I am starting to come to grips with const in terms of a reference parameter. The way I see it is that a constant reference parameter basically sets the parameter in question to the original memory space that calls the function into question. And since it is const, the value in itself cannot change.
I have found a solution with regards to a code that delivers matrix multiplication A=BC:
vector<vector<double> > mult(const vector<vector<double> >& B, const vector<vector<double> >& C)
{ ...;
return A;
}
int main()
{
vector<vector<double> > B, C;
cout << mult(B,C) << endl;
return 0;
}
I agree with the structure of the code but I am confused about the neccessity of "const" and "&". Surely the code would be exactly the same if I excluded both from the above? For "&" one could perhaps that we use less memory space by not creating an extra space for the parameters of "mult". But the use of const seems unnecessary to me.
The '&' prevents the copy constructor from being called, i.e., prevents a duplicate copy being made. It is more efficient this way because you avoid the constructor on the invocation and the destructor on the exit.
The 'const' keyword communicates to the caller that the object to which the reference refers will not be changed in the function. It also allows the function to be called with constant vectors as input. In other words, if B and C are constant, you couldn't call mult() without the const keyword in the signature.
It's been a while in C++ for me, but I think that's the gist. I'm certainly open to corrections on my answer.
There are only a few times when a const reference is, strictly-speaking, necessary. The most common is when you need to pass a const object by reference. The type system will prevent this unless the function promises not to modify the object. It can also make a difference when a function is overloaded to do something different when the object is const, and you specifically want the const version. (The latter is probably bad design!)
It would alternatively be possible to remove the const qualifier from the function argument, and to give any overloaded functions different names. In fact, references in C++ are syntactic sugar for C-style pointers, and it would be possible to replace void foo (T& x) with void foo(T* x) and every occurrence of x inside foo with (*x). Adding const T& or T* const simply means that the program will not be able to modify the object through that reference or pointer.
C had no const keyword until 1989, and you could do all the same things without it, but it’s present in order to help developers avoid bugs related to modifying the wrong variable.
It is not really necessary. As long as you pass in non-const parameters to your function, the program will not behave differently.
I can state a few examples in your case:
If one of the parameters you have to pass is const, it will not work.
Furthermore, you won't be able to do something like mult({{1, 2}, {3, 4}}, b); because that temporary object can only implicitly convert into a const reference.
If you put the definition and declaration in separate translation units (i.e. .cpp files) then the compiler might miss some optimization potential, because it wouldn't be able to assume that mult() doesn't modify its parameters.
Another argument is simply that const shows your intents more clearly.
See a few more reasons on isocpp.
The reference & prevents an unnecessary copy. Your parameter type is a std::vector, which means that copying will involve memory allocations, and for performance reasons you do not want that.
On a side note, if your code is meant to manipulate matrices, then a std::vector of std::vector is very inappropriate for performance reasons, as it makes it extremely cache inefficient and causes unnecessary dynamic allocations. You would rather use a 1D std::array and wrap it to handle 2D indices nicely. std::array has sizes known as compile time, which means that every function you pass a specific std::array to knows its size on compile-time which is good for performance, especially as std::array makes it possible to avoid dynamic allocation.
int sum(const int a,int b)
{
b=10;
// a=5; error
return (a+b);
}
in above example a is const and not b.
So a as read-only variable. If you try to change the value of a you get error. That means you have to use value of a which pass when function is call.
I have the following function declaration:
void fn(int);
This function has a single integral type parameter. Now, I could call this function by passing a non const or a const integral object. In either case, this function is going to copy the object to its local int parameter. Therefore, any modifications to this parameter is going to be local to the function and is not going to affect the actual arguments of the caller in any way. Now my question is under which scenario will I declare this single int parameter to be of const type? I don't see a need to declare this function as follows.
void fn(const int);
This is because the arguments are going to be anyway passed by value and the function can in no way modify the arguments in either case. I understand that by declaring a parameter constant the function cannot modify it inside its body. However, there is no downside here even if the function modifies since the parameter is local to the function.
