Difference between value parameter and reference parameter ? This question is asked sometime by interviewers during my interviews. Can someone tell me the exact difference that is easy to explain with example? And is reference parameter and pointer parameter are same thing ?
Thanks
Changes to a value parameter are not visible to the caller (also called "pass by value").
Changes to a reference parameter are visible to the caller ("pass by reference").
C++ example:
void by_value(int n) { n = 42; }
void by_ref(int& n) { n = 42; }
void also_value(int const& n); // Even though a reference is used, this is
// semantically a value parameter---though there are implementation
// artifacts, like not being able to write "n = 42" (it's const) and object
// identity (&n here has different ramifications than for by_value above).
One use of pointers is to implement "reference" parameters without using a special reference concept, which some languages, such as C, don't have. (Of course you can also treat pointers as values themselves.)
The main difference is whether the object passed is copied. If it's a value parameter the compiler must generate such code that altering the function parameter inside the function has no effect on the original object passsed, so it will usually copy the object. In case of reference parameters the compiler must generate such code taht all operations are done on the original object being passed.
A pointer is a low-level way of representing a reference, so passing a pointer (by value) is how languages like C typically achieve pass by reference semantics.
The difference is pretty simple: direct parameters are passed by value, and the receiver receives a copy of what is passed; meaning that if the parameter is modified by the receiver, these changes will not be reflected back to the caller. (This is often called, appropriately enough, pass by value, or by copy.
There are basically three kinds of parameters; pointer, reference and direct.
The difference is pretty simple: direct parameters are passed by value, and the receiver receives a copy of what is passed; meaning that if the parameter is modified by the receiver, these changes will not be reflected back to the caller. (This is often called, appropriately enough, pass by value, or bycopy.
Pointers are also passed by value, but rather than sending the actual value, the caller sends the address of the value. This means that by following this pointer, the receiver can modify the argument. Note that changes made to the actual pointer still aren't reflected back to the caller.
The final form, call-by-reference, is sort of a middle ground between these two approaches. Essentially it can be thought of as a pointer that looks like a value.
It is worth mentioning that at the core of it all, parameters are always passed by value, but different languages have different ways of implementing reference semantics (see Kylotans answer).
// Example using C
// bycopy
int multiply(int x, int y) {
return x * y;
}
void multiply_p(int *x, int y) {
*x *= y;
}
int main () {
int i, j, k;
i = 20;
j = 10;
k = multiply(i,j); // k is now 200
multiply_p(&i, k); // i is now 4000 (200 * 20)
return 0;
}
Pseudocode:
Pass by Value:
void setTo4(value) { // value is passed by value
value = 4;
}
int x = 1;
setTo4(x);
// x is still 1
Pass by Reference:
void setTo4(value) { // value is passed by reference
value = 4;
}
int x = 1;
setTo4(x);
// x is 4
Related
If a pointer is passed to a function for read only, then this pointer is an IN parameter.
If a pointer is passed to a function for read only, but this function makes a copy of the pointer to have access to it in module related functions for read only operations, this pointer is still IN.
If the function still uses the pointer as read only, but the other module related functions use the pointer for write operations, what does that make the pointer?
An IN parameter, but without const? An in/out parameter?
Example of what I mean:
class SteeringWheel {
public: float rotation;
public: SteeringWheel(void) {
this->rotation = 0.f;
}
};
class Car {
private: SteeringWheel *steeringWheel;
public:
/**
* #param[?] steeringWheel Is the steering wheel in or in/out?
*/
Car (SteeringWheel *steeringWheel) {
this->steeringWheel = steeringWheel;
}
/**
* #param[in] degree Steering amount in degrees.
*/
void steer(float degree)
{
this->steeringWheel->rotation += degree;
}
};
int main(int argc, char **argv)
{
SteeringWheel steeringWheel();
/* car() uses steeringWheel as read only. */
Car car(&steeringWheel);
/* steer() uses steeringWheel from car() to write. */
car.steer(50.f);
return 0;
}
I believe that the in and out specifiers do not exactly mean what you think. From the doxygen documentation of the param tag:
The \param command has an optional attribute, (dir), specifying the
direction of the parameter. Possible values are "[in]", "[in,out]",
and "[out]", note the [square] brackets in this description. When a
parameter is both input and output, [in,out] is used as attribute.
