I was watching a video and saw this code:
class Dog {
public:
Dog() : age(3), name("dummy") {}
void setAge(const int &a) { age = a; }
private:
int age;
std::string name;
};
I was curious about the function signature for setAge because I've never used const as a function parameter. I've looked at several related answers, but none that seemed to answer my question.
In such an elementary example, it's hard to see the benefit of passing a const reference to a function.
Is there any reason why you'd want to make the reference a const? The only application I could think of is in embedded programming when making a copy of a variable could waste precious space.
Are there any simple examples, perhaps, where the impact is seen easily of passing a const reference?
Consider the following three examples:
(i) void setAge(int &a) { age = a; }
(ii) void setAge(const int &a) { age = a; }
(iii) void setAge(int a) { age = a; }
Further think of your class as an encapsulated object, that is the outside world in general doesn't know what's going on inside.
Then, using case (i), the caller cannot know that a has not been changed afterwards.
int a=3;
dog.setAge(a); //case (i): what is "a" afterwards?
One does not know what value a holds after the function call -- in fact, the function signature tells the caller that a change of a is likely to occur.
On the other hand, by using variant (ii), you again pass the object via reference, that is you do not make a copy, but tell the function the memory address where it can go to accss the parameter. In contrast to case (i), now you ensure the caller "nothing is going to happen to your parameter". That is, you can safely work with the parameter afterwards and be assured it has still the same value as before (--at least in principle, as also bad things like a const_cast might happen inside the function).
Finally, in case (iii), one makes a copy of the int and uses that inside the function. For built-in types like int, that is in fact the preferred way to pass function parameters. However, it might be uneffective if the object is expensive-to-copy.
With regard to the whole const correctness-topic, see here.
Making a copy of a variable wastes precious space and wastes precious time, which is why avoiding unnecessary copies is something to worry about in any kind of programming, not just embedded one. For this reason using const T & references for passing read-only parameters has always been an idiom in C++.
However, it is usually reserved for passing in "heavy" objects, i.e. objects that that are relatively expensive to copy. Using it with scalar types (as in your example) makes little or no sense.
Consider an object which has a large memory footprint. You need to pass it to a function - a function which will only extract some information without changing the object in anyway possible. A const reference is a good candidate in such a case.
std::vector can be an example, a Matrix object is another example.
void setAge(const int &a) { age = a; }
You are passing a const int reference, you don't intend to modify the value a inside function setAge.
void setAge(int &a) { age = a; }
You are passing a int reference, you don't care does function setAge modify the value a inside function setAge.
Doing const correctness is a good programming practice here as a practice of better interface definition. First function signature convey the message very clearly, and second one is vague compared to first one.
You can read more about const correctness here https://isocpp.org/wiki/faq/const-correctness#overview-const
Related
When learning C++, one of the first functions one gets for learning the concept of a function is something like
int add(int a, int b)
{
return a+b;
}
Now I was wondering: Should I use the const-keyword here, or rather not, resulting in
int add(const int a, const int b)
{
return a+b;
}
But would that make sense? Would it speed my program up, do some other important stuff, or just increase confusion?
From the caller's perspective, both the first and the second form are the same.
Since the integers are passed by value, even if the function modifies a and b, the modified values are copies of the original and won't be visible to the caller.
However, from the function implementer's perspective there's a difference. In fact, in the second form:
int add(const int a, const int b)
you will get a compiler error if you try to modify the values of a and b inside the function's body, as they are marked as const.
Instead, you can change those values if you omit the const.
Again, those modifications will not be visible to the caller.
If you are actually struggling with correctness bugs in your codebase where const would have helped, then add const.
