Considering the following code:
template<typename T>
struct A
{
void f(){...}
friend T;
};
template<typename T>
struct B
{
void f(T){...}//code depends on T
void g(){...}//code doesn't depends on T
}
As you see, the "code" in the struct A doesn't depend on T.
Will the compiler generate different code in the final binary for every T used to instantiate A?
Same question for B::g() function, will the compiler use the same function for all instances of B<T> when it's possible, for example this is not used in g(), so no dependency on T? Does the standard have any specification for this case?
If you want to be sure what the compiler generates, why not write a non-template struct implementing the code that doesn't depend on T, and then derive your template from the non-template? You get one copy of the non-template code, which each instance of the template inherits.
At least, that's what I've done in the past when I found that template instantiation was making my compiled objects very large.
Related
The access to members of a template base class requires the syntax this->member or the using directive. Does this syntax extends also to base template classes which are not directly inherited?
Consider the following code:
template <bool X>
struct A {
int x;
};
template <bool X>
struct B : public A<X> {
using A<X>::x; // OK even if this is commented out
};
template <bool X>
struct C : public B<X> {
// using B<X>::x; // OK
using A<X>::x; // Why OK?
C() { x = 1; }
};
int main()
{
C<true> a;
return 0;
}
Since the declaration of the template class B contains using A<X>::x, naturally the derived template class C can access to x with a using B<X>::x. Nevertheless, on g++ 8.2.1 and clang++ 6.0.1 the above code compiles fine, where x is accessed in C with a using that picks up x directly from A
I would have expected that C can not access directly to A. Also, commenting out the using A<X>::x in B still makes the code to compile. Even the combination of commenting out using A<X>::x in B and at the same time employ in C using B<X>::x instead of using A<X>::x gives a code that compiles.
Is the code legal?
Addition
To be more clear: the question arises on template classes and it is about the visibility of members inherited by template classes.
By standard public inheritance, the public members of A are accessible to C, so using the syntax this->x in C one does indeed get access to A<X>::x. But what about the using directive? How does the compiler correctly resolve the using A<X>::x if A<X> is not a direct base of C?
You are using A<X> where a base class is expected.
[namespace.udecl]
3 In a using-declaration used as a member-declaration, each
using-declarator's nested-name-specifier shall name a base class of
the class being defined.
Since this appears where a class type is expected, it is known and assumed to be a type. And it is a type that is dependent on the template arguments, so it's not looked up immediately.
[temp.res]
9 When looking for the declaration of a name used in a template
definition, the usual lookup rules ([basic.lookup.unqual],
[basic.lookup.argdep]) are used for non-dependent names. The lookup of
names dependent on the template parameters is postponed until the
actual template argument is known ([temp.dep]).
So it's allowed on account of the compiler not being able to know any better. It will check the using declaration when the class is instantiated. Indeed, one can put any dependent type there:
template<bool> struct D{};
template <bool X>
struct C : public B<X> {
using D<X>::x;
C() { x = 1; }
};
This will not be checked until the value of X is known. Because B<X> can bring with it all sorts of surprises if it's specialized. One could for instance do this:
template<>
struct D<true> { char x; };
template<>
struct B<true> : D<true> {};
Making the above declaration be correct.
Is the code legal?
Yes. This is what public inheritance does.
Is it possible to allow a template class derived from B to access to x only via this->x, using B::x or B::x? ...
You can use private inheritance (i.e. struct B : private A<X>), and arrange access to A<X>::x only through B's public/protected interface.
Also, if you're worried about having hidden members, you should use class instead of struct and specify the desired visibility explicitly.
Regarding the addition, note that:
(1) the compiler knows what object A<X>::x refers to given some instance of A<X> (because A is defined in the global scope, and X is the template parameter of C).
(2) You do indeed have an instance of A<X> - this is a ponter to a derived class (it doesn't matter if A<X> is a direct base class or not).
(3) The object A<X>::x is visible in the current scope (because the inheritances and the object itself are public).
The using statement is merely syntactic sugar. Once all types are resolved, the compiler replaces following use of x with the appropriate memory address in the instance, not unlike writing this->x directly.
