I have a class which I am using to try and help understand the concept of const. For example.
Class class1:
{
protected:
double * info;
public:
void set(const int i, const int j, const double val) const
{
info[i + j] = val;
}
}
I have read multiple articles that say you write const after a class function if that function isn't going to change any member variables. With this class however it works is it correct or incorrect?
Also my understanding of const variables is that you make them const if they are never going to change, but is there any benefit to this?
Thank you, apologies if this has been answered elsewhere i just can't seem to get my head around it.
Just to add some color to already provided answers. Withing const qualified functions, all mebers effectively becoming const. For example, your info is becoming const and behaves as if it was defined like this:
double* const info;
Note, that this IS different from
const double* info;
First makes info itself unmodifiable (you can't say info = &d, but you can say *info = 42. On the other hand, the later means info is modifieable, but not where it points to. You can write info = &d, but not *info = 42.
Obviously, there is a way to have both qualities, by saying const double* const info - that would prevent modifications of both pointer and the value. However, this is not what const-qualification does for function. Is it intuitive? I dare say, it is not. However, this is what it is, so we have to live with that.
Obviously, one could argue that const-qualified functions could make all member pointers double-const, but what about pointer-like objects (say, unique_ptr)? There is no way compiler would be able to make those double const, so it is better to be consistant.
This function doesn't change any member variables. The only member variable is info, which it doesn't change. It does change the thing info points to, but that's not a member variable. So it's legal as far as the compiler's concerned.
However, note that it's likely to confuse other programmers (including yourself in six months) - generally, const functions shouldn't modify the object's state.
It is sort of correct, a const member function doesn't alter internal state. set obviously doesn't alter the pointer info, but does alter what it points to. Now your have to decide (the compiler can't) if this is const behavior or not. In most cases it wouldn't be const behavior.
Related
today I have learned about the mutable keyword in C++ and would like to use it in my code.
I have a class with many const methods and one of them should be able to modify some of the object's variables (conserving the logical state of the object). However I don't want to let all the const methods to modify the variable, only the selected one. Is there any way of doing that? Maybe with const_cast?
(The code I am talking about is an implementation of the Union-Find structure. The Find operation does not change the logical state of the structure (it only searches for a root of a tree), but changes the physical state by doing so-called path compression)
Thanks!
EDIT: I have added an excerpt from the code I am referring to:
class UnionFind {
public:
void Union(int a, int b) {...}
int Find(int x) const {
// logically, this method is const
while(x != parents[x]) {
// path compression
// the next three lines modify parents and sizes,
// but the logical state of the object is not changed
sizes[parents[x]] -= sizes[x];
sizes[parents[parents[x]]] += sizes[x];
parents[x] = parents[parents[x]];
x = parents[x];
}
return x;
}
int someOtherMethodThatAccessesParents() const {
// this method does access parents, but read only.
// I would prefer if parents behaved like if it was
// not 'mutable' inside this method
...
}
private:
// these have to be mutable if I want the Find method
// to be marked const (as it should be)
// but making them mutable then does not enforce
// the physical non-mutability in other const methods :(
mutable std::vector<int> parents;
mutable std::vector<int> sizes;
};
On first glance this can't be achieved unless you use a nasty const_cast. But don't do that since the behaviour on attempting to modify a variable following a const_cast that was originally declared as const is undefined.
However, it might be feasible to achieve what you want using friendship since that can be controlled on a function by function basis whereas mutability, as you correctly point out, cannot be.
Put the variable you want to modify in a base class and mark it private. Perhaps provide a "getter" function to that member. That function would be const and would probably return a const reference to the member. Then make your function a friend of that base class. That function will be able to change the value of that private member.
If you can afford to use mutable, that is the right way to do it.
Still, it's possible to do what you are asking for. Normally this is done via the “fake this” idiom:
MyClass *mutableThis = const_cast<MyClass*>(this);
Then access your field normally through the new pointer. This is also the way to do it if you have to support some old compiler with no mutable support.
