FAQs in section [35]:
[35.1] What's the idea behind templates?
A template is a cookie-cutter that specifies how to cut cookies that all look
pretty much the same (although the cookies can be made of various kinds of
dough, they'll all have the same basic shape). In the same way, a class
template is a cookie cutter for a description of how to build a family of
classes that all look basically the same, and a function template describes
how to build a family of similar looking functions.
Class templates are often used to build type safe containers (although this
only scratches the surface for how they can be used).
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[35.2] What's the syntax / semantics for a "class template"?
Consider a container class Array that acts like an array of integers:
// This would go into a header file such as "Array.h"
class Array {
public:
Array(int len=10) : len_(len), data_(new int[len]) { }
~Array() { delete[] data_; }
int len() const { return len_; }
const int& operator[](int i) const { return data_[check(i)]; } ← subscript operators often come in pairs
int& operator[](int i) { return data_[check(i)]; } ← subscript operators often come in pairs
Array(const Array&);
Array& operator= (const Array&);
private:
int len_;
int* data_;
int check(int i) const
{ if (i < 0 || i >= len_) throw BoundsViol("Array", i, len_);
return i; }
};
Repeating the above over and over for Array of float, of char, of
std::string, of Array-of-std::string, etc, will become tedious.
// This would go into a header file such as "Array.h"
template<typename T>
class Array {
public:
Array(int len=10) : len_(len), data_(new T[len]) { }
~Array() { delete[] data_; }
int len() const { return len_; }
const T& operator[](int i) const { return data_[check(i)]; }
T& operator[](int i) { return data_[check(i)]; }
Array(const Array<T>&);
Array<T>& operator= (const Array<T>&);
private:
int len_;
T* data_;
int check(int i) const
{ if (i < 0 || i >= len_) throw BoundsViol("Array", i, len_);
return i; }
};
Unlike template functions, template classes
(instantiations of class templates) need to be explicit about the parameters
over which they are instantiating:
int main()
{
Array<int> ai;
Array<float> af;
Array<char*> ac;
Array<std::string> as;
Array< Array<int> > aai;
...
}
Note the space between the two >'s in the last example. Without this
space, the compiler would see a >> (right-shift) token instead of two
>'s.
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[35.3] What's the syntax / semantics for a "function template"?
Consider this function that swaps its two integer arguments:
void swap(int& x, int& y)
{
int tmp = x;
x = y;
y = tmp;
}
If we also had to swap floats, longs, Strings, Sets, and FileSystems, we'd get
pretty tired of coding lines that look almost identical except for the type.
Mindless repetition is an ideal job for a computer, hence a function
template:
template<typename T>
void swap(T& x, T& y)
{
T tmp = x;
x = y;
y = tmp;
}
Every time we used swap() with a given pair of types, the compiler will go to
the above definition and will create yet another "template function" as an
instantiation of the above. E.g.,
int main()
{
int i,j; /*...*/ swap(i,j); // Instantiates a swap for int
float a,b; /*...*/ swap(a,b); // Instantiates a swap for float
char c,d; /*...*/ swap(c,d); // Instantiates a swap for char
std::string s,t; /*...*/ swap(s,t); // Instantiates a swap for std::string
...
}
Note: A "template function" is the instantiation of a "function template".
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[35.4] How do I explicitly select which version of a function template should get called?
When you call a function template, the compiler tries to deduce the
template type. Most of the time it can do that successfully, but every once
in a while you may want to help the compiler deduce the right type — either
because it cannot deduce the type at all, or perhaps because it would deduce
the wrong type.
For example, you might be calling a function template that doesn't have any
parameters of its template argument types, or you might want to force the
compiler to do certain promotions on the arguments before selecting the
correct function template. In these cases you'll need to explicitly tell the
compiler which instantiation of the function template should be called.
Here is a sample function template where the template parameter T does
not appear in the function's parameter list. In this case the compiler
cannot deduce the template parameter types when the function is
called.
template<typename T>
void f()
{
...