Maybe this example could give you some idea as to why should it be legal:
template <bool X>
struct A {
int x;
};
template <bool X>
struct B : public A<X> {
int x;
};
template <bool X>
struct C : public B<X> {
//it won't work without this
using A<X>::x;
//or
//using B<X>::x;
C() { x = 1; }
// or
//C() { this -> template x = 1; }
//C() { this -> x = 1; }
};
In case of choosing C() { this -> template x = 1; } the last inherited x (B::x) would be assigned to 1 not the A::x.
It can simply be tested by:
C<false> a;
std::cout << a.x <<std::endl;
std::cout << a.A::x <<std::endl;
std::cout << a.B::x <<std::endl;
Assuming that the programmer for struct B was not aware of struct A members, but the programmer of struct c was aware of members of both, it seems very reasonable for this feature to be allowed!
As to why should compiler be able to recognize using A<X>::x; when it is used in C<X> , consider the fact that within the definition of a class/class template all the direct/indirect inherited bases are visible regardless of the type of inheritance. But only the publicly inherited ones are accessible!
For example if it was like:
using A<true>::x;
//or
//using B<true>::x;
Then there would be a problem by doing:
C<false> a;
Or wise versa. since neither A<true> or B<true> is a base for C<false>, therefor visible. But since it is like:
using A<X>::x;
Because the generic term X is used in order to define the term A<X>, it is first deducible second recognizable, since any C<X> (if is not specialized later) is indirectly based on A<X> !
Good Luck!
template <bool X>
struct C : public B<X> {
// using B<X>::x; // OK
using A<X>::x; // Why OK?
C() { x = 1; }
};
The question is why wouldn't that be supported? Because the constrain that A<X> is a base of a specialization of the main template definition of C is a question that can only be answered, and that only makes sense for a particular template argument X?
To be able to check templates at definition time was never a design goal of C++. Many well formed-ness constrains are checked at instanciation time and this is fine.
[Without a true concept (necessary and sufficient template parameter contracts) support no variant of C++ would do significantly better, and C++ is probably too complicated and irregular to have true concepts and true separate checking of templates, ever.]
The principles that makes it necessary to qualify a name to make it dependent does not have anything with early diagnostic of errors in template code; the way name lookup works in template was considered necessary by the designers to support "sane" (actually slightly less insane) name lookup in template code: a use of a non local name in a template shouldn't bind too often to a name declared by the client code, as it would break encapsulation and locality.
Note that for any unqualified dependent name you can end up accidentally calling an unrelated clashing user function if it's a better match for overloading resolution, which is another problem that would be fixed by true concept contracts.
Consider this "system" (i.e. not part of current project) header:
// useful_lib.hh _________________
#include <basic_tool.hh>
namespace useful_lib {
template <typename T>
void foo(T x) { ... }
template <typename T>
void bar(T x) {
...foo(x)... // intends to call useful_lib::foo(T)
// or basic_tool::foo(T) for specific T
}
} // useful_lib
And that project code:
// user_type.hh _________________
struct UserType {};
// use_bar1.cc _________________
#include <useful_lib.hh>
#include "user_type.hh"
void foo(UserType); // unrelated with basic_tool::foo
void use_bar1() {
bar(UserType());
}
// use_bar2.cc _________________
#include <useful_lib.hh>
#include "user_type.hh"
void use_bar2() {
bar(UserType()); // ends up calling basic_tool::foo(UserType)
}
void foo(UserType) {}
I think that code is pretty realistic and reasonable; see if you can see the very serious and non local issue (an issue that can only be found by reading two or more distinct functions).
The issue is caused by the use of an unqualified dependent name in a library template code with a name that isn't documented (intuitivement shouldn't have to be) or that is documented but that the user wasn't interested in, as he never needed to override that part of the library behavior.
void use_bar1() {
bar(UserType()); // ends up calling ::foo(UserType)
}
That wasn't intended and the user function might have a completely different behavior and fails at runtime. Of course it could also have an incompatible return type and fail for that reason (if the library function returned a value unlike in that example, obviously). Or it could create an ambiguity during overload resolution (more involved case possible if the function takes multiple arguments and both library and user functions are templates).