The direction of the parameter usually mean the following:
in: The parameter is injected into the function as input, but not written to.
out: The parameter is injected into the function, but not as input. Rather, it is written to by the function.
in, out: The parameter is injected into the function as input and is eventually written to by the function.
In your example:
/**
* #param[?] steeringWheel Is the steering wheel in or in/out?
*/
Car (SteeringWheel *steeringWheel) {
this->steeringWheel = steeringWheel;
}
I think the steeringWheel parameter is in because you inject it and use it in your method. However, you never write to it (i.e. to the parameter itself), so it is not out. In other words, you only use your method to inject an address to your function, nothing else. The same apply for your second method, where you inject the degree parameter, but never write to it.
To clarify a bit more on the meaning of in and out, here is an example of an out parameter:
/**
* #param[out] p_param We write to the parameter!
*/
void makeFour(int * p_param)
{
*p_param = 4; // Out because of this line!
}
Notice that we write a new value directly into the parameter. This is the meaning of out: information comes out of the method through the parameter. You can now write:
int main()
{
int myInt = 0;
std::cout << myInt; // prints 0.
makeFour(&myInt); // p_param == &myInt here.
std::cout << myInt; // prints 4: the method wrote directly
// in the parameter (out)!
return 0;
}
Hope this helps!
It is not easy to decide, but I would still mark your parameter as in,out (or out), as it is a pointer to a non-const object, and you may change the state of that outside object directly or indirectly later - as in your example.
Marking it in hides the detail that the pointed SteeringWheel object may change later upon usage of Car.
Also, it can puzzle users why an input only pointer parameter is not marked const.
Making it in,out may not be accurate completely, but is surely more error prone.
An alternative could be something like the following (a note regarding the lifetime of the SteeringWheel should come handy here anyway):
/**
* #param[in] steeringWheel Pointer to the SteeringWheel object.
* #warning The memory address of the pointed object is saved.
* It must outlive this object, and can change upon usage of this object.
*/
Car (SteeringWheel *steeringWheel) {
this->steeringWheel = steeringWheel;
}
But I would just probably stick with marking it in,out.
Specifying the direction of parameters in C++ may be complicated, and frankly speaking, I am not too much in favor of them, as having tokens for pointers, references, and the keyword for constness provide enough information in the signature on how a parameter may be used. Thus, marking it in the DoxyPress documentation is a bit redundant, not expressive enough (as your example shows), and may get out of sync with the implementation. Documenting parameter directions may play a bigger role in case of other languages that lack these additional constructs in function signatures.
How does scope interact with variable declaration, initialisation, and assignment? The definition of those terms based on what I have learned so far is listed below:
Declaration: States the type of a variable, and it's name/identifier. Variables must have been declared before they can be assigned or read.
Assignment: Throws away the existing value of a variable and replaces it with a new one, the old value is thrown away at the end of the assignment statement, so the value can be incremented or otherwise adjusted, for example: x = x + y;
Initialisation: The name used for the first assignment of a variable, before initialisation, a variable has a default value, in the case of objects, those objects have a null value. Initialisation can be done in conjunction with declaration.
Scope: The "lifespan" of a variable, a variable is in scope until the end of the code block, at which point the memory used to store that variable is freed up. In effect, the variable is deleted or "killed", when the code block ends.
What I don't know is how scope interacts with Declaration and assignment. While the scope of a variable seems to be based solely on the code block in which it is Declared, I don't know how assignment interacts with scope. For example:
public class exampleClass
{
public static void main(String[] args) // using java for example
{
int x = 5; // x is declared here, and initialised with a value of 5
for (int i = 0; i < 10; i++) // i is declared and initialised here
{
x = i; // x is assigned the value of i each iteration
} // i goes out of scope here
System.out.println(x); // the value of x is printed
} // x goes out of scope here
}
In this example, x is declared and initialised (do we just say initialised?) in the main method, and is in scope for that method. However, x is assigned a value in the while loop. What will be printed when this code is executed, but more importantly why? Will it print "5", or "9"?
I have seen code throw up compiler exceptions because of syntax that would imply that x should print 5. However when I run this example code, I get "9".
One final question, why is it that multiple variables can be declared and initialised inline:
int x = 1, y = 4, z = 6;
But variables cannot be assigned inline:
x = 1, y = 4, z = 6;
The distinction between declaration and initialisation can be blurry; some languages make a clear distinction between these actions, in others initialisation is declaration. If a variable is initialised while it's being declared, it doesn't matter which you call it.
