Item 21: Use const
whenever possible.
The wonderful thing about const
is that it allows you to specify a certain semantic constraint a particular object should not be modified and compilers will enforce that constraint. It allows you to communicate to both compilers and other programmers that a value should remain invariant. Whenever that is true, you should be sure to say so explicitly, because that way you enlist your compilers' aid in making sure the constraint isn't
The const
keyword is remarkably versatile. Outside of classes, you can use it for global or namespace constants (see Items 1 and 47) and for static objects (local to a file or a block). Inside classes, you can use it for both static and nonstatic data members (see also Item 12).
For pointers, you can specify whether the pointer itself is const
, the data it points to is const
, both, or
char *p = "Hello"; // non-const pointer, // non-const data5
const char *p = "Hello"; // non-const pointer, // const data
char * const p = "Hello"; // const pointer, // non-const data
const char * const p = "Hello"; // const pointer, // const data
This syntax isn't quite as capricious as it looks. Basically, you mentally draw a vertical line through the asterisk of a pointer declaration, and if the word const
appears to the left of the line, what's pointed to is constant; if the word const
appears to the right of the line, the pointer itself is constant; if const
appears on both sides of the line, both are
When what's pointed to is constant, some programmers list const
before the type name. Others list it after the type name but before the asterisk. As a result, the following functions take the same parameter
class Widget { ... };
void f1(const Widget *pw); // f1 takes a pointer to a // constant Widget object
void f2(Widget const *pw); // so does f2
Because both forms exist in real code, you should accustom yourself to both of
Some of the most powerful uses of const
stem from its application to function declarations. Within a function declaration, const
can refer to the function's return value, to individual parameters, and, for member functions, to the function as a
Having a function return a constant value often makes it possible to reduce the incidence of client errors without giving up safety or efficiency. In fact, as Item 29 demonstrates, using const
with a return value can make it possible to improve the safety and efficiency of a function that would otherwise be
For example, consider the declaration of the operator*
function for rational numbers that is introduced in Item 19:
const Rational operator*(const Rational& lhs, const Rational& rhs);
Many programmers squint when they first see this. Why should the result of operator*
be a const
object? Because if it weren't, clients would be able to commit atrocities like
Rational a, b, c;
...
(a * b) = c; // assign to the product // of a*b!
I don't know why any programmer would want to make an assignment to the product of two numbers, but I do know this: it would be flat-out illegal if a
, b
, and c
were of a built-in type. One of the hallmarks of good user-defined types is that they avoid gratuitous behavioral incompatibilities with the built-ins, and allowing assignments to the product of two numbers seems pretty gratuitous to me. Declaring operator*
's return value const
prevents it, and that's why It's The Right Thing To
There's nothing particularly new about const
parameters they act just like local const
objects. (See Item M19, however, for a discussion of how const
parameters can lead to the creation of temporary objects.) Member functions that are const
, however, are a different
The purpose of const
member functions, of course, is to specify which member functions may be invoked on const
objects. Many people overlook the fact that member functions differing only in their constness can be overloaded, however, and this is an important feature of C++. Consider the String
class once
class String { public:
...
// operator[] for non-const objects char& operator[](int position) { return data[position]; }
// operator[] for const objects const char& operator[](int position) const { return data[position]; }
private: char *data; };
String s1 = "Hello"; cout << s1[0]; // calls non-const // String::operator[] const String s2 = "World"; cout << s2[0]; // calls const // String::operator[]
By overloading operator[]
and giving the different versions different return values, you are able to have const
and non-const
String
s handled
String s = "Hello"; // non-const String object
cout << s[0]; // fine reading a // non-const String
s[0] = 'x'; // fine writing a // non-const String
const String cs = "World"; // const String object
cout << cs[0]; // fine reading a // const String
cs[0] = 'x'; // error! writing a // const String
By the way, note that the error here has only to do with the return value of the operator[]
that is called; the calls to operator[]
themselves are all fine. The error arises out of an attempt to make an assignment to a const
char&
, because that's the return value from the const
version of operator[]
.
Also note that the return type of the non-const
operator[]
must be a reference to a char
a char
itself will not do. If operator[]
did return a simple char
, statements like this wouldn't
s[0] = 'x';
That's because it's never legal to modify the return value of a function that returns a built-in type. Even if it were legal, the fact that C++ returns objects by value (see Item 22) would mean that a copy of s.data[0]
would be modified, not s.data[0]
itself, and that's not the behavior you want,
Let's take a brief time-out for philosophy. What exactly does it mean for a member function to be const
? There are two prevailing notions: bitwise constness and conceptual
The bitwise const
camp believes that a member function is const
if and only if it doesn't modify any of the object's data members (excluding those that are static), i.e., if it doesn't modify any of the bits inside the object. The nice thing about bitwise constness is that it's easy to detect violations: compilers just look for assignments to data members. In fact, bitwise constness is C++'s definition of constness, and a const
member function isn't allowed to modify any of the data members of the object on which it is
Unfortunately, many member functions that don't act very const
pass the bitwise test. In particular, a member function that modifies what a pointer points to frequently doesn't act const
. But if only the pointer is in the object, the function is bitwise const
, and compilers won't complain. That can lead to counterintuitive
class String { public: // the constructor makes data point to a copy // of what value points to String(const char *value);
...
