences, if you use call by reference instead of call by pointer, you can reduce the possibility of NULL pointer error kind of problems.
Let's clear your confusion.
In C, there are no references. So I guess pass by reference means passing a pointer. But then why not call it pass by pointer?
Because every argument passing in C is pass-by-value. Even a pointer argument is a copy. But it contains (or points to, if you prefer) the same value -- memory address. That is how you can change the variable it points to, but not the pointer itself. Since it's a copy, whatever you do will not affect the pointer on the caller level.
In C++, we have both pointers and references (and stuff like iterators that lies close). So what does pass by reference mean here?
It means, that the argument is an alias of a variable on the caller level, not a copy, which allows us to change it.
Hope that helped.
A reference in general is an instance that is referencing something else. Thus in a wider sense, also a pointer can be considered as one possible implementation of a reference. References in C++ are just called references, because apart from referencing something they offer no other features.
Pass-by-reference is used in general to distinguish from pass-by-value. Whether it is via pointer or via a reference is often just a minor detail. However, with C++ references it is imho more clear what is the purpose of the function parameter. Eg:
int foo(int& a); // pass-by-reference
int foo(const int& a); // is pratically pass-by-value
// (+ avoiding the copy of the parameter)
on the other hand, with references (as compared to pointers) it is not so obvious at the call site if it is pass-by-value or pass-by-reference. E.g.
int x;
int y = foo(x); // could be pass-by-value or pass-by-reference
int z = foo(&x); // obviously pass-by-reference (as a pointer)
Imagine you have to paint your house...
by value: you bring a copy of your house to the painter => much effort (maybe on rails)
by reference: you give your house address to the painter so he can come and paint it
"Pass by reference" (or "call by reference") is a term for a type of parameter passing when calling a function, and the idea is older than C++. It does not necessarily have to be done using C++ "references". C doesn't have a built-in mechanism to do this, so you have to use pointers.
Just to add to the answers, referencing does not mean reference by address. The compiler may use any method to reference to a variable.
when you pass something by reference you're working with the address and not the value of a variable directly, If you use a reference parameter you're getting the address of the variable you pass in.
From there you can manipulate it how ever you want as the variable you passed in WILL change if you change the reference in the function. It's an easier way to work with large amounts of a data it really just saves on memory etc..
In C there are two concepts
1. Call by value - Here copy of values are passed so actual values will not change outside the function.
2. Call by reference - but here actual values (Address of actual operands) are passed so it will change the values globally.
Where in C++ there are two concepts
1. Pass by value - it is same as c, actual values will not change, scope of this values are of function only.
2. Pass by Reference - actual values (Address of actual operands) are passed so it will change the values globally, it means if values gets changed then it will affect in whole program.
In Pass by Reference, the address of operands are passed that's why it is called as Pass By Reference not as pointer.
Related
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What's the difference between passing by reference vs. passing by value?
I read that in C arguments are passed by value, but what's is the difference between passing arguments by value (like in C) or by refencence (like C++ - C#)?
What's the difference between a pointer and a reference?
void with_ptr(int *i)
{ *i = 0; }
void with_ref(int &i)
{ i = 0; }
In these cases are modified both value? If yes, why C++ allows to pass arguments by reference? I think it is not clear inside the function that the i value could be modified.
what's is the difference between passing arguments by value or by reference
If you pass by value, changes to the variable will be local to the function, since the value is copied when calling the function. Modifications to reference arguments will propagate to the original value.
What's the difference between a pointer and a reference?
The difference is largely syntactic, as you have seen in your code. Furthermore, a pointer can be reassigned to point to something else (unless it’s declared const), while a reference can’t; instead, assigning to a reference is going to assign to the referenced value.
I think it is not clear inside the function that the i value could be modified.
On the contrary, it’s absolutely clear: the function signature tells you so.
There’s actually a case to be made that it’s not clear outside the function. That’s why original versions of C# for instance mandated that you explicitly annotate any by-reference calling with ref (i.e. f(ref x) instead of plain f(x)). This would be similar to calling a function in C++ using f(&x) to make it clear that a pointer is passed.
But in recent versions of C#, the use of ref for calling was made optional since it didn’t confer enough of an advantage after all.