That said, there are related issues that you should consider. Top-level qualifiers of function parameters aren't part of the function type, so your function type is still just int(int, int). (The qualifiers only affect the parameter variables in the function definition.) That means that any declaration of the function also ignores qualifiers, so int add(int, int) and int add(const int, int) and int add(int, const int) all declare the same function. So you have to decide on a policy for how you write header files. And now you have three essential positions you can take:
Always qualify in both declaration and definition. The upside is that this perhaps keeps code looking "consistent" (think copy/pasting when creating implementations). The downside is that the qualifiers have nothing to do with the interface, and are not at all enforcible (you can change the definition later!), so at best it's noise, at worst it's wrong.
Qualify in the definition, but not in the other declarations. The upside is that this communicates the interface correctly and you still get const checking in the definition. The downside is that some people might be confused by the discrepancies in spelling. (Just like people may be confused that a static constexpr T foo; class member can be defined with const T C::foo;.)
Don't qualify either. The upside is that it's consistent, clean, easy to remember and minimal. The downside is that you're missing out on correctness checks in your definition.
There's no right answer. If you're the codebase owner or project lead, you should decide based on what the biggest problems in your codebase and team are. (My personal position is to stick with (3) until you have a good reason to change.)
Using int add(const int a, const int b) means that you cannot modify the input parameters within the function body.
Perhaps the compiler can make some optimisations based on that, so it ought therefore to be never slower than the non-const equivalent. But I have never observed this effect in practice.
Passing const parameters can also increase the stability of your code base, particularly in a collaborative project.
All this said though, I find it too verbose, rather unnecessary, and would use int add(int a, int b). Very occasionally, for particularly long functions, I exploit the fact that you can declare the function with non-const parameters, and define it with the parameters const.
I feel like everyone is dancing around this part of the answer...
It's true that using const will keep the function from modifying the value of your int a & b while inside the function. This can be extremely useful, so use it as you wish, of the compiler allows it. But, the function caller will never know about any changes to a & b once the function finishes. So even if a & b are changed, no one except the defined function will know their updated values.
int funcB(int a, int b)
{
a = a+1;
b = b*b;
return a+b;
}
void funcA()
{
int s = 5;
int t = 6;
int result = funcB(s, t);
printf("%f + %f = %f", s,t, result);
}
funcA prints: "5 + 6 = 42"
Const protection is often used when passing values by reference, ie:
int function(const int &a, const int &b) {}
This passes a reference of a and b to the function (ie, does not make copies of a and b but passes only a memory address of that variable, aka: the handle). When passing a variable by reference, any changes made to the variable are remembered outside the scope of the function and can change the way your program runs. This is generally undesired behavior.
So if you rework funcB from above and pass by reference:
int funcB(int &a, int &b)
{
a = a+1;
b = b*b;
return a+b;
}
funcA prints: "6 + 36 = 42"
If you add const correctness to funcB:
int funcB(const int &a, const int &b)
{
a = a+1;
b = b*b;
return a+b;
}
I don't think the compiler will let you even do this since you would be explicitly trying to modify values that you've protected via const.
Another time when it may be really important to use const is when you're passing by pointer, instead of reference or copy...
int funcB(int *a, int *b)
{
a = a+1;
b = b*b;
return a+b;
}
Unless you're a pointer expert, avoid passing pointers without const pretention. This func will likely attempt to iterate the index of your pointer arrays and you'd open yourself up to run time errors related to out of bound memory. You could accidently see memory from an entirely different program... but probably not.
Lastly, since you're just passing int, there's no practical need to pass by reference (which is done often to keep from adding complex data into memory because each non-reference or non-pointer pass to functions copies the value into memory for the life of the function being called) since the memory footprint of int is so small. Unless, you're working with specialized hardware that has extremely limited memory, then it may be useful; this won't apply to most standard computers and desktops made within the past 20 years, or smart phones.
int add(const int a, const int b)
{
return a+b;
}
Here const in used so that the function does not modify the original values of a and b by any chance.
In the above example it does not make sense. But if it would have been an example like
int add(const int *a, const int *b)
{
//*a = 10; //This will throw error.
return a+b;
}
In functions where objects,arrays or such data types are passed its a good approach to use const to avoid mutation of original data structure.
