I would like to have a class implement operator() several different ways based on an option set in the class. Because this will be called a large number of times, I don't want to use anything that branches. Ideally, the operator() would be a function pointer that can be set with a method. However, I'm not sure what this would actually look like. I tried:
#include <iostream>
class Test {
public:
int (*operator())();
int DoIt1() {
return 1;
}
int DoIt2() {
return 2;
}
void SetIt(int i) {
if(i == 1) {
operator() = &Test::DoIt1;
} else {
operator() = &Test::DoIt2;
}
}
};
int main()
{
Test t1;
t1.SetIt(1);
std::cout << t1() << std::endl;
t1.SetIt(2);
std::cout << t1() << std::endl;
return 0;
}
I know it will work if I create another function pointer and call that from the operator() function. But is it possible to have the operator() function itself be a function pointer? Something along the lines of what I posted (which doesn't compile)?
The above code gives:
test.cxx:5:21: error: declaration of ‘operator()’ as non-function
test.cxx: In member function ‘void Test::SetIt(int)’:
test.cxx:17:16: error: ‘operator()’ not defined
test.cxx:19:16: error: ‘operator()’ not defined
test.cxx: In function ‘int main()’:
test.cxx:30:19: error: no match for call to ‘(Test) ()’
test.cxx:34:19: error: no match for call to ‘(Test) ()’
Your class needs to somehow remember what function pointer to use. Store it as a class member:
class Test
{
public:
Test() : func(0) {}
int operator()() {
// Note that pointers to Test member functions need a pointer to Test to work.
return (this->*func)(); // undefined behavior if func == 0
}
void SetIt(int i) {
if(i == 1) {
func = &Test::DoIt1;
} else {
func = &Test::DoIt2;
}
}
private:
int DoIt1() {
return 1;
}
int DoIt2() {
return 2;
}
// Typedef of a pointer to a class method.
typedef int (Test::*FuncPtr)();
FuncPtr func;
};
However, before you go into the effort of doing this, profile your code first and see if branching via switch or if is actually a bottleneck (it might not be!). Modern processors have very counterintuitive performance characteristics, so compilers may be able to generate better code than you think it does. The only way to make sure that branching is actually too costly for you to use is to profile your code. (And by "profiling" I mean "run well-designed experiments", not "come up with a hunch without testing".)
You can make your operator() an inline function which calls another pointer. The optimizer should take the extra indirection out completely.
One solution is provided by #In silico which is valid in both C++03 and C++11.
Here is another solution for C++11 only:
std::function<int(Test*)> func;
func = &Test::DoIt1;
func(this); //this syntax is less cumbersome compared to C++03 solution
A quick online full demo
Related
#include <iostream>
template <int N>
class X {
public:
using I = int;
void f(I i) {
std::cout << "i: " << i << std::endl;
}
};
template <int N>
void fppm(void (X<N>::*p)(typename X<N>::I)) {
p(0);
}
int main() {
fppm(&X<33>::f);
return 0;
}
I just don't understand the compile error message of the code.
error: called object type 'void (X<33>::*)(typename X<33>::I)' is not a function or function pointer
p(0);
I think p is a function which returns void and takes int as its argument. But apparently, it's not. Could somebody give me clue?
Since p is a pointer to a nonstatic member function, you need an instance to call it with. Thus, first instantiate an object of X<33> in main:
int main() {
X<33> x;
fppm(x, &X<33>::f); // <-- Signature changed below to accept an instance
Then in your function, change the code to accept an instance of X<N> and call the member function for it:
template <int N>
void fppm(X<N> instance, void (X<N>::*p)(typename X<N>::I)) {
(instance.*p)(0);
}
The syntax may look ugly but the low precedence of the pointer to member operator requires the need for the parentheses.