Consider this:
1) Passing by reference provides more simple element access i instead of *i
2) Generally you cannot pass null reference to a method, but can pass a null pointer
3) You can't change the address of reference, but can change it for a pointer(although, as pointer itself passed by value, this change will be discarded upon function exit)
Hope, this helped a bit
Actually, in the first case, you can't modify the argument. The pointer itself is immutable, you can only modify the value it points to.
If yes, why C++ allows to pass arguments by reference?
Because pointers can very easily be miss-used. References should almost always be prefered. For your case, what if you pass a NULL to with_ptr? You'll get undefined behavior, which is not possible if you use with_ref.
I think it is not clear inside the function that the i value could be modified.
It is very clear. If you see a function that takes a parameter by non-const reference, you can assume it will be changed.
I think that a method can only change an argument's value, if this is passed by reference. If you pass a argument by value in a method, then whatever change you make to its value, this will no be available in the parent method.
As far as I know, I think the reference is safer to use in a sense that it can't be modified (always points to the same thing), and should be initialized if it's a local variable. Pointer, however, can be change to point to somewhere else.
int x = 10;
int &y = x;
int *p = &x;
p++; //Legal if you know what's next
y++; // Increases the value of x. Now x = y = 11;
As my two cents, I think reference variables are mere alternative names for the same memory address by which it was initialized. This also explains pretty nice:
http://www.dgp.toronto.edu/~patrick/csc418/wi2004/notes/PointersVsRef.pdf
Pass by reference:
I have learned that when a variables is passed as a reference to a function then instead of copy, the actual data is passed to the function but i think that if it is really the case then we shouldn't be able to access that data again once the program execution returns to main() after the stack frame of that function gets destroyed and leave that reference variable with zero or null value in main() but it is not the case and we can still access it in the main() so i think that in pass by reference, the memory address of that variable is passed to that function reference variable parameter and then we use that memory in that function with another name(reference variable) and when that function gets destroyed then the reference variables get destroyed rather than the actual data.
is my thinking towards this concept is right or am i doing some mistake in understanding this concept ?
Return By Reference
When a variable which is passed as a reference to another function returns back as a reference to main() then is the memory address passed back to main() or what actually is returned ?
I have learned that when a variables is passed as a reference to a function then instead of copy, the actual data is passed to the function
This is wrong. Passing the actual data would mean the value is passed (either by copy or by move). When passing by reference, a reference to the actual data is passed to the function.
so i think that in pass by reference, the memory address of that variable is passed to that function reference variable parameter and then we use that memory in that function with another name(reference variable)
This is right.
When a variable which is passed as a reference to another function returns back as a reference to main() then is the memory address passed back to main() or what actually is returned ?
A reference can be thought of as a memory address, so returning a reference is similar to returning a pointer (memory address).
When returning a reference from a function, you need to be careful that you don't return a reference to a variable that's local to the function, because that variable no longer exists after the function returns, and you are left with a dangling reference that may crash your program when you try to use it.
passed as a reference to a function then instead of copy, the actual data is passed to the function
It is unclear what you mean by "actual" data. A copy is actual data as much as the origianl object is.
References are a form of indirection. The reference variable indirectly refers to the object to which it was bound. An object variable is distinct from other objects, and its value may be copied from another.
if it is really the case then we shouldn't be able to access that data again once the program execution returns to main()
Binding a reference to an object does not make the object or its data disappear. Example:
void foo(int& ref);
int main()
{
int obj;
foo(obj);
// obj still exists here
}
If you bind a reference to an object defined in main, then the object still exists in main regardless of where that reference was bound.
i doing some mistake in understanding this concept ?
Yes. You did not understand yet what a reference is, or at least were not able to describe them correctly.
Your concept of pass by reference is wrong.
Passing a variable to a function by reference means that something else (let's call it a handle) is passed that acts an alias of the original variable. Every operation that can be performed on the handle within the function (e.g. assigning it a value, retrieving its value, calculating its address, calling a member function if it is an object) are referred to the original variable that was passed by the caller. How those affects are achieved depends on the implementation (e.g. the compiler).
The called function, by performing operations on the handle, does not cause the original variable to cease to exist. (Except in some very specific circumstances, which I won't go into).
