In lesson 4.13 -- Const, constexpr, and symbolic constants, you learned that fundamental data types (int, double, char, etc…) can be made const via the const keyword, and that all const variables must be initialized at time of creation.
In the case of const fundamental data types, initialization can be done through copy, direct, or uniform initialization:
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const int value1 = 5; // copy initialization const int value2(7); // direct initialization const int value3 { 9 }; // uniform initialization (C++11) |
Const classes
Similarly, instantiated class objects can also be made const by using the const keyword. Initialization is done via class constructors:
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const Date date1; // initialize using default constructor const Date date2(2020, 10, 16); // initialize using parameterized constructor const Date date3 { 2020, 10, 16 }; // initialize using parameterized constructor (C++11) |
Once a const class object has been initialized via constructor, any attempt to modify the member variables of the object is disallowed, as it would violate the const-ness of the object. This includes both changing member variables directly (if they are public), or calling member functions that set the value of member variables. Consider the following class:
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class Something { public: int m_value; Something(): m_value(0) { } void setValue(int value) { m_value = value; } int getValue() { return m_value ; } }; int main() { const Something something; // calls default constructor something.m_value = 5; // compiler error: violates const something.setValue(5); // compiler error: violates const return 0; } |
Both of the above lines involving variable something are illegal because they violate the constness of something by either attempting to change a member variable directly, or by calling a member function that attempts to change a member variable.
Just like with normal variables, you’ll generally want to make your class objects const when you need to ensure they aren’t modified after creation.
Const member functions
Now, consider the following line of code:
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std::cout << something.getValue(); |
Perhaps surprisingly, this will also cause a compile error, even though getValue() doesn’t do anything to change a member variable! It turns out that const class objects can only explicitly call const member functions, and getValue() has not been marked as a const member function.
A const member function is a member function that guarantees it will not modify the object or call any non-const member functions (as they may modify the object).
To make getValue() a const member function, we simply append the const keyword to the function prototype, after the parameter list, but before the function body:
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class Something { public: int m_value; Something(): m_value(0) { } void resetValue() { m_value = 0; } void setValue(int value) { m_value = value; } int getValue() const { return m_value; } // note addition of const keyword after parameter list, but before function body }; |
Now getValue() has been made a const member function, which means we can call it on any const objects.
For member functions defined outside of the class definition, the const keyword must be used on both the function prototype in the class definition and on the function definition:
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class Something { public: int m_value; Something(): m_value(0) { } void resetValue() { m_value = 0; } void setValue(int value) { m_value = value; } int getValue() const; // note addition of const keyword here }; int Something::getValue() const // and here { return m_value; } |
Futhermore, any const member function that attempts to change a member variable or call a non-const member function will cause a compiler error to occur. For example:
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class Something { public: int m_value ; void resetValue() const { m_value = 0; } // compile error, const functions can't change member variables. }; |
In this example, resetValue() has been marked as a const member function, but it attempts to change m_value. This will cause a compiler error.
Note that constructors cannot be marked as const. This is because constructors need to be able to initialize their member variables, and a const constructor would not be able to do so. Consequently, the language disallows const constructors.
Rule: Make any member function that does not modify the state of the class object const, so that it can be called by const objects
Const references
Although instantiating const class objects is one way to create const objects, a more common way is by passing an object to a function by const reference.
In the lesson on passing arguments by reference, we covered the merits of passing class arguments by const reference instead of by value. To recap, passing a class argument by value causes a copy of the class to be made (which is slow) -- most of the time, we don’t need a copy, a reference to the original argument works just fine, and is more performant because it avoids the needless copy. We typically make the reference const in order to ensure the function does not inadvertently change the argument, and to allow the function to work with R-values (e.g. literals), which can be passed as const references, but not non-const references.
Can you figure out what’s wrong with the following code?
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#include <iostream> class Date { private: int m_year; int m_month; int m_day; public: Date(int year, int month, int day) { setDate(year, month, day); } void setDate(int year, int month, int day) { m_year = year; m_month = month; m_day = day; } int getYear() { return m_year; } int getMonth() { return m_month; } int getDay() { return m_day; } }; // note: We're passing date by const reference here to avoid making a copy of date void printDate(const Date &date) { std::cout << date.getYear() << "/" << date.getMonth() << "/" << date.getDay() << '\n'; } int main() { Date date(2016, 10, 16); printDate(date); return 0; } |
The answer is that inside of the printDate function, date is treated as a const object. And with that const date, we’re calling functions getYear(), getMonth(), and getDay(), which are all non-const. Since we can’t call non-const member functions on const objects, this will cause a compile error.