Note however that this is generally a dangerous practice, as it can easily lead you into the dreaded realm of undefined behavior. If the original object was actually declared const (as opposed to just being accessed via a const pointer/reference), you're asking for trouble.
In short: use mutable when you can, use fake this when you can't, but only when you know what you're doing.
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;
}
As far as I know , making constant functions in a class is useful for read/write compiler optimizations.
A constant function within a class means that the class members will remain constant during the execution of the function.
However, you can bypass this by const casting the implicit parameter (ofc this is a very bad practice).
My questions is as follows :
What pitfalls can the following code cause (especially in terms of performance unrelated to thread synchronization) ?
int myClass::getSomething() const
{
myClass* writableThis = const_cast<myClass*>(this);
writableThis->m_nMemberInt++;
...
return m_nSomeOtherUnchangedMember;
}
Another related question :
Is the behavior compiler/platform/os specific ?
I would also very much appreciate if someone could explain the magic under the hood when such a code is compiled/executed (I'm speculating that the CPU is making out-of-order optimizations based on the fact that the function is const , and not respecting this during actual execution should have some side effects).
EDIT :
Thank you for clarifying this for me. After further research all the received answers are correct but I can accept only one :).
Regarding the const qualifier being used solely for syntax corectness , I believe this answer is both right and wrong, the correct way to state this (imho) would be that it is used mostly for syntax corectness (in a very limited number of scenarios it can produce different / better code ). References : SO Related question , related article
The const_cast<T>(this) trick is potentially unsafe, because the user of your member function may run into undefined behavior without doing anything wrong on their side.
The problem is that casting away const-ness is allowed only when you start with a non-const object. If your object is constant, the function that casts away its const-ness and uses the resultant pointer to change object's state triggers undefined behavior:
struct Test {
int n;
Test() : n(0) {}
void potentiallyUndefinedBehavior() const {
Test *wrong = const_cast<Test*>(this);
wrong->n++;
}
};
int main() {
Test t1;
// This call is OK, because t1 is non-const
t1.potentiallyUndefinedBehavior();
const Test t2;
// This triggers undefined behavior, because t2 is const
t2.potentiallyUndefinedBehavior();
return 0;
}
The trick with const_cast<T>(this) has been invented for caching values inside member functions with const qualifier. However, it is no longer useful, because C++ added a special keyword for this sort of things: by marking a member mutable you make that member writable inside const-qualified methods:
struct Test {
mutable int n;
Test() : n(0) {}
void wellDefinedBehavior() const {
n++;
}
};
Now the const member function will not trigger undefined behavior regardless of the context.
The CPU doesn't know anything about const, which is a C++ keyword. By the time the compiler has transformed the C++ code to assembly, there's not much left of that.
Of course, there's a real possibility that the generated code is entirely different because of the const keyword. For instance, the const version of some operator[] may return a T object by value whereas the non-const version must return a T&. A CPU doesn't even know what function it's in, or even assume the existence of functions.
My answer is to use the storage class mutable for any thing which need to be modified in const methods.
It's built into the language, so there are several benefits. It's a tighter control for how const methods modify data members. Other developers will know these data members will change in const methods. If there are any compiler optimizations, the compiler will know to do the right thing.
class myClass {
private:
int m_nSomeOtherUnchangedMember;
mutable int m_nMemberInt;
…
public:
int getSomething() const;
…
};
int myClass::getSomething() const
{
m_nMemberInt++;
…
return m_nSomeOtherUnchangedMember;
}
As far as I know , making constant functions in a class is useful for read/write compiler optimizations.
No. We use const methods to enforce semantic guarantees, not to allow optimizations (with the possible exception of avoiding copies).
What pitfalls can the following code cause
Firstly, it can break program semantics.
For example, std::map nodes store std::pair<const Key, T>, because the Key shouldn't mutate after it has been inserted. If the key changes value, the map sorting invariant is incorrect, and subsequent find/insert/rebalance operations will misbehave.
If you call a const-qualified method on this const key, and that method changes the Key in a way that affects how it compares, then you've cunningly broken the map.