You're right that to the caller there is no difference -- it only matters inside the function. I prefer to add the const whenever I can. As I'm writing the function, I'm thinking "I want this parameter but have no intention of modifying it (even just locally)" and the compiler will keep me honest.
Also, in some cases the compiler may be able to do more optimizations if it knows the variable is const (as in loop bounds).
Just because it's allowed, doesn't mean there's a point to it.
I wouldn't be surprised if this could cause overloading to behave slightly differently, but I think you're basically right - there's no good outside-the-function reason to do it.
One possible problem is confusing readers who might think you intended a reference instead, but forgot the "&".
Sometimes templates are written in a general way, and end up doing things like that. With function template parameter deduction, though, that const tends to get thrown away.
this is effectively a const argument. It is a pointer, but it's immutable in the sense of your example. (Of course, this is a keyword, not a variable, and unary & doesn't work with it.) If you want an argument to behave like that, declare it const.
From the C++ Spec: http://www.kuzbass.ru:8086/docs/isocpp/over.html
Parameter declarations that differ
only in the presence or absence of
const and/or volatile are equivalent.
That is, the const and volatile
type-specifiers for each parameter
type are ignored when determining
which function is being declared,
defined, or called.
Example:
typedef const int cInt;
int f (int);
int f (const int); // redeclaration of f(int)
int f (int) { ... } // definition of f(int)
int f (cInt) { ... } // error: redefinition of f(int)
A favorite interview question in C++ interview is what is the difference between passing by values, passing by pointer and passing by reference.
In this case we are passing by value and that too int and hence it will not matter. But I am not sure about the user created classes. In that case when the compiler see the object is passed by const value it may decide to pass it by the const reference. Nowdays compilers are intelligent and I don't see why they can not do it.
How far do you go with const? Do you just make functions const when necessary or do you go the whole hog and use it everywhere? For example, imagine a simple mutator that takes a single boolean parameter:
void SetValue(const bool b) { my_val_ = b; }
Is that const actually useful? Personally I opt to use it extensively, including parameters, but in this case I wonder if it's worthwhile?
I was also surprised to learn that you can omit const from parameters in a function declaration but can include it in the function definition, e.g.:
.h file
void func(int n, long l);
.cpp file
void func(const int n, const long l)
Is there a reason for this? It seems a little unusual to me.
const is pointless when the argument is passed by value since you will
not be modifying the caller's object.
Wrong.
It's about self-documenting your code and your assumptions.
If your code has many people working on it and your functions are non-trivial then you should mark const any and everything that you can. When writing industrial-strength code, you should always assume that your coworkers are psychopaths trying to get you any way they can (especially since it's often yourself in the future).
Besides, as somebody mentioned earlier, it might help the compiler optimize things a bit (though it's a long shot).
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. For copied objects it doesn't really matter, although it can be safer as it signals intent within the function. It's really a judgement call. I do tend to use const_iterator though when looping on something and I don't intend on modifying it, so I guess to each his own, as long as const correctness for reference types is rigorously maintained.
Sometimes (too often!) I have to untangle someone else's C++ code. And we all know that someone else's C++ code is a complete mess almost by definition :) So the first thing I do to decipher local data flow is put const in every variable definition until compiler starts barking. This means const-qualifying value arguments as well, because they are just fancy local variables initialized by caller.
Ah, I wish variables were const by default and mutable was required for non-const variables :)
Extra Superfluous const are bad from an API stand-point:
Putting extra superfluous const's in your code for intrinsic type parameters passed by value clutters your API while making no meaningful promise to the caller or API user (it only hampers the implementation).
Too many 'const' in an API when not needed is like "crying wolf", eventually people will start ignoring 'const' because it's all over the place and means nothing most of the time.
The "reductio ad absurdum" argument to extra consts in API are good for these first two points would be is if more const parameters are good, then every argument that can have a const on it, SHOULD have a const on it. In fact, if it were truly that good, you'd want const to be the default for parameters and have a keyword like "mutable" only when you want to change the parameter.