If this wasn't bad enough, now consider linking use_bar1.cc and use_bar2.cc; now we have two uses of the same template function in different contexts, leading to different expansions (in macro-speak, as templates are only slightly better than glorified macros); unlike preprocessor macros, you are not allowed to do that as the same concrete function bar(UserType) is being defined in two different ways by two translation units: this is an ODR violation, the program is ill formed no diagnostic required. That means that if the implementation doesn't catch the error at link time (and very few do), the behavior at runtime is undefined from start: no run of the program has defined behavior.
If you are interested, the design of name lookup in template, in the era of the "ARM" (Annotated C++ Reference Manual), long before ISO standardization, is discussed in D&E (Design and Evolution of C++).
Such unintentional binding of a name was avoided at least with qualified names and non dependent names. You can't reproduce that issue with a non dependent unqualified names:
namespace useful_lib {
template <typename T>
void foo(T x) { ... }
template <typename T>
void bar(T x) {
...foo(1)... // intends to call useful_lib::foo<int>(int)
}
} // useful_lib
Here the name binding is done such that no better overload match (that is no match by a non template function) can "beat" the specialization useful_lib::foo<int> because the name is bound in the context of the template function definition, and also because useful_lib::foo hides any outside name.
Note that without the useful_lib namespace, another foo that happened to be declared in another header included before could still be found:
// some_lib.hh _________________
template <typename T>
void foo(T x) { }
template <typename T>
void bar(T x) {
...foo(1)... // intends to call ::foo<int>(int)
}
// some_other_lib.hh _________________
void foo(int);
// user1.cc _________________
#include <some_lib.hh>
#include <some_other_lib.hh>
void user1() {
bar(1L);
}
// user2.cc _________________
#include <some_other_lib.hh>
#include <some_lib.hh>
void user2() {
bar(2L);
}
You can see that the only declarative difference between the TUs is the order of inclusion of headers:
user1 causes the instanciation of bar<long> defined without foo(int) visible and name lookup of foo only finds the template <typename T> foo(T) signature so binding is obviously done to that function template;
user2 causes the instanciation of bar<long> defined with foo(int) visible so name lookup finds both foo and the non template one is a better match; the intuitive rule of overloading is that anything (function template or regular function) that can match less argument lists wins: foo(int) can only match exactly an int while template <typename T> foo(T) can match anything (that can be copied).
So again the linking of both TUs causes an ODR violation; the most likely practical behavior is that which function is included in the executable is unpredictable, but an optimizing compiler might assume that the call in user1() does not call foo(int) and generate a non inline call to bar<long> that happens to be the second instanciation that ends up calling foo(int), which might cause incorrect code to be generated [assume foo(int) could only recurse through user1() and the compiler sees it doesn't recurse and compile it such that recursion is broken (this can be the case if there is a modified static variable in that function and the compiler moves modifications across function calls to fold successive modifications)].
This shows that templates are horribly weak and brittle and should be used with extreme care.
But in your case, there is no such name binding issue, as in that context a using declaration can only name a (direct or indirect) base class. It doesn't matter that the compiler cannot know at definition time whether it's a direct or indirect base or an error; it will check that in due time.
While early diagnostic of inherently erroneous code is permitted (because sizeof(T()) is exactly the same as sizeof(T), the declared type of s is illegal in any instantiation):
template <typename T>
void foo() { // template definition is ill formed
int s[sizeof(T) - sizeof(T())]; // ill formed
}
diagnosing that at template definition time is not practically important and not required for conforming compilers (and I don't believe compiler writers try to do it).
Diagnostic only at the point of instantiation of issues that are guaranteed to be caught at that point is fine; it does not break any design goal of C++.