}
To call this function with T being an int or a std::string, you
could say:
#include <string>
void sample()
{
f<int>(); // type T will be int in this call
f<std::string>(); // type T will be std::string in this call
}
Here is another function whose template parameters appear in the function's
list of formal parameters (that is, the compiler can deduce the
template type from the actual arguments):
template<typename T>
void g(T x)
{
...
}
Now if you want to force the actual arguments to be promoted before the
compiler deduces the template type, you can use the above technique. E.g., if
you simply called g(42) you would get g<int>(42), but if you
wanted to pass 42 to g<long>(), you could say this:
g<long>(42). (Of course you could also promote the parameter
explicitly, such as either g(long(42)) or even g(42L), but
that ruins the example.)
Similarly if you said g("xyz") you'd end up calling
g<char*>(char*), but if you wanted to call the std::string
version of g<>() you could say g<std::string>("xyz"). (Again
you could also promote the argument, such as g(std::string("xyz")),
but that's another story.)
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[35.5] What is a "parameterized type"?
Another way to say, "class templates."
A parameterized type is a type that is parameterized over another type or some
value. List<int> is a type (List) parameterized over another type (int).
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[35.6] What is "genericity"?
Yet another way to say, "class templates."
Not to be confused with "generality" (which just means avoiding solutions which
are overly specific), "genericity" means class templates.
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[35.7] Why can't I separate the definition of my templates class from it's declaration and put it inside a .cpp file?
If all you want to know is how to fix this situation, read
the next FAQ. But in order to
understand why things are the way they are, first accept these facts:
- A template is not a class or a function. A template is a
"pattern" that the compiler uses to generate a family of
classes or functions.
- In order for the compiler to generate the code, it must see
both the template definition (not just declaration) and the specific
types/whatever used to "fill in" the template. For example, if you're
trying to use a Foo<int>, the compiler must see both the Foo
template and the fact that you're trying to make a specific
Foo<int>.
- Your compiler probably doesn't remember the details of one
.cpp file while it is compiling another .cpp file. It
could, but most do not and if you are reading this FAQ, it
almost definitely does not. BTW this is called the "separate
compilation model."
Now based on those facts, here's an example that shows why things are the way
they are. Suppose you have a template Foo defined like this:
template<typename T>
class Foo {
public:
Foo();
void someMethod(T x);
private:
T x;
};
Along with similar definitions for the member functions:
template<typename T>
Foo<T>::Foo()
{
...
}
template<typename T>
void Foo<T>::someMethod(T x)
{
...
}
Now suppose you have some code in file Bar.cpp that uses
Foo<int>:
// Bar.cpp
void blah_blah_blah()
{
...
Foo<int> f;
f.someMethod(5);
...
}
Clearly somebody somewhere is going to have to use the "pattern" for
the constructor definition and for the someMethod() definition and
instantiate those when T is actually int. But if you had put the
definition of the constructor and someMethod() into file
Foo.cpp, the compiler would see the template code when it
compiled Foo.cpp and it would see Foo<int> when it
compiled Bar.cpp, but there would never be a time when it saw
both the template code and Foo<int>. So by rule #2 above, it
could never generate the code for Foo<int>::someMethod().
A note to the experts: I have obviously made several simplifications
above. This was intentional so please don't complain too loudly. If you know
the difference between a .cpp file and a compilation unit, the
difference between a class template and a template class, and the fact that
templates really aren't just glorified macros, then don't complain: this
particular question/answer wasn't aimed at you to begin with. I simplified
things so newbies would "get it," even if doing so offends some experts.
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[35.8] How can I avoid linker errors with my template functions?
Tell your C++ compiler which instantiations to make while it is compiling your
template function's .cpp file.
As an example, consider the header file foo.h which contains the
following template function declaration:
// file "foo.h"
template<typename T>
extern void foo();
Now suppose file foo.cpp actually defines that template function:
// file "foo.cpp"
#include <iostream>
#include "foo.h"
template<typename T>
void foo()
{
std::cout << "Here I am!\n";
}
Suppose file main.cpp uses this template function by calling
foo<int>():
// file "main.cpp"
#include "foo.h"
int main()
{
foo<int>();
...