So lets try putting in const whereever we can:
void mungerum(char * buffer, const char * mask, int count);
void mungerum(char * const buffer, const char * const mask, const int count);
Consider the line of code above. Not only is the declaration more cluttered and longer and harder to read but three of the four 'const' keywords can be safely ignored by the API user. However, the extra use of 'const' has made the second line potentially DANGEROUS!
Why?
A quick misread of the first parameter char * const buffer might make you think that it will not modify the memory in data buffer that is passed in -- however, this is not true! Superfluous 'const' can lead to dangerous and incorrect assumptions about your API when scanned or misread quickly.
Superfluous const are bad from a Code Implementation stand-point as well:
#if FLEXIBLE_IMPLEMENTATION
#define SUPERFLUOUS_CONST
#else
#define SUPERFLUOUS_CONST const
#endif
void bytecopy(char * SUPERFLUOUS_CONST dest,
const char *source, SUPERFLUOUS_CONST int count);
If FLEXIBLE_IMPLEMENTATION is not true, then the API is “promising” not to implement the function the first way below.
void bytecopy(char * SUPERFLUOUS_CONST dest,
const char *source, SUPERFLUOUS_CONST int count)
{
// Will break if !FLEXIBLE_IMPLEMENTATION
while(count--)
{
*dest++=*source++;
}
}
void bytecopy(char * SUPERFLUOUS_CONST dest,
const char *source, SUPERFLUOUS_CONST int count)
{
for(int i=0;i<count;i++)
{
dest[i]=source[i];
}
}
That’s a very silly promise to make. Why should you make a promise that gives no benefit at all to your caller and only limits your implementation?
Both of these are perfectly valid implementations of the same function though so all you’ve done is tied one hand behind your back unnecessarily.
Furthermore, it’s a very shallow promise that is easily (and legally circumvented).
inline void bytecopyWrapped(char * dest,
const char *source, int count)
{
while(count--)
{
*dest++=*source++;
}
}
void bytecopy(char * SUPERFLUOUS_CONST dest,
const char *source,SUPERFLUOUS_CONST int count)
{
bytecopyWrapped(dest, source, count);
}
Look, I implemented it that way anyhow even though I promised not to – just using a wrapper function. It’s like when the bad guy promises not to kill someone in a movie and orders his henchman to kill them instead.
Those superfluous const’s are worth no more than a promise from a movie bad-guy.
But the ability to lie gets even worse:
I have been enlightened that you can mismatch const in header (declaration) and code (definition) by using spurious const. The const-happy advocates claim this is a good thing since it lets you put const only in the definition.
// Example of const only in definition, not declaration
struct foo { void test(int *pi); };
void foo::test(int * const pi) { }
However, the converse is true... you can put a spurious const only in the declaration and ignore it in the definition. This only makes superfluous const in an API more of a terrible thing and a horrible lie - see this example:
struct foo
{
void test(int * const pi);
};
void foo::test(int *pi) // Look, the const in the definition is so superfluous I can ignore it here
{
pi++; // I promised in my definition I wouldn't modify this
}
All the superfluous const actually does is make the implementer's code less readable by forcing him to use another local copy or a wrapper function when he wants to change the variable or pass the variable by non-const reference.
Look at this example. Which is more readable ? Is it obvious that the only reason for the extra variable in the second function is because some API designer threw in a superfluous const ?
struct llist
{
llist * next;
};
void walkllist(llist *plist)
{
llist *pnext;
while(plist)
{
pnext=plist->next;
walk(plist);
plist=pnext; // This line wouldn't compile if plist was const
}
}
void walkllist(llist * SUPERFLUOUS_CONST plist)
{
llist * pnotconst=plist;
llist *pnext;
while(pnotconst)
{
pnext=pnotconst->next;
walk(pnotconst);
pnotconst=pnext;
}
}
Hopefully we've learned something here. Superfluous const is an API-cluttering eyesore, an annoying nag, a shallow and meaningless promise, an unnecessary hindrance, and occasionally leads to very dangerous mistakes.