If I have a class template and I use a smart pointer to a dynamically allocated instance of a specialized instance, does that cause the entire class template to be defined by the complier or will it also wait for a member function to be called from the pointer before it is instantiated?
template <class T>
class Test {
public:
void nothing();
void operation();
static const int value;
};
template <class T>
const int Test<T>::value = 100;
template <class T>
void Test<T>::nothing() {
/* invalid code */
int n = 2.5f;
}
template <class T>
void Test<T>::operation() {
double x = 2.5 * value;
}
int main() {
std::unique_ptr<Test<int>> ptr = new Test<int>(); // mark1
ptr->operation(); // mark2
return 0;
}
Does the entire class template get instantiated at mark1?
If not does that mean this code will compile correctly and the member function Test::nothing() not be instantiated?
Does the entire class template get instantiated at mark1?
Yes. The class template is implicitly instantiated — only the class template, not all its members.
If not does that mean this code will compile correctly and the member function Test::nothing() not be instantiated?
The not doesn't imply that, rather if nothing() is not used, it is not instantited.
The full answer to this probably depends highly on what compiler you are using.
At //mark1, the compiler will notice that at least portions of the Test<int> class need to instantiated. Whether or not it does it right then, or in a later phase, is up to the compiler designer.
At //mark2, Test<int>.operation() is obviously needed, it is either marked for later instantiation or created on the spot, again depending on what the compiler designers decided.
Since Test<int>.nothing() is never referenced, the compiler has the freedom to instantiate it or not. Some older compilers blindly instantiated the entire class, but I suspect the majority of modern compilers will only instantiate what they can prove to be necessary (or at least that they can't prove is not necessary). Again, though, where that happens within the compiler depends on the way the compiler designers built the compiler.
So as it turns out for the compiler I'm using (MS Visual C++), my supposition was correct that, for the code as presented in the question, the class template member instantiation would not take place at //mark1 but rather at //mark2 and Test<int>.nothing() would not be created by the compiler.
However, it seems I left out a critical part of the issue that I was experiencing. My actual class was a part of a virtual hierarchy, and according to the MSDN help library all virtual members are instantiated at object creation. So in the example above, if both member functions, i.e. operation() and nothing(), are virtual then at //mark2 the compiler would try to generate code for both functions and the validation of nothing() would fail.
http://msdn.microsoft.com/en-us/library/7y5ca42y.aspx
http://wi-fizzle.com/howtos/vc-stl/templates.htm#t9
We have following class definition
template<typename T>
class Klass {...}
and we also have below two instantiations
Klass<int> i;
Klass<double> d;
how many copies of Klass' methods are generated by the C++ compiler?
Can somebody explain it? Thanks!
Klass isn't a type, so it doesn't make sense to talk of Klass's methods. Kalss<int> is a type with it's own methods, and so is Klass<double>. In your example there would be one set of methods for each type.
Edit in real life, it isn't as simple as that. The question of the actual existence of the methods also depends on other factors, see #KerrekSB's answer to this question.
Each template instance is an entirely separate, distinct and independent type of its own. However, the code for class template member functions is only generated if the member function is actually used for a given template instantiation (unless you instantiate the template explicitly for some set of parameters). Among other things, this means that if some class template member function's body doesn't actually make sense for a given template parameter, then you can still use the overall template as long as you don't invoke that member function, since the code for the member function will never get compiled.
Also bear in mind that templates can be specialized:
template <typename T> struct Foo {
int bar;
void chi();
};
template <> struct Foo<int> {
double bar(bool, char) const;
typedef std::vector<bool> chi;
bool y;
};
As you can see, there's a lot you cannot just tell from a template itself until you see which actual instantiations you'll be talking about.
I have a class Base in base.h, which has a template function
class Base {
template <typename T> void test(T a);
}
this template is supposed to read in int or double type, and I have class Derived, which is derived from class Base
I tried to call function test in class Derived, but I have the linker error.