The fix is simple: make getYear(), getMonth(), and getDay() const:
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class Date { private: int m_year; int m_month; int m_day; public: Date(int year, int month, int day) { setDate(year, month, day); } // setDate() cannot be const, modifies member variables void setDate(int year, int month, int day) { m_year = year; m_month = month; m_day = day; } // The following getters can all be made const int getYear() const { return m_year; } int getMonth() const { return m_month; } int getDay() const { return m_day; } }; |
Now in function printDate(), const date will be able to successfully call getYear(), getMonth(), and getDay().
Overloading const and non-const function
Finally, although it is not done very often, it is possible to overload a function in such a way to have a const and non-const version of the same function:
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#include <string> class Something { private: std::string m_value; public: Something(const std::string &value="") { m_value= value; } const std::string& getValue() const { return m_value; } // getValue() for const objects std::string& getValue() { return m_value; } // getValue() for non-const objects }; |
The const version of the function will be called on any const objects, and the non-const version will be called on any non-const objects:
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int main() { Something something; something.getValue() = "Hi"; // calls non-const getValue(); const Something something2; something2.getValue(); // calls const getValue(); return 0; } |
Overloading a function with a const and non-const version is typically done when the return value needs to differ in constness. In the example above, the non-const version of getValue() will only work with non-const objects, but is more flexible in that we can use it to both read and write m_value (which we do by assigning the string “Hi”).
The const version of getValue() will work with either const or non-const objects, but returns a const reference, to ensure we can’t modify the const object’s data.
This works because the const-ness of the function is considered part of the function’s signature, so a const and non-const function which differ only in const-ness are considered distinct.
Summary
Because passing objects by const reference is common, your classes should be const-friendly. That means making any member function that does not modify the state of the class object const!
 8.11 -- Static member variables |
 Index |
 8.9 -- Class code and header files |
QUESTION: I'm making an integer class that uses one byte of storage. It can be signed, so my question is this: how do I make the name of the class indicate that it is signed or unsigned?
You don't include "unsigned" in the name
under const references example, I think you missed the printDate Function.
printDate() is a non-member function, so it's defined outside the class. Am I missing something?
Just a quick sum up question.
int getYear() const { return m_year; } // Is used when we are working with a const class, but returns a normal int variable?
const int getYear() const { return m_year; } // Is used when we are working with a const class and returns a const int variable?
Also, const in the middle of the function allows us to use non-const class, but treats the function as const and does not allow any changes to be made?
The member function `getYear` is `const`. It can be used on `const` objects.
`getYear` is non-const, it can only be used on non-const objects. The returned `int` is `const`. That doesn't make sense, because `getYear` is returning by value. `const` return types should be used for references, pointers, etc.
why is 'getYear' non-const in the second example.
It's not, thanks for pointing out my mistake! I wrote another reply to Ged's comment for clarification.
Sorry for my last reply, I didn't read your code properly.
Both functions are `const`, they can be used on `const` as well as non-`const` objects. Neither is allowed to modify members of `Date`.
The second function returns a `const int`, this has no practical use and can prevent return-value-optimization.
Hi, Alex and Nascardriver! What is 'const' actually in terms of class. We make the class const or we make the object const? const class or const object?
There is no "const class", only an object of the class or the class's members can be `const`.
I understand this example is for demonstration, but (at line 11) isn't returning a reference potentially unsafe? The reference can be used to modify the state of the object perhaps putting it in an inconsistent state. If I really wanted to return a reference, it always be by const reference so that I am confident that all state changes to the object are through the methods I write, correct?
Yes it is potentially unsafe. A better example would be array element access, but then the example would be bigger.
Thank you!
you mean line 12 as at line 11 the function is returning by const reference
Yes!
I just don't Understand the need of the & sign after std::string , because if you ommit it , it is giving the same results. could you explain please ?
It may look like it's doing the same, but it's not. See lesson 6.11, 7.3 and 7.4a.
Thus return is giving back the original value( reference ) instead of making a copy ? Am I Right ?