}
If you compile and (try to) link these two .cpp files, most compilers will
generate linker errors. There are three solutions for this. The first
solution is to physically move the definition of the template function into
the .h file, even if it is not an inline function. This solution may
(or may not!) cause significant code bloat, meaning your executable size may
increase dramatically (or, if your compiler is smart enough, may not; try it
and see).
The other solution is to leave the definition of the template function in the
.cpp file and simply add the line template void foo<int>(); to that
file:
// file "foo.cpp"
#include <iostream>
#include "foo.h"
template<typename T> void foo()
{
std::cout << "Here I am!\n";
}
template void foo<int>();
If you can't modify foo.cpp, simply create a new .cpp file such as
foo-impl.cpp as follows:
// file "foo-impl.cpp"
#include "foo.cpp"
template void foo<int>();
Notice that foo-impl.cpp #includes a .cpp file, not a .h file.
If that's confusing, click your heels twice, think of Kansas, and repeat after
me, "I will do it anyway even though it's confusing." You can trust me on
this one. But if you don't trust me or are simply curious,
the rationale is given earlier.
If you are using Comeau
C++, you probably want to check out the export keyword.
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[35.9] How can I avoid linker errors with my template classes?
Tell your C++ compiler which instantiations to make while it is compiling your
template class's .cpp file.
(If you've already read the previous FAQ, this answer is completely symmetric
with that one, so you can probably skip this answer.)
As an example, consider the header file Foo.h which contains the
following template class. Note that method Foo<T>::f() is inline and
methods Foo<T>::g() and Foo<T>::h() are not.
// file "Foo.h"
template<typename T>
class Foo {
public:
void f();
void g();
void h();
};
template<typename T>
inline
void Foo<T>::f()
{
...
}
Now suppose file Foo.cpp actually defines the non-inline
methods Foo<T>::g() and Foo<T>::h():
// file "Foo.cpp"
#include <iostream>
#include "Foo.h"
template<typename T>
void Foo<T>::g()
{
std::cout << "Foo<T>::g()\n";
}
template<typename T>
void Foo<T>::h()
{
std::cout << "Foo<T>::h()\n";
}
Suppose file main.cpp uses this template class by creating a
Foo<int> and calling its methods:
// file "main.cpp"
#include "Foo.h"
int main()
{
Foo<int> x;
x.f();
x.g();
x.h();
...
}
If you compile and (try to) link these two .cpp files, most compilers will
generate linker errors. There are three solutions for this. The first
solution is to physically move the definition of the template functions into
the .h file, even if they are not inline functions. This solution may
(or may not!) cause significant code bloat, meaning your executable size may
increase dramatically (or, if your compiler is smart enough, may not; try it
and see).
The other solution is to leave the definition of the template function in the
.cpp file and simply add the line template class Foo<int>; to that file:
// file "Foo.cpp"
#include <iostream>
#include "Foo.h"
...definition of Foo<T>::f() is unchanged -- see above...
...definition of Foo<T>::g() is unchanged -- see above...
template class Foo<int>;
If you can't modify Foo.cpp, simply create a new .cpp file such as
Foo-impl.cpp as follows:
// file "Foo-impl.cpp"
#include "Foo.cpp"
template class Foo<int>;
Notice that Foo-impl.cpp #includes a .cpp file, not a .h file.
If that's confusing, click your heels twice, think of Kansas, and repeat after
me, "I will do it anyway even though it's confusing." You can trust me on
this one. But if you don't trust me or are simply curious,
the rationale is given earlier.
If you are using Comeau
C++, you probably want to check out the export keyword.
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[35.10] Why do I get linker errors when I use template friends?
Ah, the intricacies of template friends. Here's an example of what people
often want to do:
#include <iostream>
template<typename T>
class Foo {
public:
Foo(const T& value = T());
friend Foo<T> operator+ (const Foo<T>& lhs, const Foo<T>& rhs);
friend std::ostream& operator<< (std::ostream& o, const Foo<T>& x);
private:
T value_;
};
Naturally the template will need to actually be used somewhere:
int main()
{
Foo<int> lhs(1);
Foo<int> rhs(2);
Foo<int> result = lhs + rhs;
std::cout << result;
...