The following two lines are functionally equivalent:
int foo (int a);
int foo (const int a);
Obviously you won't be able to modify a in the body of foo if it's defined the second way, but there's no difference from the outside.
Where const really comes in handy is with reference or pointer parameters:
int foo (const BigStruct &a);
int foo (const BigStruct *a);
What this says is that foo can take a large parameter, perhaps a data structure that's gigabytes in size, without copying it. Also, it says to the caller, "Foo won't* change the contents of that parameter." Passing a const reference also allows the compiler to make certain performance decisions.
*: Unless it casts away the const-ness, but that's another post.
const should have been the default in C++.
Like this :
int i = 5 ; // i is a constant
var int i = 5 ; // i is a real variable
When I coded C++ for a living I consted everything I possibly could. Using const is a great way to help the compiler help you. For instance, const-ing your method return values can save you from typos such as:
foo() = 42
when you meant:
foo() == 42
If foo() is defined to return a non-const reference:
int& foo() { /* ... */ }
The compiler will happily let you assign a value to the anonymous temporary returned by the function call. Making it const:
const int& foo() { /* ... */ }
Eliminates this possibility.
There is a good discussion on this topic in the old "Guru of the Week" articles on comp.lang.c++.moderated here.
The corresponding GOTW article is available on Herb Sutter's web site here.
1. Best answer based on my assessment:
The answer by #Adisak is the best answer here based on my assessment. Note that this answer is in part the best because it is also the most well-backed-up with real code examples, in addition to using sound and well-thought-out logic.
2. My own words (agreeing with the best answer):
For pass-by-value there is no benefit to adding const. All it does is:
limit the implementer to have to make a copy every time they want to change an input param in the source code (which change would have no side effects anyway since what's passed in is already a copy since it's pass-by-value). And frequently, changing an input param which is passed by value is used to implement the function, so adding const everywhere can hinder this.
and adding const unnecessarily clutters the code with consts everywhere, drawing attention away from the consts that are truly necessary to have safe code.
When dealing with pointers or references, however, const is critically important when needed, and must be used, as it prevents undesired side effects with persistent changes outside the function, and therefore every single pointer or reference must use const when the param is an input only, not an output. Using const only on parameters passed by reference or pointer has the additional benefit of making it really obvious which parameters are pointers or references. It's one more thing to stick out and say "Watch out! Any param with const next to it is a reference or pointer!".
What I've described above has frequently been the consensus achieved in professional software organizations I have worked in, and has been considered best practice. Sometimes even, the rule has been strict: "don't ever use const on parameters which are passed by value, but always use it on parameters passed by reference or pointer if they are inputs only."
3. Google's words (agreeing with me and the best answer):
(From the "Google C++ Style Guide")
For a function parameter passed by value, const has no effect on the caller, thus is not recommended in function declarations. See TotW #109.
Using const on local variables is neither encouraged nor discouraged.
Source: the "Use of const" section of the Google C++ Style Guide: https://google.github.io/styleguide/cppguide.html#Use_of_const. This is actually a really valuable section, so read the whole section.
Note that "TotW #109" stands for "Tip of the Week #109: Meaningful const in Function Declarations", and is also a useful read. It is more informative and less prescriptive on what to do, and based on context came before the Google C++ Style Guide rule on const quoted just above, but as a result of the clarity it provided, the const rule quoted just above was added to the Google C++ Style Guide.
Also note that even though I'm quoting the Google C++ Style Guide here in defense of my position, it does NOT mean I always follow the guide or always recommend following the guide. Some of the things they recommend are just plain weird, such as their kDaysInAWeek-style naming convention for "Constant Names". However, it is still nonetheless useful and relevant to point out when one of the world's most successful and influential technical and software companies uses the same justification as I and others like #Adisak do to back up our viewpoints on this matter.