Secondly, it can kill your program. If you have a const object overlaid on a genuinely read-only address range, or you have a statically-initialized const object in the read-only initialized data segment, then writing to it will cause some kind of protection error
As other stated the const-correctness was designed as a help for the programmers and not an help for the optimizer. You should remember 4 things:
1. const references and const methods are not faster
2. const references and const methods are not faster
3. const references and const methods are not faster
4. const references and const methods are not faster
More specifically the optimizer simply completely ignores const-ness of references or of methods because const doesn't really mean in that context what you are thinking.
A const reference to an object doesn't mean that for example during the execution of a method the object will remain constant. Consider for example:
struct MyObject {
int x;
void foo() const {
printf("%i\n", x);
char *p = new char[10];
printf("%i\n", x);
delete[] p;
}
};
the compiler cannot assume that x member didn't mutate between the two calls to printf. The reason is that std::operator new global allocator could have been overloaded and the code could have regular non-const pointer to the instance. Therefore it's perfectly legal for the global allocator to change x during the execution of foo. The compiler cannot know this is not going to happen (the global allocator could be overloaded in another
compilation unit).
Calling any unknown code (i.e. basically any non-inlined function) can mutate any part of an object, being in a const method or not. The const method simply means that you cannot use this to mutate the object, not that the object is constant.
If const correctness is really an help for the programmers is another question on which I personally have a quite heretic point of view, but that's another story...
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Why isn't the const qualifier working on pointer members on const objects?
Consider the following class that has a pointer member int *a. The const method constMod is allowed by the compiler even though it modifies the pointer data. Why doesn't the compiler make the pointer data const in the context of the const method? If a was just an int we wouldn't be allowed to modify it in a const method.
class ConstTest
{
public:
ConstTest(int *p): a(p) {}
void constMod() const {
++(*a);
}
int *a;
};
I'm using g++ on linux.
Inside constMod(), the declaration of a is treated as:
int *const a;
which means the pointer has a constant value, not what it points to. It sounds like you are expecting it to be treated as:
const int *a;
which is different.
The pointer itself is not modified, only the data pointed. Even if it sounds strange from a human point of view, from a compiler point of view, none of the member of the class are altered.
It's just an ownership issue... there's no way the compiler can know if the pointed-to object is logically part of the object or not, so it's left for the programmer to police such issues. const members are allowed to perform operations with side effects, as long as they don't modify their own apparent value. It's not really any different from letting them call say std::cout::operator<<() or some other non-const function....
What is const in the above case is the pointer itself. You are not allowed to do ++a;. However, it doesn't prevent you from modifying the data being pointed to.
I believe because here you are trying to change the value where datamember is pointing to.
If you try to modify the data member it would be error.
Meaning, you'll get error if you make a point to something else instead of changing the value
const member functions will not allow you to modify it's members.
In your example, you have a pointer that points an int.
And in the const method, you are modifying the value that it is pointed to but not the pointer itself.
Try giving, ++a, which will actually modify the pointer value and will not be allowed in your const method.
Consider
const T immutable_object0;
const T immutable_object1;
const T* mutable_view = ...a condition... ? &immutable_object0 : &immutable_object1;
// then later you reseat
mutable_view = ...another condition... ? &immutable_object0 : &immutable_object1;
or
const int immutable_data[120]
const int* mutable_view = immutable_data;
for(const int* end = immutable_data + 120; mutable_view != end; ++mutable_view) {
// can't modify *mutable_view
}
It's because pointers don't always have ownership. In some circumstances pointers are views to an object (first example) or are iterators into a raw array (second example). For those cases it doesn't make sense to restrict the operations available on the pointers just because the data pointed to is immutable (or seen as immutable).
The first example is a bit contrived but a valid version of it is when you use a pointer member to implement object association and you don't want the hassle of a reference member. Sometimes the pointed-to types happen to be const.
I know there are few question about const correctness where it is stated that the declaration of a function and its definition do not need to agree for value parameters. This is because the constness of a value parameter only matters inside the function. This is fine:
// header
int func(int i);
// cpp
int func(const int i) {
return i;
}
Is doing this really a best practice? Because I've never seen anyone do it. I've seen this quotation (not sure of the source) in other places this has been discussed:
"In fact, to the compiler, the function signature is the same whether you include this const in front of a value parameter or not."