For example;
#include <iostream>
void foo(int &x)
{
x = 42; // x is the handle I refer to above
}
int main()
{
int y = 16;
std::cout << y << '\n';
foo(y);
std::cout << y << '\n';
}
will print the values 16 and 42, in that order. The assignment x = 42 in foo() has the effect of changing the value of y in main(). The lifetime of y is unaffected, so y continues to exist until the end of main().
That thing I have referred to as a "handle" is more commonly known as a reference.
Similarly, your concept of returning a reference is wrong. Returning a reference gives a handle to the caller, and the caller can then use the returned handle to access (and perform operations on) whatever variable is returned by the function.
Note that, in the above, I have said nothing about passing the address of a variable around. Because the description above focuses on the observable effect of using references, not on how that observable effect is achieved (e.g. by the compiler).
Behind the scenes, your compiler MIGHT implement all of the magic of references by passing the address of the affected variables around (in fact, most modern compilers do). That is one possible implementation approach. But there are, technically, other ways of achieving that effect - the C++ standard does not require the address of variables to be passed around to achieve the behaviours associated with passing them around by reference.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What's the difference between passing by reference vs. passing by value?
I read that in C arguments are passed by value, but what's is the difference between passing arguments by value (like in C) or by refencence (like C++ - C#)?
What's the difference between a pointer and a reference?
void with_ptr(int *i)
{ *i = 0; }
void with_ref(int &i)
{ i = 0; }
In these cases are modified both value? If yes, why C++ allows to pass arguments by reference? I think it is not clear inside the function that the i value could be modified.
what's is the difference between passing arguments by value or by reference
If you pass by value, changes to the variable will be local to the function, since the value is copied when calling the function. Modifications to reference arguments will propagate to the original value.
What's the difference between a pointer and a reference?
The difference is largely syntactic, as you have seen in your code. Furthermore, a pointer can be reassigned to point to something else (unless it’s declared const), while a reference can’t; instead, assigning to a reference is going to assign to the referenced value.
I think it is not clear inside the function that the i value could be modified.
On the contrary, it’s absolutely clear: the function signature tells you so.
There’s actually a case to be made that it’s not clear outside the function. That’s why original versions of C# for instance mandated that you explicitly annotate any by-reference calling with ref (i.e. f(ref x) instead of plain f(x)). This would be similar to calling a function in C++ using f(&x) to make it clear that a pointer is passed.
But in recent versions of C#, the use of ref for calling was made optional since it didn’t confer enough of an advantage after all.
Consider this:
1) Passing by reference provides more simple element access i instead of *i
2) Generally you cannot pass null reference to a method, but can pass a null pointer
3) You can't change the address of reference, but can change it for a pointer(although, as pointer itself passed by value, this change will be discarded upon function exit)
Hope, this helped a bit
Actually, in the first case, you can't modify the argument. The pointer itself is immutable, you can only modify the value it points to.
If yes, why C++ allows to pass arguments by reference?
Because pointers can very easily be miss-used. References should almost always be prefered. For your case, what if you pass a NULL to with_ptr? You'll get undefined behavior, which is not possible if you use with_ref.
I think it is not clear inside the function that the i value could be modified.
It is very clear. If you see a function that takes a parameter by non-const reference, you can assume it will be changed.
I think that a method can only change an argument's value, if this is passed by reference. If you pass a argument by value in a method, then whatever change you make to its value, this will no be available in the parent method.
As far as I know, I think the reference is safer to use in a sense that it can't be modified (always points to the same thing), and should be initialized if it's a local variable. Pointer, however, can be change to point to somewhere else.
int x = 10;
int &y = x;
int *p = &x;
p++; //Legal if you know what's next
y++; // Increases the value of x. Now x = y = 11;
As my two cents, I think reference variables are mere alternative names for the same memory address by which it was initialized. This also explains pretty nice:
http://www.dgp.toronto.edu/~patrick/csc418/wi2004/notes/PointersVsRef.pdf
I'm new to the C++ community, and just have a quick question about how C++ passes variables by reference to functions.
When you want to pass a variable by reference in C++, you add an & to whatever argument you want to pass by reference. How come when you assign a value to a variable that is being passed by reference why do you say variable = value; instead of saying *variable = value?
void add_five_to_variable(int &value) {
// If passing by reference uses pointers,
// then why wouldn't you say *value += 5?