4. Clang's linter, clang-tidy, has some options for this:
A. It's also worth noting that Clang's linter, clang-tidy, has an option, readability-avoid-const-params-in-decls, described here, to support enforcing in a code base not using const for pass-by-value function parameters:
Checks whether a function declaration has parameters that are top level const.
const values in declarations do not affect the signature of a function, so they should not be put there.
Examples:
void f(const string); // Bad: const is top level.
void f(const string&); // Good: const is not top level.
And here are two more examples I'm adding myself for completeness and clarity:
void f(char * const c_string); // Bad: const is top level. [This makes the _pointer itself_, NOT what it points to, const]
void f(const char * c_string); // Good: const is not top level. [This makes what is being _pointed to_ const]
B. It also has this option: readability-const-return-type - https://clang.llvm.org/extra/clang-tidy/checks/readability-const-return-type.html
5. My pragmatic approach to how I'd word a style guide on the matter:
I'd simply copy and paste this into my style guide:
[COPY/PASTE START]
Always use const on function parameters passed by reference or pointer when their contents (what they point to) are intended NOT to be changed. This way, it becomes obvious when a variable passed by reference or pointer IS expected to be changed, because it will lack const. In this use case const prevents accidental side effects outside the function.
It is not recommended to use const on function parameters passed by value, because const has no effect on the caller: even if the variable is changed in the function there will be no side effects outside the function. See the following resources for additional justification and insight:
"Google C++ Style Guide" "Use of const" section
"Tip of the Week #109: Meaningful const in Function Declarations"
Adisak's Stack Overflow answer on "Use of 'const' for function parameters"
"Never use top-level const [ie: const on parameters passed by value] on function parameters in declarations that are not definitions (and be careful not to copy/paste a meaningless const). It is meaningless and ignored by the compiler, it is visual noise, and it could mislead readers" (https://abseil.io/tips/109, emphasis added).
The only const qualifiers that have an effect on compilation are those placed in the function definition, NOT those in a forward declaration of the function, such as in a function (method) declaration in a header file.
Never use top-level const [ie: const on variables passed by value] on values returned by a function.
Using const on pointers or references returned by a function is up to the implementer, as it is sometimes useful.
TODO: enforce some of the above with the following clang-tidy options:
https://clang.llvm.org/extra/clang-tidy/checks/readability-avoid-const-params-in-decls.html
https://clang.llvm.org/extra/clang-tidy/checks/readability-const-return-type.html
Here are some code examples to demonstrate the const rules described above:
const Parameter Examples:
(some are borrowed from here)
void f(const std::string); // Bad: const is top level.
void f(const std::string&); // Good: const is not top level.
void f(char * const c_string); // Bad: const is top level. [This makes the _pointer itself_, NOT what it points to, const]
void f(const char * c_string); // Good: const is not top level. [This makes what is being _pointed to_ const]
const Return Type Examples:
(some are borrowed from here)
// BAD--do not do this:
const int foo();
const Clazz foo();
Clazz *const foo();
// OK--up to the implementer:
const int* foo();
const int& foo();
const Clazz* foo();
[COPY/PASTE END]
Keywords: use of const in function parameters; coding standards; C and C++ coding standards; coding guidelines; best practices; code standards; const return values
I use const on function parameters that are references (or pointers) which are only [in] data and will not be modified by the function. Meaning, when the purpose of using a reference is to avoid copying data and not to allow changing the passed parameter.
Putting const on the boolean b parameter in your example only puts a constraint on the implementation and doesn't contribute for the class's interface (although not changing parameters is usually advised).
The function signature for
void foo(int a);
and
void foo(const int a);
is the same, which explains your .c and .h
Asaf
I say const your value parameters.
Consider this buggy function:
bool isZero(int number)
{
if (number = 0) // whoops, should be number == 0
return true;
else
return false;
}
If the number parameter was const, the compiler would stop and warn us of the bug.
If you use the ->* or .* operators, it's a must.
It prevents you from writing something like
void foo(Bar *p) { if (++p->*member > 0) { ... } }
which I almost did right now, and which probably doesn't do what you intend.