In the end, I realised that if in base.cpp, I add
void test(int a);
void test(double a);
there will be no compiler error. This solution seems awkward, is there a better solution? Thank you
C++ templates must be defined (given a complete function body) in the same translation unit (.CPP file plus all included header files) where they are used. In your header file, all you have done is declared (given the name and signature of) the function. The result is that when you include base.h, all the compiler sees is:
class Base {
template <typename T> void test(T a);
}
This declares but does not define the function. To define it, you must include a function body:
class Base {
template <typename T> void test(T a)
{
// do something cool with a here
}
}
The reason why this is required is that the C++ compiler generates code for templates on an "as-needed" basis. For example, if you call:
Base obj;
obj.test< int >( 1 );
obj.test< char >( 'c' );
The compiler will generate two sets of machine code based on the Base::test template, one for an int and one for a char. The limitation here is that the definition of the Base::test template must be in the same translation unit (.CPP file), or else the compiler will not know how to build the machine code for each version of the Base::test function. The compiler only operates on one translation unit at a time, so it has no idea whether or not you've defined Base::test< T > in some other CPP file. It can only work with what it has at hand.
This is quite different from the way generics work in C#, Java, and similar languages. Personally I like to think of templates as a text macro that gets expanded by the compiler as needed. That forces me to keep in mind that the complete body of the template function needs to be included in any CPP file where it is used.
You must fully define the template function test before it can be used. The easiest way to do that is just to write your function body in the header in base.h:
class Base {
template <typename T> void test(T a)
{
... function body here
}
}
if you declare template function in base class which means it take the template argument at compile time but if u try to access through derived class which is runtime implementation so template request at compile time which ur providing at runtime is not possible ,and main thing c++ does not support this.
I heard C++ templates wont generate errors until they are used. Is it true ? Can someone explain me how they work ?
Templates follow two phase compilation model.
struct X{
private:
void f(){}
};
template<class T> void f(T t){
int; // gives error in phase 1 (even if f(x) call is commented in main)
t.f(); // gives error only when instantiated with T = X, as x.f() is private, in phase 2
}
int main(){
X x;
f(x);
}
They generate compiler errors when they are compiled. They are compiled separately for each actual parameter passed as the template argument(s) (this is unlike Java Generics), e.g., if I have:
template <typename T> class foo { ... }
and
int main() {
foo<char> c;
foo<int> i ;
}
the template foo gets compiled twice, once for chars, once for ints.
If you never (directly or indirectly) instantiated or used template foo, it wouldn't be compiled and you'd not see any compiler errors.
Once compiled, they're just "normal" C++ code, and like any code, can generate runtime errors.
From here,
From the point of view of the
compiler, templates are not normal
functions or classes. They are
compiled on demand, meaning that the
code of a template function is not
compiled until an instantiation with
specific template arguments is
required. At that moment, when an
instantiation is required, the
compiler generates a function
specifically for those arguments from
the template.
Hope it helps..
Conceptually, at the highest level
template <Type value, class Y, ...>
...fn-or-class...
may be usefully compared to
#define FN_OR_CLASS(VALUE, TYPE_Y, ...) \
...fn-or-class...
Both basically wait until called/instantiated then substitute the specified types and values to produce tailored code with full compile-time optimisation for those values. But, templates differ from #defines in that they're proper compile-stage constructs that can be enclosed in namespaces, must satisfy the lexer, and not all of a class template is generated when the first instantiation is seen - rather, functions are generated on an as-needed basis.
When the compiler first encounters a template, it does a rough check that the template's content could make sense for some hypothetical instantiation. Later, when it encounters a specific instantiation, then for class templates only the functions that are used are further checked to make sure they can be compiled with the specific parameters in use. This does mean that a class template can appear - for some limited usage - to support instantiation with specific parameters, but if you start using some other functions in the template API then suddenly you can find that it can't be compiled with that presumed-suitable parameter... can force you to redesign your usage rather late in the day. That is one of the reasons that C++0x had planned to introduce Concepts: they elegantly allow templates to check that parameters meet all the template's expectations - if they allow any instantiation, then the user can assume that the full API of the template can be used.
template <class T>
struct X
{
void f() { }
void g() { T::whatever(); } // only error if g() called w/o T::whatever
};
int main()
{
X<int> x;
x.f();
// x.g(); // would cause an error as int::whatever() doesn't exist...
}
The SFINAE (substitution failure is not an error) technique can then allow the compiler to select between multiple nearly-matching functions based on the actual instantiating template parameters. This can be used to implement basic compile-time introspection, such as "does this class have a member function fn(int)?".