}
And of course the various member and friend functions will need to be defined
somewhere:
template<typename T>
Foo<T>::Foo(const T& value = T())
: value_(value)
{ }
template<typename T>
Foo<T> operator+ (const Foo<T>& lhs, const Foo<T>& rhs)
{ return Foo<T>(lhs.value_ + rhs.value_); }
template<typename T>
std::ostream& operator<< (std::ostream& o, const Foo<T>& x)
{ return o << x.value_; }
The snag happens when the compiler sees the friend lines way up in the
class definition proper. At that moment it does not yet know the
friend functions are themselves templates; it assumes they are
non-templates like this:
Foo<int> operator+ (const Foo<int>& lhs, const Foo<int>& rhs)
{ ... }
std::ostream& operator<< (std::ostream& o, const Foo<int>& x)
{ ... }
When you call the operator+ or operator<< functions, this
assumption causes the compiler to generate a call to the non-template
functions, but the linker will give you an "undefined external" error because
you never actually defined those non-template functions.
The solution is to convince the compiler while it is examining the class
body proper that the operator+ and operator<< functions
are themselves templates. There are several ways to do this; one simple
approach is pre-declare each template friend function above the
definition of template class Foo:
template<typename T> class Foo; // pre-declare the template class itself
template<typename T> Foo<T> operator+ (const Foo<T>& lhs, const Foo<T>& rhs);
template<typename T> std::ostream& operator<< (std::ostream& o, const Foo<T>& x);
Also you add <> in the friend lines, as shown:
#include <iostream>
template<typename T>
class Foo {
public:
Foo(const T& value = T());
friend Foo<T> operator+ <> (const Foo<T>& lhs, const Foo<T>& rhs);
friend std::ostream& operator<< <> (std::ostream& o, const Foo<T>& x);
private:
T value_;
};
After the compiler sees that magic stuff, it will be better informed about the
friend functions. In particular, it will realize that the
friend lines are referring to functions that are themselves templates.
That eliminates the confusion.
Another approach is to define the friend function within the
class body at the same moment you declare it to be a friend. For
example:
#include <iostream>
template<typename T>
class Foo {
public:
Foo(const T& value = T());
friend Foo<T> operator+ (const Foo<T>& lhs, const Foo<T>& rhs)
{
...
}
friend std::ostream& operator<< (std::ostream& o, const Foo<T>& x)
{
...
}
private:
T value_;
};
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[35.11] How can any human hope to understand these overly verbose template-based error messages?
Here's a free tool that
transforms error messages
into something more understandable. At the time of this writing, it
works with the following compilers: Comeau C++, Intel C++, CodeWarrior C++,
gcc, Borland C++, Microsoft Visual C++, and EDG C++.
Here's an example showing some unfiltered gcc error messages:
rtmap.cpp: In function `int main()':
rtmap.cpp:19: invalid conversion from `int' to `
std::_Rb_tree_node<std::pair<const int, double> >*'
rtmap.cpp:19: initializing argument 1 of `std::_Rb_tree_iterator<_Val, _Ref,
_Ptr>::_Rb_tree_iterator(std::_Rb_tree_node<_Val>*) [with _Val =
std::pair<const int, double>, _Ref = std::pair<const int, double>&, _Ptr =
std::pair<const int, double>*]'
rtmap.cpp:20: invalid conversion from `int' to `
std::_Rb_tree_node<std::pair<const int, double> >*'
rtmap.cpp:20: initializing argument 1 of `std::_Rb_tree_iterator<_Val, _Ref,
_Ptr>::_Rb_tree_iterator(std::_Rb_tree_node<_Val>*) [with _Val =
std::pair<const int, double>, _Ref = std::pair<const int, double>&, _Ptr =
std::pair<const int, double>*]'
E:/GCC3/include/c++/3.2/bits/stl_tree.h: In member function `void
std::_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::insert_unique(_II,
_II) [with _InputIterator = int, _Key = int, _Val = std::pair<const int,
double>, _KeyOfValue = std::_Select1st<std::pair<const int, double> >,
_Compare = std::less<int>, _Alloc = std::allocator<std::pair<const int,
double> >]':