What I intended to say was
void foo(Bar *p) { if (++(p->*member) > 0) { ... } }
and if I had put a const in between Bar * and p, the compiler would have told me that.
Ah, a tough one. On one side, a declaration is a contract and it really does not make sense to pass a const argument by value. On the other hand, if you look at the function implementation, you give the compiler more chances to optimize if you declare an argument constant.
const is pointless when the argument is passed by value since you will not be modifying the caller's object.
const should be preferred when passing by reference, unless the purpose of the function is to modify the passed value.
Finally, a function which does not modify current object (this) can, and probably should be declared const. An example is below:
int SomeClass::GetValue() const {return m_internalValue;}
This is a promise to not modify the object to which this call is applied. In other words, you can call:
const SomeClass* pSomeClass;
pSomeClass->GetValue();
If the function was not const, this would result in a compiler warning.
Marking value parameters 'const' is definitely a subjective thing.
However I actually prefer to mark value parameters const, just like in your example.
void func(const int n, const long l) { /* ... */ }
The value to me is in clearly indicating that the function parameter values are never changed by the function. They will have the same value at the beginning as at the end. For me, it is part of keeping to a very functional programming sort of style.
For a short function, it's arguably a waste of time/space to have the 'const' there, since it's usually pretty obvious that the arguments aren't modified by the function.
However for a larger function, its a form of implementation documentation, and it is enforced by the compiler.
I can be sure if I make some computation with 'n' and 'l', I can refactor/move that computation without fear of getting a different result because I missed a place where one or both is changed.
Since it is an implementation detail, you don't need to declare the value parameters const in the header, just like you don't need to declare the function parameters with the same names as the implementation uses.
I tend to use const wherever possible. (Or other appropriate keyword for the target language.) I do this purely because it allows the compiler to make extra optimizations that it would not be able to make otherwise. Since I have no idea what these optimizations may be, I always do it, even where it seems silly.
For all I know, the compiler might very well see a const value parameter, and say, "Hey, this function isn't modifying it anyway, so I can pass by reference and save some clock cycles." I don't think it ever would do such a thing, since it changes the function signature, but it makes the point. Maybe it does some different stack manipulation or something... The point is, I don't know, but I do know trying to be smarter than the compiler only leads to me being shamed.
C++ has some extra baggage, with the idea of const-correctness, so it becomes even more important.
May be this wont be a valid argument. but if we increment the value of a const variable inside a function compiler will give us an error:
"error: increment of read-only parameter". so that means we can use const key word as a way to prevent accidentally modifying our variables inside functions(which we are not supposed to/read-only). so if we accidentally did it at the compile time compiler will let us know that. this is specially important if you are not the only one who is working on this project.
In the case you mention, it doesn't affect callers of your API, which is why it's not commonly done (and isn't necessary in the header). It only affects the implementation of your function.
It's not particularly a bad thing to do, but the benefits aren't that great given that it doesn't affect your API, and it adds typing, so it's not usually done.
I do not use const for value-passed parametere. The caller does not care whether you modify the parameter or not, it's an implementation detail.
What is really important is to mark methods as const if they do not modify their instance. Do this as you go, because otherwise you might end up with either lots of const_cast<> or you might find that marking a method const requires changing a lot of code because it calls other methods which should have been marked const.
I also tend to mark local vars const if I do not need to modify them. I believe it makes the code easier to understand by making it easier to identify the "moving parts".
On compiler optimizations: http://www.gotw.ca/gotw/081.htm
I use const were I can. Const for parameters means that they should not change their value. This is especially valuable when passing by reference. const for function declares that the function should not change the classes members.
To summarize:
"Normally const pass-by-value is unuseful and misleading at best." From GOTW006
But you can add them in the .cpp as you would do with variables.
Note that the standard library doesn't use const. E.g. std::vector::at(size_type pos). What's good enough for the standard library is good for me.