As denoted in the comments already, p is a pointer to member function, but you call it like a static function (p(0);). You need a concrete object to call p on:
X<N> x;
(x.*p)(0);
// or:
X<N>* xx = new X<N>();
(xx->*p)(0);
delete xx;
Be aware that the .*/->* operators have lower precedence than the function call operator, thus you need the parentheses.
Side note: Above is for better illustration, modern C++ might use auto keyword and smart pointers instead, which could look like this:
auto x = std::make_unique<X<N>>();
(x.get()->*p)(0);
I have a class MyClass. In it, I want to create an array of function pointer and initialize the array with the functions of another class (MemberClass).
MyClass has a pointer to MemberClass as member.
but I get this compile time error
error C2440: 'initializing' : cannot convert from 'void (__thiscall MyMemberClass::* )(void)' to 'F'
//This class has the function that I want to create a function pointer to
class MemberClass
{
private:
int myValue1;
int myValue2;
public:
int GetValue1() //I want to point to this function from a function pointer in another class
{
return myvalue1;
}
int GetValue2() //I want to point to this function from a function pointer in another class
{
return myvalue2;
}
}
//This has array of function pointer that I want to point to member function of another class
Class MyClass
{
typedef void(*F)();
private:
MemberClass* mclass;
F[] f;
Process();
}
.cpp
MyClass::MyClass()
{
f[2] = {&MemberClass::GetValue1, &MemberClass::GetValue2} //This line throws error
//error C2440: 'initializing' : cannot convert from 'void (__thiscall MyMemberClass::* )(void)' to 'F'
}
void MyClass::Processing()
{
//This is how I am hoping to use the function pointer array
F[Index]();
}
F is declared as pointer to function with no parameters returning void. But your functions return int, and are member functions of MemberClass rather than plain ordinary functions. So the type you need is
typedef int (MemberClass::*F)();
Calling it is also more interesting:
int result = (mclass->*f[index])();
Suggestion: rather than a method pointer, use C++11's functional library.
I'm butchering OP's sample code slightly to simplify the example.
MemberClass stays mostly the same. I removed the member variables because the methods are now hard-coded to return 1 and 2 to make them easy to tell apart.
#include <iostream>
#include <functional>
class MemberClass
{
public:
int GetValue1()
{
return 1;
}
int GetValue2()
{
return 2;
}
};
myClass gets a rip-up because this is where the action is.
class MyClass
{
private:
I'm using an array of std::function instead of a typedef and an array of the typedef. Note the template argument int(). This is an array of functions that takes nothing and returns an int. Magic in std::bind will provide the hidden this parameter required by methods. If the function has parameters that are not known at the time of binding, use std::placeholders to save room to them in the method's parameter list.
Since the methods are bound to their object, there is no longer any need to store MemberClass* mclass;
std::function<int()> f[2];
public:
Calling the function is simple: index the array and stick the brackets on.
int Process(int index)
{
return f[index]();
}
The constructor is either a bit trickier, or less tricky, depending on your school of thought. I'm using an initializer list because it is cleaner (to me, at anyrate) and often has performance advantages. For one thing, you can swap out the array for a std::vector or most other containers without having to change a line of code other than the variable definition.
f[0] = std::bind(&MemberClass::GetValue1, mem);
f[1] =...
inside the body of the constructor is still an option.
MyClass(MemberClass * mem):
f{std::bind(&MemberClass::GetValue1, mem),
std::bind(&MemberClass::GetValue2, mem)}
{
}
};
And a silly little bit of test code to make sure this all works. Why? Because every time you don't test code in it's simplest form you're taking an unnecessary risk. If it doesn't work small, it won't work big.
int main()
{
MemberClass c;
MyClass d(&c);
std::cout << d.Process(0) << std::endl;
std::cout << d.Process(1) << std::endl;
}
All together for one cut and paste-able block:
#include <iostream>
#include <functional>
class MemberClass
{
public:
int GetValue1()
{
return 1;
}
int GetValue2()
{
return 2;
}
};
class MyClass
{
private:
std::function<int()> f[2];
public:
int Process(int index)
{
return f[index]();
}
MyClass(MemberClass * mem):
f{std::bind(&MemberClass::GetValue1, mem),
std::bind(&MemberClass::GetValue2, mem)}
{
}
};
int main()
{
MemberClass c;
MyClass d(&c);
std::cout << d.Process(0) << std::endl;
std::cout << d.Process(1) << std::endl;
}
When should I explicitly write this->member in a method of
a class?