E:/GCC3/include/c++/3.2/bits/stl_map.h:272: instantiated from `void std::map<_
Key, _Tp, _Compare, _Alloc>::insert(_InputIterator, _InputIterator) [with _Input
Iterator = int, _Key = int, _Tp = double, _Compare = std::less<int>, _Alloc = st
d::allocator<std::pair<const int, double> >]'
rtmap.cpp:21: instantiated from here
E:/GCC3/include/c++/3.2/bits/stl_tree.h:1161: invalid type argument of `unary *
'
Here's what the filtered error messages look like (note: you can configure the
tool so it shows more information; this output was generated with settings to
strip things down to a minimum):
rtmap.cpp: In function `int main()':
rtmap.cpp:19: invalid conversion from `int' to `iter'
rtmap.cpp:19: initializing argument 1 of `iter(iter)'
rtmap.cpp:20: invalid conversion from `int' to `iter'
rtmap.cpp:20: initializing argument 1 of `iter(iter)'
stl_tree.h: In member function `void map<int,double>::insert_unique(_II, _II)':
[STL Decryptor: Suppressed 1 more STL standard header message]
rtmap.cpp:21: instantiated from here
stl_tree.h:1161: invalid type argument of `unary *'
Here is the source code to generate the above example:
#include <map>
#include <algorithm>
#include <cmath>
const int values[] = { 1,2,3,4,5 };
const int NVALS = sizeof values / sizeof (int);
int main()
{
using namespace std;
typedef map<int, double> valmap;
valmap m;
for (int i = 0; i < NVALS; i++)
m.insert(make_pair(values[i], pow(values[i], .5)));
valmap::iterator it = 100; // error
valmap::iterator it2(100); // error
m.insert(1,2); // error
return 0;
}
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[35.12] Why am I getting errors when my template-derived-class accesses something it inherited from its template-base-class?
Perhaps surprisingly, the following code is not valid C++, even though some compilers accept it:
template<typename T>
class B {
public:
void f() { }
};
template<typename T>
class D : public B<T> {
public:
void g()
{
f(); ← compiler might give an error here
}
};
This might hurt your head; better if you sit down.
Within D<T>::g(), the name f does not depend on template
parameter T, so f is known as a nondependent name.
B<T>, on the other hand, is dependent on template parameter T
so B<T> is called a dependent name.
Here's the rule: the compiler does not look in dependent base classes (like
B<T>) when looking up nondependent names (like f).
This doesn't mean that inheritance doesn't work. Class D<int> is
still derived from class B<int>, the compiler still lets you
implicitly do the is-a conversions (e.g., D<int>* to B<int>*),
dynamic binding still works when virtual functions are invoked, etc. But
there is an issue about how names are looked up.
Workarounds:
- Change the call from f() to this->f(). Since this
is always implicitly dependent in a template, this->f is dependent and
the lookup is therefore deferred until the template is actually instantiated,
at which point all base classes are considered.
- Insert using B<T>::f; just prior to calling f().
- Change the call from f() to B<T>::f(). Note however that
this might not give you what you want if f() is virtual, since it
inhibits the virtual dispatch mechanism.
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[35.13] Can the previous problem hurt me silently? Is it possible that the compiler will silently generate the wrong code? Updated!
[Recently fixed a typo thanks to Assaf Lavie (in 5/05). Click here to go to the next FAQ in the "chain" of recent changes.]
Yes.
Since the non-dependent name f is not looked up in the dependent
base-class B<T>, the compiler will search the enclosing scope (such as
the enclosing namespace) for the name f. This can cause it to
silently(!) do the wrong thing.
For example:
void f() { } // a global ("namespace scope") function
template<typename T>
class B {
public:
void f() { }
};
template<typename T>
class D : public B<T> {
public:
void g()
{
f();
}
};
Here the call within D<T>::g() will silently(!) call ::f()
rather than B<T>::f().
You have been warned.
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Revised Jun 1, 2005