"Avoid const pass-by-value parameters in function declarations. Still make the parameter const in the same function's definition if it won't be modified."
The second paragraph says to not put the const in the declaration. I assume this is because the constness of a value parameter is meaningless as part of a interface definition. It is an implementation detail.
Based on this recommendation, is it also recommended for the pointer values of pointer parameters? (It is meaningless on a reference parameter since you can't reassign a reference.)
// header
int func1(int* i);
int func2(int* i);
// cpp
int func1(int* i) {
int x = 0;
*i = 3; // compiles without error
i = &x; // compiles without error
return *i;
}
int func2(int* const i) {
int x = 0;
*i = 3; // compiles without error
i = &x; // compile error
return *i;
}
Summary: Making value parameters is useful to catch some logic errors. Is it a best practice? Do you go to the extreme of leaving the const out of the header file? Is it just as useful to const pointer values? Why or why not?
Some references:
C++ const keyword - use liberally?
Use of 'const' for function parameters
An example of when const value parameters are useful:
bool are_ints_equal(const int i, const int j) {
if (i = j) { // without the consts this would compile without error
return true;
} else {
return false;
}
// return i = j; // I know it can be shortened
}
I've read many times that making value parameters in a function const is a bad thing to do because it's unnecessary.
However, I find it occasionally helpful to me as a check that my implementation doesn't do something I don't intend (as in the example at the end of your question).
So, while it may not add value to the caller, it does sometimes add a small bit of value to me as an implementer, and it doesn't take anything away from the caller. So I see no harm using it.
For example, I may be implementing a C function that takes a couple pointers to a buffer - a pointer to the start, and a pointer to the end. I'm going to put data in the buffer, but want to ensure that I don't overrun the end. So inside the function there's code that will increment a pointer as I'm adding data to it. Making the pointer to the end of the buffer a const parameter will ensure that I don't code up a bug that accidentally increments the end boundary pointer instead of the pointer I really should be incrementing.
So a fillArray function with a signature like this:
size_t fillArray( data_t* pStart, data_t* const pEnd);
will prevent me from accidentally incrementing pEnd when I really mean to increment pStart. It's not a huge thing, but I'm pretty sure everyone who has programmed for any period of time in C has run across such a bug.
My take on it:
It's not a bad idea, but the issue is minor and your energy might be better spent on other things.
In your question you provided a good example of when it might catch an error, but occasionally you also end up doing something like this:
void foo(const int count /* … */)
{
int temp = count; // can't modify count, so we need a copy of it
++temp;
/* … */
}
The pros and cons are minor either way.
Unfortunately, some compilers (I'm looking at you, Sun CC!) incorrectly differentiate between arguments declared const and ones not declared so, and you can get errors about undefined functions.
I think this is dependent on your personal style.
It doesn't add or subtract to what clients can pass to your function. In essence it's like a compile-time assertion. If it helps you to know that value won't change, go ahead and do it, but I don't see a big reason for others to do it.
One reason I might not do it is that the const-ness of the value parameter is an implementation detail that your clients don't need to know about. If you later (purposely) change your function so that it actually does change that value, you will need to change the signature of your function, which will force your clients to re-compile.
This is similar to why some people recommend having no public virtual methods (the functions virtual-ness is an implementation detail that should be hidden from clients), but I'm not in that particular camp.
If there is const keyword present; it means value of 'i' (which is const type) can not be modified.
If value of 'i' is changed inside foo function compiler will throw error: "
Can not modify const object
But changing '*i' (i.e. *i = 3;)means you are not changing value of 'i' but value of address pointed by 'i'
Actually,the const function is appropriate for large objects that should not be altered by function.
I like const correctness for situations like this:
void foo(const Bar &b) //I know b cannot be changed
{
//do something with b
}
This lets me use b without fear of modifying it, but I don't have to pay the cost of a copy constructor.