If the parameter is passed by value (and is not a reference), usually there is not much difference whether the parameter is declared as const or not (unless it contains a reference member -- not a problem for built-in types). If the parameter is a reference or pointer, it is usually better to protect the referenced/pointed-to memory, not the pointer itself (I think you cannot make the reference itself const, not that it matters much as you cannot change the referee).
It seems a good idea to protect everything you can as const. You can omit it without fear of making a mistake if the parameters are just PODs (including built-in types) and there is no chance of them changing further along the road (e.g. in your example the bool parameter).
I didn't know about the .h/.cpp file declaration difference, but it does make some sense. At the machine code level, nothing is "const", so if you declare a function (in the .h) as non-const, the code is the same as if you declare it as const (optimizations aside). However, it helps you to enlist the compiler that you will not change the value of the variable inside the implementation of the function (.ccp). It might come handy in the case when you're inheriting from an interface that allows change, but you don't need to change to parameter to achieve the required functionality.
I wouldn't put const on parameters like that - everyone already knows that a boolean (as opposed to a boolean&) is constant, so adding it in will make people think "wait, what?" or even that you're passing the parameter by reference.
the thing to remember with const is that it is much easier to make things const from the start, than it is to try and put them in later.
Use const when you want something to be unchanged - its an added hint that describes what your function does and what to expect. I've seen many an C API that could do with some of them, especially ones that accept c-strings!
I'd be more inclined to omit the const keyword in the cpp file than the header, but as I tend to cut+paste them, they'd be kept in both places. I have no idea why the compiler allows that, I guess its a compiler thing. Best practice is definitely to put your const keyword in both files.
As parameters are being passed by value,it doesnt make any difference if you specify const or not from the calling function's perspective.It basically does not make any sense to declare pass by value parameters as const.
All the consts in your examples have no purpose. C++ is pass-by-value by default, so the function gets copies of those ints and booleans. Even if the function does modify them, the caller's copy is not affected.
So I'd avoid extra consts because
They're redudant
They clutter up
the text
They prevent me from
changing the passed in value in
cases where it might be useful or efficient.
There's really no reason to make a value-parameter "const" as the function can only modify a copy of the variable anyway.
The reason to use "const" is if you're passing something bigger (e.g. a struct with lots of members) by reference, in which case it ensures that the function can't modify it; or rather, the compiler will complain if you try to modify it in the conventional way. It prevents it from being accidentally modified.
Const parameter is useful only when the parameter is passed by reference i.e., either reference or pointer. When compiler sees a const parameter, it make sure that the variable used in the parameter is not modified within the body of the function. Why would anyone want to make a by-value parameter as constant? :-)
I know the question is "a bit" outdated but as I came accross it somebody else may also do so in future... ...still I doubt the poor fellow will list down here to read my comment :)
It seems to me that we are still too confined to C-style way of thinking. In the OOP paradigma we play around with objects, not types. Const object may be conceptually different from a non-const object, specifically in the sense of logical-const (in contrast to bitwise-const). Thus even if const correctness of function params is (perhaps) an over-carefulness in case of PODs it is not so in case of objects. If a function works with a const object it should say so. Consider the following code snippet
#include <iostream>
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
class SharedBuffer {
private:
int fakeData;
int const & Get_(int i) const
{
std::cout << "Accessing buffer element" << std::endl;
return fakeData;
}
public:
int & operator[](int i)
{
Unique();
return const_cast<int &>(Get_(i));
}
int const & operator[](int i) const
{
return Get_(i);
}
void Unique()
{
std::cout << "Making buffer unique (expensive operation)" << std::endl;
}
};
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
void NonConstF(SharedBuffer x)
{
x[0] = 1;
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
void ConstF(const SharedBuffer x)
{
int q = x[0];
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
int main()
{
SharedBuffer x;
NonConstF(x);
std::cout << std::endl;
ConstF(x);
return 0;
}
ps.: you may argue that (const) reference would be more appropriate here and gives you the same behaviour. Well, right. Just giving a different picture from what I could see elsewhere...