If you want to be truly const correct, then you should add it, but in reality all it will do is make you type and read more.
Nothing will go faster, and while it could mean that your variables go into another location in memory, it's unlikely to on most modern machines.
What it will stop you doing is accidentally assigning them values, but as they're local to the function, it's relatively unimportant. What would matter is if they were references, as this demonstrates your intent to the caller to not change them.
You can use const there if you like.
It is unlikely to speed up your program because any reasonable compiler can already see that no code path alters the value of a or b and make any optimisations it needs.
a and b are int, which are passed by value so making them const has no impact on the users of this function.
The only possible advantage is where your function is long and more complex and you want to avoid possible programming errors where the original values of a or b are changed during the function.
If you will use const then you cannot modify the value of a,b. Thats why we don't use const.
The primary purpose of constness is to provide documentation and prevent programming mistakes. Const allows you to make it clear to yourself and others that something should not be changed. Moreover, it has the added benefit that anything that you declare const will in fact remain const short of the use of forceful methods. It's particularly useful to declare reference parameters to functions as const references:
bool SomeFunction (const myObj& obj);
Here, a myObj object is passed by reference into SomeFunction. For safety's sake, const is used to ensure that SomeFunction cannot change the object--after all, it's just supposed to make sure that the object is in a valid state. This can prevent silly programming mistakes that might otherwise result in damaging the object (for instance, by setting a field of the class for testing purposes, which might result in the field's never being reset). Moreover, by declaring the argument const, users of the function can be sure that their object will not be changed and not need to worry about the possible side effects of making the function call.
moreover, The reason is that const for the parameter only applies locally within the function, since it is working on a copy of the data. This means the function signature is really the same anyways. It's probably bad style to do this a lot though. I personally tend to not use const except for reference and pointer parameters.
as a law of thumb we are advised to use const as much as possible because we get a great help from the compiler so whenever we try to change a constant the mistake will be caught by the compiler; it's a good thing avoiding being caught in error-prones.
the second thing: a and b passed by value so they are created as local constants But as long as they are not changeable why we need copies??
the good thing is to pass a const reference which means never change parameters and avoid copies (by value):
int add(const int& a, const int& b) // now there are no copies of a and b in addition we cannot modify a and be
{
return a+b;
}
Prior to C++11, if I had a function that operated on large objects, my instinct would be to write functions with this kind of prototype.
void f(A &return_value, A const ¶meter_value);
(Here, return_value is just a blank object which will receive the output of the function. A is just some class which is large and expensive to copy.)
In C++11, taking advantage of move semantics, the default recommendation (as I understand it) is the more straightforward:
A f(A const ¶meter_value);
Is there ever still a need to do it the old way, passing in an object to hold the return value?
Others have covered the case where A might not have a cheap move constructor. I'm assuming your A does. But there is still one more situation where you might want to pass in an "out" parameter:
If A is some type like vector or string and it is known that the "out" parameter already has resources (such as memory) that can be reused within f, then it makes sense to reuse that resource if you can. For example consider:
void get_info(std::string&);
bool process_info(const std::string&);
void
foo()
{
std::string info;
for (bool not_done = true; not_done;)
{
info.clear();
get_info(info);
not_done = process_info(info);
}
}
vs:
std::string get_info();
bool process_info(const std::string&);
void
foo()
{
for (bool not_done = true; not_done;)
{
std::string info = get_info();
not_done = process_info(info);
}
}
In the first case, capacity will build up in the string as the loop executes, and that capacity is then potentially reused on each iteration of the loop. In the second case a new string is allocated on every iteration (neglecting the small string optimization buffer).
Now this isn't to say that you should never return std::string by value. Just that you should be aware of this issue and apply engineering judgment on a case by case basis.
It is possible for an object to be large and expensive to copy, and for which move semantics cannot improve on copying. Consider:
struct A {
std::array<double,100000> m_data;
};
It may not be a good idea to design your objects this way, but if you have an object of this type for some reason and you want to write a function to fill the data in then you might do it using an out param.