However, x is assigned a value in the while loop. What will be printed when this code is executed, but more importantly why? Will it print "5", or "9"?
9, because that's the last value that has been assigned to it before you print it.
Scope: The "lifespan" of a variable, a variable is in scope until the end of the code block, at which point the memory used to store that variable is freed up. In effect, the variable is deleted or "killed", when the code block ends.
Yes and no. Scope defines in what parts of code a particular variable is available. Different languages can have very different definitions of their scoping rules. A variable is typically garbage collected (in languages where that's applicable) when it goes out of scope, when no piece of code has any further access to it. In a simple function block, that happens when the function ends.
However, see this Javascript example:
function foo() {
var bar = 'baz';
return function () {
alert(bar);
};
}
The inner function which is returned from this function still holds a reference to bar. Even if foo ends, bar is being closed over by a closure and is still in scope within the inner function. As long as a reference to that returned function exists, bar still exists.
Say we have this block:
int (^aBlock)(BOOL) = ^(BOOL param) { ...
My current understanding of this is: the first int is the return type, (^aBlock)(BOOL) gives the name of the method and the type of its parameter, and = ^(BOOL param) is the parameter's name inside the block ... plus the parameter's type again?
Why is the syntax such that we have to list the parameter type twice? Could the two types ever be different?
It is not quite "listing the parameter type twice", you are in the first case declaring the type of a block variable, and in the second case you are defining a block literal.
Then you are assigning the literal to the value of the variable. You could even do something like this, which is equivalent and better illustrates the fact that these are really two totally independent declarations, despite being associated with an assign statement:
id thisBlock = ^id (id x, NSUInteger idx) {
NSLog(#"x = %#",x);
return x;
};
id (^thatBlock)(id obj, NSUInteger index) = thisBlock;
The fact that they are independent of each other means it's probably not even correct to attempt to provide some kind of transference or inheritance of typing information from the left hand side of the expression to the right. And yes, the types can be different - consider this code compiles and executes just fine:
id (^thatBlock)(NSArray *, NSDictionary *, NSString *) = ^id (id x, id y, id z) {
NSLog(#"x = %#",x);
return x;
};
thatBlock(#[],#{},#"");
Hope this helps!
Why is the syntax such that we have to list the parameter type twice?
The block is designed in this way, and so you can do it like this:
int (^aBlock)(BOOL);
aBlock = ^(BOOL param) {
...
};
It just likes
- (int)aMethodWithParam:(BOOL)param;
- (int)aMethodWithParam:(BOOL)param {
...
}
Could the two types ever be different?
Nope, and what's more, the order of the types should be the same, i.e.:
int (^aBlock)(BOOL, NSString*) = ^(BOOL param, NSString *aString) {
...
};
And here's a clear figure for block:
The code snippet you have given isn't a block declaration: it's a block declaration and a block definition. First, you declare an identifier named aBlock:
int (^aBlock)(BOOL)
then you define a block:
^(BOOL param) { ...
Both of these are parsed and evaluated separately. Since you are assigning one to the other, the compiler does a type-check to make sure that the expression on the left hand side (your aBlock declaration) is of the same type as the expression on the right hand side (the block definition).
So, the answer is, these parts need to be evaluated separately. Since the block definition is being compiled on its own, you have to include the type of param, because otherwise the compiler won't know what type it should be. (Yes, you could make an exception in this case, and look across the assignment, but everywhere else in the C language you declare an identifier by giving a type first, so why do it differently here?)
But -- you may say -- if that's true why didn't I have to define the return type (int) on the right-hand side?
Very astute of you. The answer is that, when writing a block expression, you don't need to define the return type, because the compiler can infer it from the return statement (or lack thereof) inside the block.
(So why do you have to include return types with function definitions? Well, history, I guess. Programming language definitions were created by imperfect humans.)
According to Ry's Objective C Tutorial:
Blocks use all the same mechanics as normal functions. You can declare a block variable just like you would declare a function.
NSInteger (^BlocksAddition)(NSInteger x,NSInteger y)=^NSInteger(NSInteger x, NSInteger y){
return x+y;
};
NSUInteger result=BlocksAddition(4,5);
NSLog(#"Addition Result:%d",result);
Not sure what I'm doing wrong here. I have a struct that is used heavily through my program.
typedef struct _MyStruct {
// ... handful of non-trivial fields ...