So it’s faster ?
You're right. It's faster. And since the reference is non-const, the caller can modify `m_value` when they call `getValue`.
Thank you very much really !!!
What is "Hi" being assigned to?
Because `something.getValue()` return a reference to `something.m_value`.
hi, I was trying to write my own matrix class using two dimensional array but getting error
error:
main.cc: In constructor ‘Array::Array(int)’:
main.cc:11:30: error: array size in new-expression must be constant
ptr=new int[m_size][m_size];
^
main.cc:11:30: error: ‘this’ is not a constant expression
Makefile:3: recipe for target 'a.out' failed
make: *** [a.out] Error 1
Allocate the out array first, then allocate the inner arrays.
`ptr` should be `int**`.
Hello,
I have a question: Why can we access to a const member function which the object isn't const, or vice versa ?
A non-const object is allowed to be modified, so it doesn't matter whether or not the function is const.
The other way around doesn't work. If `Something::getValue` is non-const, line 20 will cause an error.
Hi,
In this example, resetValue() has been marked as a const member function, but it attempts to change m_value. This will cause a compiler error.
_____________________________________________________________
m_value(0) in default constructor
m_value = 0 in void resetValue() const
Looks like the value of m_value is still zero in both.
Does it consider as a change to m_value?
Yes, the value is irrelevant.
Hello,
"A const member function is a member function that guarantees it will not modify the object or call any non-const member functions (as they may modify the object)."
Could you please give me an example for "a const member function is a member function that guarantees it will not call any non-const member functions"?
Thanks
In class Something, getValue() is const, but resetValue() is not. Therefore, getValue() could not call resetValue(), as a const function can not call a non-const function.
What will be the working of the class if it has a constant function and instantiated to a static object??
Excuse me but how did we create a constant object using the second way (the constant reference way), if I can still change the object named date with the void function named setDate()?
Thanks in advance
You can't change a const object.
"that is, it’s legal for a const class object to call a constructor that initializes all, some, or none of the member variables" Can you explain this with an example?
I tried to interprete this with the following code but it generated the error message "error: invalid use of ‘myClass::myClass’"
* Line 6, 8, 17: Initialize your variables with uniform initialization.
You cannot call a constructor on an existing object. A constructor can only be called during initialization.
The sentence only says, that @myClass::myClass can initialize @m_value1, but it doesn't have to. If there were more member variables, the constructor could initialize all, some, or none of those variables.
Quick question:
Why is this allowed:
but this not:
Error C2440 'return': cannot convert from 'const int' to 'int &'
The return value must be a const int& to work. Why? In the first (working) example the return value is also not a const int.
Version 1 creates a copy of @number. The caller can modify that number without affecting the @number object in the class.
Version 2 returns a reference to @number, giving the caller direct access to a local variable. Since that reference isn't const, the caller could modify the value, but the object @getNumber was called on is const, so it's illegal to modify the object.
Hi Alex,
From what I've learned about the hidden "*this" pointer. The compiler does this conversion:
Now I think making a member function const does something to "*this". Possibly this conversion?
Is this correct?
In the final section on "Overloading const and non-const function" we also have a const return type of std::string&. Meaning at the end of the day, the compiler compiles this:
I found this part of the chapter a bit confusing as the const return type had nothing to do with the overloading of const and non-const function itself. Doing this for a function that returns by value
makes it easier to understand the overloading concept at first.
> Is this correct?
Correct
Hi Alex,
Can you please mention the following?:
1) When would we generally want to declare a function as "const". I mean example, the lesson covers why we would need to.
2) Example of when we would want to declare a const class object, besides passing a const reference to a function.
I am having trouble thinking of examples for this in real life. All the other chapter subsections have these examples/tips which makes it easier to understand and remember when to do what.
I think this Subsection would need an update on that. The "What, why, how?" is covered but the "when" may need an update.
Love the lessons, thank you for your hardwork!
1) Const class objects can only explicitly call const member functions, therefore you should mark
2) I added the following line, "Just like with normal variables, you'll generally want to make your class objects const when you need to ensure they aren't modified after creation."
Examples are fairly easy to imagine -- basically, any class object that might be used as a constant value would benefit from being made const. e.g:
"The const version of getValue() will work with either const or non-const objects",
suppose both getvalue(const and non const) not changing member vars , and are little different ,one print "const" other "non const" ,
which will get called from non const object.
thanks.