// Or does C++ do some behind the scene stuff here?
value += 5;
}
int main() {
int i = 1;
add_five_to_variable(i);
cout << i << endl; // i = 6
return 0;
}
If C++ is using pointers to do this with behind the scenes magic, why aren't dereferences needed like with pointers? Any insight would be much appreciated.
When you write,
int *p = ...;
*p = 3;
That is syntax for assigning 3 to the object referred to by the pointer p. When you write,
int &r = ...;
r = 3;
That is syntax for assigning 3 to the object referred to by the reference r. The syntax and the implementation are different. References are implemented using pointers (except when they're optimized out), but the syntax is different.
So you could say that the dereferencing happens automatically, when needed.
C++ uses pointers behind the scenes but hides all that complication from you. Passing by reference also enables you to avoid all the problems asssoicated with invalid pointers.
When you pass an object to a function by reference, you manipulate the object directly in the function, without referring to its address like with pointers. Thus, when manipulating this variable, you don't want to dereference it with the *variable syntax. This is good practice to pass objects by reference because:
A reference can't be redefined to point to another object
It can't be null. you have to pass a valid object of that type to the function
How the compiler achieves the "pass by reference" is not really relevant in your case.
The article in Wikipedia is a good ressource.
There are two questions in one, it seems:
one question is about syntax: the difference between pointer and reference
the other is about mechanics and implementation: the in-memory representation of a reference
Let's address the two separately.
Syntax of references and pointers
A pointer is, conceptually, a "sign" (as road sign) toward an object. It allows 2 kind of actions:
actions on the pointee (or object pointed to)
actions on the pointer itself
The operator* and operator-> allow you to access the pointee, to differenciate it from your accesses to the pointer itself.
A reference is not a "sign", it's an alias. For the duration of its life, come hell or high water, it will point to the same object, nothing you can do about it. Therefore, since you cannot access the reference itself, there is no point it bothering you with weird syntax * or ->. Ironically, not using weird syntax is called syntactic sugar.
Mechanics of a reference
The C++ Standard is silent on the implementation of references, it merely hints that if the compiler can it is allowed to remove them. For example, in the following case:
int main() {
int a = 0;
int& b = a;
b = 1;
return b;
}
A good compiler will realize that b is just a proxy for a, no room for doubts, and thus simply directly access a and optimize b out.
As you guessed, a likely representation of a reference is (under the hood) a pointer, but do not let it bother you, it does not affect the syntax or semantics. It does mean however that a number of woes of pointers (like access to objects that have been deleted for example) also affect references.
The explicit dereference is not required by design - that's for convenience. When you use . on a reference the compiler emits code necessary to access the real object - this will often include dereferencing a pointer, but that's done without requiring an explicit dereference in your code.
What would be better practice when giving a function the original variable to work with:
unsigned long x = 4;
void func1(unsigned long& val) {
val = 5;
}
func1(x);
or:
void func2(unsigned long* val) {
*val = 5;
}
func2(&x);
IOW: Is there any reason to pick one over another?
My rule of thumb is:
Use pointers if you want to do pointer arithmetic with them (e.g. incrementing the pointer address to step through an array) or if you ever have to pass a NULL-pointer.
Use references otherwise.
I really think you will benefit from establishing the following function calling coding guidelines:
As in all other places, always be const-correct.
Note: This means, among other things, that only out-values (see item 3) and values passed by value (see item 4) can lack the const specifier.
Only pass a value by pointer if the value 0/NULL is a valid input in the current context.
Rationale 1: As a caller, you see that whatever you pass in must be in a usable state.
Rationale 2: As called, you know that whatever comes in is in a usable state. Hence, no NULL-check or error handling needs to be done for that value.
Rationale 3: Rationales 1 and 2 will be compiler enforced. Always catch errors at compile time if you can.
If a function argument is an out-value, then pass it by reference.
Rationale: We don't want to break item 2...
Choose "pass by value" over "pass by const reference" only if the value is a POD (Plain old Datastructure) or small enough (memory-wise) or in other ways cheap enough (time-wise) to copy.
Rationale: Avoid unnecessary copies.