operator char *() const { return data;}
private: char *data; };
const String s = "Hello"; // declare constant object
char *nasty = s; // calls op char*() const
*nasty = 'M'; // modifies s.data[0]
cout << s; // writes "Mello"
Surely there is something wrong when you create a constant object with a particular value and you invoke only const
member functions on it, yet you are still able to change its value! (For a more detailed discussion of this example, see Item 29.)
This leads to the notion of conceptual constness. Adherents to this philosophy argue that a const
member function might modify some of the bits in the object on which it's invoked, but only in ways that are undetectable by clients. For example, your String
class might want to cache the length of the object whenever it's requested (see Item M18):
class String { public: // the constructor makes data point to a copy // of what value points to String(const char *value): lengthIsValid(false) { ... }
...
size_t length() const;
private: char *data;
size_t dataLength; // last calculated length // of string
bool lengthIsValid; // whether length is // currently valid };
size_t String::length() const { if (!lengthIsValid) { dataLength = strlen(data); // error! lengthIsValid = true; // error! }
return dataLength; }
This implementation of length
is certainly not bitwise const both dataLength
and lengthIsValid
may be modified yet it seems as though it should be valid for const
String
objects. Compilers, you will find, respectfully disagree; they insist on bitwise constness. What to
The solution is simple: take advantage of the const
-related wiggle room the mutable
. When applied to nonstatic data members, mutable
frees those members from the constraints of bitwise
class String { public:
... // same as above
private: char *data;
mutable size_t dataLength; // these data members are // now mutable; they may be mutable bool lengthIsValid; // modified anywhere, even // inside const member }; // functions
size_t String::length() const { if (!lengthIsValid) { dataLength = strlen(data); // now fine lengthIsValid = true; // also fine }
return dataLength; }
mutable
is a wonderful solution to the bitwise-constness-is-not-quite-what-I-had-in-mind problem, but it was added to C++ relatively late in the standardization process, so your compilers may not support it yet. If that's the case, you must descend into the dark recesses of C++, where life is cheap and constness may be cast
Inside a member function of class C
, the this
pointer behaves as if it had been declared as
C * const this; // for non-const member // functions
const C * const this; // for const member // functions
That being the case, all you have to do to make the problematic version of String::length
(i.e., the one you could fix with mutable
if your compilers supported it) valid for both const
and non-const
objects is to change the type of this
from const
C
*
const
to C
*
const
. You can't do that directly, but you can fake it by initializing a local pointer to point to the same object as this
does. Then you can access the members you want to modify through the local
size_t String::length() const { // make a local version of this that's // not a pointer-to-const String * const localThis = const_cast<String * const>(this);
if (!lengthIsValid) { localThis->dataLength = strlen(data); localThis->lengthIsValid = true; }
return dataLength; }
Pretty this ain't, but sometimes a programmer's just gotta do what a programmer's gotta
Unless, of course, it's not guaranteed to work, and sometimes the old cast-away-constness trick isn't. In particular, if the object this
points to is truly const
, i.e., was declared const
at its point of definition, the results of casting away its constness are undefined. If you want to cast away constness in one of your member functions, you'd best be sure that the object you're doing the casting on wasn't originally defined to be const
.
There is one other time when casting away constness may be both useful and safe. That's when you have a const
object you want to pass to a function taking a non-const
parameter, and you know the parameter won't be modified inside the function. The second condition is important, because it is always safe to cast away the constness of an object that will only be read not written even if that object was originally defined to be const
.
For example, some libraries have been known to incorrectly declare the strlen
function as
size_t strlen(char *s);
Certainly strlen
isn't going to modify what s
points to at least not the strlen
I grew up with. Because of this declaration, however, it would be invalid to call it on pointers of type const
char
*
. To get around the problem, you can safely cast away the constness of such pointers when you pass them to strlen
:
const char *klingonGreeting = "nuqneH"; // "nuqneH" is // "Hello" in // Klingon size_t length = strlen(const_cast<char*>(klingonGreeting));
Don't get cavalier about this, though. It is guaranteed to work only if the function being called, strlen
in this case, doesn't try to modify what its parameter points
"Hello"
is const char []
, a type that's almost always treated as const char*
. We'd therefore expect it to be a violation of const
correctness to initialize a char*
variable with a string literal like "Hello"
. The practice is so common in C, however, that the standard grants a special dispensation for initializations like this. Nevertheless, you should try to avoid them, because they're deprecated.