} MyStruct;
I expect (read, intend) for lots of parts of the program to return one of these structs, but many of them should be able to return a "null" struct, which is a singleton/global. The exact use case is for the implementing function to say "I can't find what you asked me to return".
I assumed this would be a simple case of defining a variable in a header file, and initializing it in the .c file.
// MyStruct.h
// ... Snip ...
MyStruct NotFoundStruct;
-
// MyStruct.c
NotFoundStruct.x = 0;
NotFoundStruct.y = 0;
// etc etc
But the compiler complains that the initialization is not constant.
Since I don't care about what this global actually references in memory, I only care that everything uses the same global, I tried just removing the initialization and simply leaving the definition in the header.
But when I do this:
MyStruct thing = give_me_a_struct(some_input);
if (thing == NotFoundStruct) {
// ... do something special
}
Th compiler complains that the operands to the binary operator "==" (or "!=") are invalid.
How does one define such as globally re-usable (always the same memory address) struct?
This doesn't directly answer your question, but it won't fit in a comment...
If you have a function that may need to return something or return nothing, there are several options that are better than returning a "null struct" or "sentinel struct," especially since structs are not equality comparable in C.
One option is to return a pointer, so that you can actually return NULL to indicate that you are really returning nothing; this has the disadvantage of having significant memory management implications, namely who owns the pointer? and do you have to create an object on the heap that doesn't already exist on the heap to do this?
A better option is to take a pointer to a struct as an "out" parameter, use that pointer to store the actual result, then return an int status code indicating success or failure (or a bool if you have a C99 compiler). This would look something like:
int give_me_a_struct(MyStruct*);
MyStruct result;
if (give_me_a_struct(&result)) {
// yay! we got a result!
}
else {
// boo! we didn't get a result!
}
If give_me_a_struct returns zero, it indicates that it did not find the result and the result object was not populated. If it returns nonzero, it indicates that it did find the result and the result object was populated.
C doesn't allow global non-const assignments. So you must do this in a function:
void init() {
NotFoundStruct.x = 0;
NotFoundStruct.y = 0;
}
As for the comparison, C doesn't know how to apply a == operator to a struct. You can overload (redefine) the operator in C++, but not in C.
So to see if a return value is empty, your options are to
Have each function return a boolean value to indicate found or not, and return the struct's values via pointers through the argument list. (eg. bool found = give_me_a_struct(some_input, &thing);)
Return a pointer to a struct, which can be NULL if nothing exists. (eg. MyStruct* thing = give_me_a_struct(some_input);)
Add an additional field to the struct that indicates whether the object is valid.
The third option is the most generic for other cases, but requires more data to be stored. The best bet for your specific question is the first option.
// MyStruct.h
typedef struct _MyStruct {
// fields
} MyStruct;
extern MyStruct NotFoundStruct;
// MyStruct.c
#include "my_struct.h"
MyStruct NotFoundStruct = {0};
But since you can't use the == operator, you will have to find another way to distinguish it. One (not ideal) way is to have a bool flag reserved to indicate validity. That way, only that must be checked to determine if it's a valid instance.
But I think you should consider James's proposed solution instead
In the header:
// Structure definition then
extern MyStruct myStruct;
In the .c that contains global data
struct MyStruct myStruct
{
initialize field 1,
initialize field 2,
// etc...
};
Can someone please explain me the following code snippet?
value struct ValueStruct {
int x;
};
void SetValueOne(ValueStruct% ref) {
ref.x = 1;
}
void SetValueTwo(ValueStruct ref) {
ref.x = 2;
}
void SetValueThree(ValueStruct^ ref) {
ref->x = 3;
}
ValueStruct^ first = gcnew ValueStruct;
first->x = 0;
SetValueOne(*first);
ValueStruct second;
second.x = 0;
SetValueTwo(second); // am I creating a copy or what? is this copy Disposable even though value types don't have destructors?
ValueStruct^ third = gcnew ValueStruct;
third->x = 0;
SetValueThree(third); // same as the first ?
And my second question is: is there any reason to have something like that?:
ref struct RefStruct {
int x;
};
RefStruct% ref = *gcnew RefStruct;
// rather than:
// RefStruct^ ref = gcnew RefStruct;
// can I retrieve my handle from ref?