Hi hrmn!
Non-const objects prefer the non-const functions.
Output
non-const
const
const
it clear things,
thanks that was helpful.
rule: Make any member function that does not modify the state of the class object const
should be
Rule: Make any member function that does not modify the state of the class object const
//difference : rule->Rule
Section: Const references
As noted previously, "performant" is not a real word. Also, saying that something performs better doesn't say in what way it performs better. Are we executing faster?
Are we saving memory? Are we making the code easier to understand?
I understand the usage of two const, as you replied to someone "The leftmost const is part of the function's return type, indicating that a const reference is returned. The rightmost const indicates that the member function itself is const. This means it won't change the state of the object and can be called by const objects of the class."
However, why will the compiler issue an error when I remove the first const? Const object could still call this function because a const is added between the function name and the function body even after i remove the first const.
Hi Aron!
Removing the first const causes an error, because it will allow direct access to @m_value.
But const functions aren't allowed to modify members so you're getting an error.
I have similar doubt(as of jules) but with const part of getvalue function i.e.
now when you call its function in main i.e
it will return const string refrence but you are not taking that refrence anywhere in any const variable isn't it wrong. I think it should be done like this
It's perfectly okay to ignore a return value. You are not obligated to print it or assign it to anything.
Is the leftmost const a must? or is the rightmost const enough to overload the function?
The leftmost const is part of the return type, and return types aren't considered when determining whether a function is unique or not.
Wouldn't this be a good place to cover the advantages of immutability, and how immutability heavily improves ones design, especially in regard to concurrency (even though you haven't covered that yet)? And also cover the tradeoffs with immutability, for instance some people might be coding for embedded devices, where the cost of extra instances etc. might not be acceptable
Yes, it might be worth a mention. I'll add it to my to-do list. Thanks for the suggestion!
Hello,
I don't understand how this:
can be called with:
With the assignation of a value?
Thanks.
getValue() returns a reference to m_value, so the return value can be treated as if it were m_value itself. So something.getValue() = "Hi" is essentially the same as something.m_value = "Hi".
Is this a flaw in C++ then? Because m_value is a private element of the class, doesn't this mean that the reference has different qualities and essentially breaks the properties of the class? Or are references supposed to work this way, forgive me, this is my second run through these tutorials and I've only made it to chapter 9 on the first pass.
This is intentional. You can limit access to the returned data using the const keyword, this will prevent the caller from making any changed to the data.
Yup, making a member private only means the member can't be accessed _directly_ from outside the class. However, it can still be accessed indirectly (through a pointer or reference) if you expose public methods which return those.
You can also limit exposure by returning members of fundamental types by value instead of reference.
"You can also limit exposure by returning members of fundamental types by value instead of reference."
Sounds to me I can't return a private member variable by value but it is ok that I can return a private member variable by reference?
In my example program, it is understood that main() is not a member of Something; therefore, it does not have access to m_value private member variable. Then, why getValue() can return something.m_value to main() without any problem?
`main` doesn't have access to `something.m_value`, `Something::getValue` returns a copy of `m_value`. `main` can't modify it and it doesn't notice if `something.m_value` gets modified.
Hi Alex,
It is mentioned there we can not create const constructor and const object can only call const function so when we create a object like ex: const Example objEx; how this is able to initialize via non const constructor?
Const variables still need to be initialized:
Const classes are no different -- when you create a const object, the constructor has to be non-const so you can initialize the class to whatever value is appropriate.
When a class object is created, the constructor is called implicitly (by the code that creates the object) -- it's smart enough to know it should always invoke a non-const constructor.
hi Alex,
there is no return in getValue() function but member variable "m_something" has value "hi",
how can "m_something" has value ?, is this default assignment to the first member variable ?,
-sorry bad english-
If your compiler doesn't prevent this (which it really should), the C++ standard has this to say: "Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function."
Hi, could you tell me why we need to use const twice in the const function?
I conjecture that the second const (right before the function body) is what distinguishes the const function from the non-const function during method overloading but I do not understand the use of the first const (right before the function type).
Thanks.
The leftmost const is part of the function's return type, indicating that a const reference is returned. The rightmost const indicates that the member function itself is const. This means it won't change the state of the object, and can be called by const objects of the class.
Thanks, got it.