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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 .. index:: C++ static data, declaring and defining

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;;        // 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 can’t 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: In member function `int Derived<T>::g()': error: there are no arguments to `f' that depend on a template
   parameter, so a declaration of `f' must be available 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 *);

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(char *n) : name(strdup(n)){}
  Base& operator= (const Base& other){
   free (name);
   name = strdup (;

struct A:virtual Base{
  int val;

struct B:virtual Base{
  int bval;

struct Derived:public A, public B{

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: val, name (via strdup), bval, and name again.

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.