// RefStruct^ myref = ???
What is more: I see no difference between value type and ref type, since both can be pointed by handler ;(
Remember that the primary use of C++/CLI is for developing class libraries for consumption by GUIs / web services built in other .NET languages. So C++/CLI has to support both reference and value types because other .NET languages do.
Furthermore, C# can have ref parameters that are value typed as well, this isn't unique to C++/CLI and it doesn't in any way make value types equivalent to reference types.
To answer the questions in your code comments:
am I creating a copy or what?
Yes, SetValueTwo takes its parameter by value, so a copy is made.
is this copy Disposable even though value types don't have destructors?
Incorrect. Value types can have destructors. Value types cannot have finalizers. Since this particular value type has a trivial destructor, the C++/CLI compiler will not cause it to implement IDisposable. In any case, if a parameter is an IDisposable value type, the C++/CLI compiler will ensure that Dispose is called when the variable goes out of scope, just like stack semantics for local variables. This includes abnormal termination (thrown exception), and allows managed types to be used with RAII.
Both
ValueStruct% ref = *gcnew ValueStruct;
and
ValueStruct^ ref = gcnew ValueStruct;
are allowed, and put a boxed value type instance on the managed heap (which isn't a heap at all, but a FIFO queue, however Microsoft chooses to call it a heap like the native memory area for dynamic allocation).
Unlike C#, C++/CLI can keep typed handles to boxed objects.
If a tracking reference is to a value type instance on the stack or embedded in another object, then the value type content has to be boxed in the process of formed the reference.
Tracking references can also be used with reference types, and the syntax to obtain a handle is the same:
RefClass^ newinst = gcnew RefClass();
RefClass% reftoinst = *newinst;
RefClass^% reftohandle = newinst;
RefClass stacksem;
RefClass^ ssh = %stacksem;
One thing that I can never seem to remember completely is that the syntax isn't 100% consistent compared to native C++.
Declare a reference:
int& ri = i; // native
DateTime% dtr = dt; // managed tracking reference
Declare a pointer:
int* pi; // native
Stream^ sh; // tracking handle
Form a pointer:
int* pi = &ri; // address-of native object
DateTime^ dth = %dtr; // address-of managed object
Note that the unary address-of operator is the same as the reference notation in both standard C++ and C++/CLI. This seems to contradict a tracking reference cannot be used as a unary take-address operator (MSDN) which I'll get back to in a second.
First though, the inconsistency:
Form a reference from a pointer:
int& iref = *pi;
DateTime% dtref = *dth;
Note that the unary dereference operator is always *. It is the same as the pointer notation only in the native world, which is completely opposite of address-of which, as mentioned above, are always the same symbol as the reference notation.
Compilable example:
DateTime^ dth = gcnew DateTime();
DateTime% dtr = *dth;
DateTime dt = DateTime::Now;
DateTime^ dtbox = %dt;
FileInfo fi("temp.txt");
// FileInfo^ fih = &fi; causes error C3072
FileInfo^ fih = %fi;
Now, about unary address-of:
First, the MSDN article is wrong when it says:
The following sample shows that a tracking reference cannot be used as a unary take-address operator.
The correct statement is:
% is the address-of operator for creation of a tracking handle. However its use is limited as follows:
A tracking handle must point to an object on the managed heap. Reference types always exist on the managed heap so there is no problem. However, value types and native types may be on the stack (for local variables) or embedded within another object (member variables of value type). Attempts to form a tracking handle will form a handle to a boxed copy of the variable: the handle is not linked to the original variable. As a consequence of the boxing process, which requires metadata which does not exist for native types, it is never possible to have a tracking handle to an instance of a native type.
Example code:
int i = 5;
// int^ ih = %i; causes error C3071
System::Int32 si = 5;
// System::Int32^ sih = %si; causes error C3071
// error C3071: operator '%' can only be applied to an instance
// of a ref class or a value-type
If System::Int32 isn't a value type then I don't know what is. Let's try System::DateTime which is a non-primitive value type:
DateTime dt = DateTime::Now;
DateTime^ dtbox = %dt;
This works!
As a further unfortunate restriction, primitive types which have dual identity (e.g. native int and managed value type System::Int32) are not handled correctly, the % (form tracking reference) operator cannot perform boxing even when the .NET name for the type is given.