Common Misunderstandings with GNU C++#
C++ is a complex language and an evolving one, and its standard definition (the ISO C++ standard) was only recently completed. As a result, your C++ compiler may occasionally surprise you, even when its behavior is correct. This section discusses some areas that frequently give rise to questions of this sort.
Declare and Define Static Members#
When a class has static data members, it is not enough to declare the static member; you must also define it. For example:
class Foo
{
...
void method();
static int bar;
};
This declaration only establishes that the class Foo
has an
int
named Foo::bar
, and a member function named
Foo::method
. But you still need to define both
method
and bar
elsewhere. According to the ISO
standard, you must supply an initializer in one (and only one) source
file, such as:
int Foo::bar = 0;
Other C++ compilers may not correctly implement the standard behavior. As a result, when you switch to g++ from one of these compilers, you may discover that a program that appeared to work correctly in fact does not conform to the standard: g++ reports as undefined symbols any static data members that lack definitions.
Name Lookup, Templates, and Accessing Members of Base Classes#
The C++ standard prescribes that all names that are not dependent on template parameters are bound to their present definitions when parsing a template function or class.The C++ standard just uses the term ‘dependent’ for names that depend on the type or value of template parameters. This shorter term will also be used in the rest of this section.
Only names that are dependent are looked up at the point of instantiation. For example, consider
void foo(double);
struct A {
template <typename T>
void f () {
foo (1); // 1
int i = N; // 2
T t;
t.bar(); // 3
foo (t); // 4
}
static const int N;
};
Here, the names foo
and N
appear in a context that does
not depend on the type of T
. The compiler will thus require that
they are defined in the context of use in the template, not only before
the point of instantiation, and will here use ::foo(double)
and
A::N
, respectively. In particular, it will convert the integer
value to a double
when passing it to ::foo(double)
.
Conversely, bar
and the call to foo
in the fourth marked
line are used in contexts that do depend on the type of T
, so
they are only looked up at the point of instantiation, and you can
provide declarations for them after declaring the template, but before
instantiating it. In particular, if you instantiate A::f<int>
,
the last line will call an overloaded ::foo(int)
if one was
provided, even if after the declaration of struct A
.
This distinction between lookup of dependent and non-dependent names is called two-stage (or dependent) name lookup. G++ implements it since version 3.4.
Two-stage name lookup sometimes leads to situations with behavior different from non-template codes. The most common is probably this:
template <typename T> struct Base {
int i;
};
template <typename T> struct Derived : public Base<T> {
int get_i() { return i; }
};
In get_i()
, i
is not used in a dependent context, so the
compiler will look for a name declared at the enclosing namespace scope
(which is the global scope here). It will not look into the base class,
since that is dependent and you may declare specializations of
Base
even after declaring Derived
, so the compiler cannot
really know what i
would refer to. If there is no global
variable i
, then you will get an error message.
In order to make it clear that you want the member of the base class,
you need to defer lookup until instantiation time, at which the base
class is known. For this, you need to access i
in a dependent
context, by either using this->i
(remember that this
is of
type Derived<T>*
, so is obviously dependent), or using
Base<T>::i
. Alternatively, Base<T>::i
might be brought
into scope by a using
-declaration.
Another, similar example involves calling member functions of a base class:
template <typename T> struct Base {
int f();
};
template <typename T> struct Derived : Base<T> {
int g() { return f(); };
};
Again, the call to f()
is not dependent on template arguments
(there are no arguments that depend on the type T
, and it is also
not otherwise specified that the call should be in a dependent context).
Thus a global declaration of such a function must be available, since
the one in the base class is not visible until instantiation time. The
compiler will consequently produce the following error message:
x.cc: In member function `int Derived<T>::g()':
x.cc:6: error: there are no arguments to `f' that depend on a template
parameter, so a declaration of `f' must be available
x.cc:6: error: (if you use `-fpermissive', G++ will accept your code, but
allowing the use of an undeclared name is deprecated)
To make the code valid either use this->f()
, or
Base<T>::f()
. Using the -fpermissive
flag will also let
the compiler accept the code, by marking all function calls for which no
declaration is visible at the time of definition of the template for
later lookup at instantiation time, as if it were a dependent call.
We do not recommend using -fpermissive
to work around invalid
code, and it will also only catch cases where functions in base classes
are called, not where variables in base classes are used (as in the
example above).
Note that some compilers (including G++ versions prior to 3.4) get these examples wrong and accept above code without an error. Those compilers do not implement two-stage name lookup correctly.
Temporaries May Vanish Before You Expect#
It is dangerous to use pointers or references to portions of a
temporary object. The compiler may very well delete the object before
you expect it to, leaving a pointer to garbage. The most common place
where this problem crops up is in classes like string classes,
especially ones that define a conversion function to type char *
or const char *
—which is one reason why the standard
string
class requires you to call the c_str
member
function. However, any class that returns a pointer to some internal
structure is potentially subject to this problem.
For example, a program may use a function strfunc
that returns
string
objects, and another function charfunc
that
operates on pointers to char
:
string strfunc ();
void charfunc (const char *);
void
f ()
{
const char *p = strfunc().c_str();
...
charfunc (p);
...
charfunc (p);
}
In this situation, it may seem reasonable to save a pointer to the C
string returned by the c_str
member function and use that rather
than call c_str
repeatedly. However, the temporary string
created by the call to strfunc
is destroyed after p
is
initialized, at which point p
is left pointing to freed memory.
Code like this may run successfully under some other compilers, particularly obsolete cfront-based compilers that delete temporaries along with normal local variables. However, the GNU C++ behavior is standard-conforming, so if your program depends on late destruction of temporaries it is not portable.
The safe way to write such code is to give the temporary a name, which forces it to remain until the end of the scope of the name. For example:
const string& tmp = strfunc ();
charfunc (tmp.c_str ());
Implicit Copy-Assignment for Virtual Bases#
When a base class is virtual, only one subobject of the base class belongs to each full object. Also, the constructors and destructors are invoked only once, and called from the most-derived class. However, such objects behave unspecified when being assigned. For example:
struct Base{
char *name;
Base(const char *n) : name(strdup(n)){}
Base& operator= (const Base& other){
free (name);
name = strdup (other.name);
return *this;
}
};
struct A:virtual Base{
int val;
A():Base("A"){}
};
struct B:virtual Base{
int bval;
B():Base("B"){}
};
struct Derived:public A, public B{
Derived():Base("Derived"){}
};
void func(Derived &d1, Derived &d2)
{
d1 = d2;
}
The C++ standard specifies that Base::Base
is only called once
when constructing or copy-constructing a Derived object. It is
unspecified whether Base::operator=
is called more than once when
the implicit copy-assignment for Derived objects is invoked (as it is
inside func
in the example).
G++ implements the ‘intuitive’ algorithm for copy-assignment: assign all
direct bases, then assign all members. In that algorithm, the virtual
base subobject can be encountered more than once. In the example, copying
proceeds in the following order: name
(via strdup
),
val
, name
again, and bval
.
If application code relies on copy-assignment, a user-defined copy-assignment operator removes any uncertainties. With such an operator, the application can define whether and how the virtual base subobject is assigned.