Note: small enough and cheap enough are not absolute measurables.
This ultimately ends up being subjective. The discussion thus far is useful, but I don't think there is a correct or decisive answer to this. A lot will depend on style guidelines and your needs at the time.
While there are some different capabilities (whether or not something can be NULL) with a pointer, the largest practical difference for an output parameter is purely syntax. Google's C++ Style Guide (https://google.github.io/styleguide/cppguide.html#Reference_Arguments), for example, mandates only pointers for output parameters, and allows only references that are const. The reasoning is one of readability: something with value syntax should not have pointer semantic meaning. I'm not suggesting that this is necessarily right or wrong, but I think the point here is that it's a matter of style, not of correctness.
Pointers
A pointer is a variable that holds a memory address.
A pointer declaration consists of a base type, an *, and the variable name.
A pointer can point to any number of variables in lifetime
A pointer that does not currently point to a valid memory location is given the value null (Which is zero)
BaseType* ptrBaseType;
BaseType objBaseType;
ptrBaseType = &objBaseType;
The & is a unary operator that returns the memory address of its operand.
Dereferencing operator (*) is used to access the value stored in the variable which pointer points to.
int nVar = 7;
int* ptrVar = &nVar;
int nVar2 = *ptrVar;
Reference
A reference (&) is like an alias to an existing variable.
A reference (&) is like a constant pointer that is automatically dereferenced.
It is usually used for function argument lists and function return values.
A reference must be initialized when it is created.
Once a reference is initialized to an object, it cannot be changed to refer to another object.
You cannot have NULL references.
A const reference can refer to a const int. It is done with a temporary variable with value of the const
int i = 3; //integer declaration
int * pi = &i; //pi points to the integer i
int& ri = i; //ri is refers to integer i – creation of reference and initialization
You should pass a pointer if you are going to modify the value of the variable.
Even though technically passing a reference or a pointer are the same, passing a pointer in your use case is more readable as it "advertises" the fact that the value will be changed by the function.
If you have a parameter where you may need to indicate the absence of a value, it's common practice to make the parameter a pointer value and pass in NULL.
A better solution in most cases (from a safety perspective) is to use boost::optional. This allows you to pass in optional values by reference and also as a return value.
// Sample method using optional as input parameter
void PrintOptional(const boost::optional<std::string>& optional_str)
{
if (optional_str)
{
cout << *optional_str << std::endl;
}
else
{
cout << "(no string)" << std::endl;
}
}
// Sample method using optional as return value
boost::optional<int> ReturnOptional(bool return_nothing)
{
if (return_nothing)
{
return boost::optional<int>();
}
return boost::optional<int>(42);
}
Use a reference when you can, use a pointer when you have to.
From C++ FAQ: "When should I use references, and when should I use pointers?"
A reference is an implicit pointer. Basically you can change the value the reference points to but you can't change the reference to point to something else. So my 2 cents is that if you only want to change the value of a parameter pass it as a reference but if you need to change the parameter to point to a different object pass it using a pointer.
Consider C#'s out keyword. The compiler requires the caller of a method to apply the out keyword to any out args, even though it knows already if they are. This is intended to enhance readability. Although with modern IDEs I'm inclined to think that this is a job for syntax (or semantic) highlighting.
Pass by const reference unless there is a reason you wish to change/keep the contents you are passing in.
This will be the most efficient method in most cases.
Make sure you use const on each parameter you do not wish to change, as this not only protects you from doing something stupid in the function, it gives a good indication to other users what the function does to the passed in values. This includes making a pointer const when you only want to change whats pointed to...
Pointers:
Can be assigned nullptr (or NULL).
At the call site, you must use & if your type is not a pointer itself,
making explicitly you are modifying your object.
Pointers can be rebound.
References:
Cannot be null.
Once bound, cannot change.
Callers don't need to explicitly use &. This is considered sometimes
bad because you must go to the implementation of the function to see if
your parameter is modified.
A reference is similar to a pointer, except that you don’t need to use a prefix ∗ to access the value referred to by the reference. Also, a reference cannot be made to refer to a different object after its initialization.
References are particularly useful for specifying function arguments.
for more information see "A Tour of C++" by "Bjarne Stroustrup" (2014) Pages 11-12