It depends: does your compiler support return-value-optimization, and is your function f designed to be able to use the RVO your compiler supports?
If so, then yes, by all means return by value. You will gain nothing at all by passing a mutable parameter, and you'll gain a great deal of code clarity by doing it this way. If not, then you have to investigate the definition of A.
For some types, a move is nothing more than a copy. If A doesn't contain anything that is actually worth moving (pointers transferring ownership and so forth), then you're not going to gain anything by moving. A move isn't free, after all; it's simply a copy that knows that anything owned by the original is being transferred to the copy. If the type doesn't own anything, then a move is just a copy.
Forgive me if this has been asked before, I am sure it has but I couldn't find an answer I was happy with.
I am coming to cpp from a heavy Java background and would like to understand when to return a reference/pointer to an object rather than a copy.
for the following class definition:
class SpaceShip {
string name;
WeaponSystem weaponSystem; //represents some object, this is just an example, I dont have this type of object at all in my program
int hull;
string GetName() const {
return name;
}
WeaponSystem GetWeaponSystem() const {
return weaponSystem;
}
int GetHull() const {
return hull;
}
};
I know that returning a copy of things is expensive, I would think this means I want to avoid returning something like a string or weaponSystem by value, but an int by value is ok.
Is this right? I also know that I need to be aware of where things live in memory, does returning a reference to something in this class mean danger down the line if this object is destroyed and something still owns a reference to it's name?
On your last point, you definitely need to be a lot more careful about resource management in C++ than in Java. In particular, you need to decide when an object is no longer needed. Returning by reference has an effect of aliasing to the returned object. It is not noticeable when the object you are sharing is immutable, but unlike Java's Strings, C++ string are mutable. Therefore if you return name by value and then rename your SpaceShip, the caller would see the old name even after the renaming. If you return by reference, however, the caller will see a change as soon as ShaceShip is renamed.
When you deal with copying complex objects, you can decide how much is copied by providing a custom implementation of a copy constructor. If you decide to provide a copy constructor, don't forget the rule of three, and override the other two.
It "works" but you should have
const string& GetName() const {
It may also be beneficial to have the following also
const WeaponSystem& GetWeaponSystem() const {
Also, class is private by default, as such, your accessor functions are private.
the thing you have to know is every getter of your class must be prototype like that :
const <type> &className::getXXX() const
{
...
}
and every setter you make like that :
void className::setXXX(const <type> &)
{
...
}
Use reference when it's possible.
Sometimes, with complex object you can use pointer. That's depend on your code structure.
In my previous app I had an object like this:
class myType
{
public:
int a;
string b;
}
It had a lot of instances scattered everywhere and passed around to nearly every function.
The app was slow. Profiling said that 95% of time is eaten by the string allocator function.
I know how to work with the object above, but not how to work with string pointers.
class myType
{
public:
int a;
string* b;
}
They told me to use pointers as above.
How much faster is it with a string pointer?
What is copied when I copy the object?
How to the following using the class with the pointer:
Access the string value
Modify the string value without modifying the one in the object (copy?)
General things that change if I use string pointers?
It will actually probably be slower - you still need to create and copy the strings, but now you have the overhead of dynamic allocation on top. My guess is that you are copying your objects around too much - whenever you call a function, your myType objects should be passed as const references, wherever possible, not by value:
void f( const myType & mt ) {
// stuff
}
If you actually need to change mt, you would have used a non-const reference - this is also less expensive than passing a value and returning a new value with modified fields.
I think using a pointer like this is a bad idea. Instead, look at how your myType is being used instead. In particular, instead of this:
void foo(myType a)
{
// ...
}
Consider this:
void foo(myType const &a)
{
// ...
}
In the former case, a copy of myType needs to be created to pass to the function foo(), in the second, no copy is needed, since a reference is passed instead (it's marked as const so that you can be sure foo() doesn't try to modify it - giving you (almost)the same behaviour as the first method).
There are probably other things you could change, but my guess is that doing this would give you the most bang for your buck (and it's a pretty mechanical change, so hopefully not too much chance of problems being introduced)
To add to the other answers, you also should be making sure any member functions of that class that pass a string are passing by const reference. For example, say your class constructor definition looks like this:
myType::myType(int a, string b)
Use this instead:
myType::myType(int a, const string& b)
So basically, go through all your function parameters throughout your project, and change string to const string&, and myType to const myType&. This alone should fix the majority of your performance issues.
Note: About dynamically allocating the string and passing as a pointer: This is not a good idea, as though it will lighten the performance load somewhat, you're going to be extremely vulnerable to memory leaks, which makes debugging a nightmare (in addition to being much more haphazardly destructive to your performance than running slow). As a general rule, I highly discourage passing naked pointers. There's almost always a better, safer alternative.
In a project I maintain, I see a lot of code like this for simple get/set methods
const int & MyClass::getFoo() { return m_foo; }
void MyClass::setFoo(const int & foo) { m_foo = foo; }
What is the point in doing that instead of the following?
int MyClass::getFoo() { return m_foo; } // Removed 'const' and '&'
void MyClass::setFoo(const int foo) { m_foo = foo; } // Removed '&'
Passing a reference to a primitive type should require the same (or more) effort as passing the type's value itself, right?
It's just a number after all...
Is this just some attempted micro-optimization or is there a true benefit?
The difference is that if you get that result into a reference yourself you can track the changes of the integer member variable in your own variable name without recalling the function.
const &int x = myObject.getFoo();
cout<<x<<endl;
//...
cout<<x<<endl;//x might have changed
It's probably not the best design choice, and it's very dangerous to return a reference (const or not), in case a variable that gets freed from scope is returned. So if you return a reference, be careful to be sure it is not a variable that goes out of scope.
There is a slight difference for the modifier too, but again probably not something that is worth doing or that was intended.
void test1(int x)
{
cout<<x<<endl;//prints 1
}
void test2(const int &x)
{
cout<<x<<endl;//prints 1 or something else possibly, another thread could have changed x
}
int main(int argc, char**argv)
{
int x = 1;
test1(x);
//...
test2(x);
return 0;
}
So the end result is that you obtain changes even after the parameters are passed.
To me, passing a const reference for primitives is a mistake. Either you need to modify the value, and in that case you pass a non-const reference, or you just need to access the value and in that case you pass a const.
Const references should only be used for complex classes, when copying objects could be a performance problem. In the case of primitives, unless you need to modify the value of the variable you shouldn't pass a reference. The reason is that references take more computation time than non-references, since with references, the program needs to look up in a table to find the address of the object. When this look-up time is shorter than the copying time, references are an improvement.
Generally, ints and addresses have the same byte length in low-level implementations. So the time of copying an int as a return value for a function is equivalent to the time of copying an address. But in the case where an int is returned, no look up is performed, therefore performance is increased.
The main difference between returning a value and returning a const reference is that you then can const_cast that reference and alter the value.
It's an example of bad design and an attempt to create a smart design where easy and concise design would be more than enough. Instead of just returning a value the author makes readers of code think what intention he might have had.
There is not much benefit. I have seen this in framework or macro generated getters and setters before. The macro code did not distinguish between primitive and non-POD types and just used const type& across the board for setters. I doubt that it is an efficiency issue or a genuine misunderstanding; chances are this is a consistency issue.
I think this type of code is written who have misunderstood the concept of references and use it for everything including primitive data types. I've also seen some code like this and can't see any benefit of doing this.
There is no point and benefit except
void MyClass::setFoo(const int foo)
void MyClass::setFoo(const int& foo)
as then you won't be able to reuse 'foo' variable inside 'setFoo' implementation. And I believe that 'int&' is just because Guy just get used to pass all things by const reference and there is nothing wrong with that.