Usually, you do not have to, this-> is implied.
Sometimes, there is a name ambiguity, where it can be used to disambiguate class members and local variables. However, here is a completely different case where this-> is explicitly required.
Consider the following code:
template<class T>
struct A {
T i;
};
template<class T>
struct B : A<T> {
T foo() {
return this->i; //standard accepted by all compilers
//return i; //clang and gcc will fail
//clang 13.1.6: use of undeclared identifier 'i'
//gcc 11.3.0: 'i' was not declared in this scope
//Microsoft C++ Compiler 2019 will accept it
}
};
int main() {
B<int> b;
b.foo();
}
If you omit this->, some compilers do not know how to treat i. In order to tell it that i is indeed a member of A<T>, for any T, the this-> prefix is required.
Note: it is possible to still omit this-> prefix by using:
template<class T>
struct B : A<T> {
int foo() {
return A<T>::i; // explicitly refer to a variable in the base class
//where 'i' is now known to exist
}
};
If you declare a local variable in a method with the same name as an existing member, you will have to use this->var to access the class member instead of the local variable.
#include <iostream>
using namespace std;
class A
{
public:
int a;
void f() {
a = 4;
int a = 5;
cout << a << endl;
cout << this->a << endl;
}
};
int main()
{
A a;
a.f();
}
prints:
5
4
There are several reasons why you might need to use this pointer explicitly.
When you want to pass a reference to your object to some function.
When there is a locally declared object with the same name as the member object.
When you're trying to access members of dependent base classes.
Some people prefer the notation to visually disambiguate member accesses in their code.
Although I usually don't particular like it, I've seen others use this-> simply to get help from intellisense!
There are few cases where using this must be used, and there are others where using the this pointer is one way to solve a problem.
1) Alternatives Available: To resolve ambiguity between local variables and class members, as illustrated by #ASk.
2) No Alternative: To return a pointer or reference to this from a member function. This is frequently done (and should be done) when overloading operator+, operator-, operator=, etc:
class Foo
{
Foo& operator=(const Foo& rhs)
{
return * this;
}
};
Doing this permits an idiom known as "method chaining", where you perform several operations on an object in one line of code. Such as:
Student st;
st.SetAge (21).SetGender (male).SetClass ("C++ 101");
Some consider this consise, others consider it an abomination. Count me in the latter group.
3) No Alternative: To resolve names in dependant types. This comes up when using templates, as in this example:
#include <iostream>
template <typename Val>
class ValHolder
{
private:
Val mVal;
public:
ValHolder (const Val& val)
:
mVal (val)
{
}
Val& GetVal() { return mVal; }
};
template <typename Val>
class ValProcessor
:
public ValHolder <Val>
{
public:
ValProcessor (const Val& val)
:
ValHolder <Val> (val)
{
}
Val ComputeValue()
{
// int ret = 2 * GetVal(); // ERROR: No member 'GetVal'
int ret = 4 * this->GetVal(); // OK -- this tells compiler to examine dependant type (ValHolder)
return ret;
}
};
int main()
{
ValProcessor <int> proc (42);
const int val = proc.ComputeValue();
std::cout << val << "\n";
}
4) Alternatives Available: As a part of coding style, to document which variables are member variables as opposed to local variables. I prefer a different naming scheme where member varibales can never have the same name as locals. Currently I'm using mName for members and name for locals.
Where a member variable would be hidden by
a local variable
If you just want
to make it explictly clear that you
are calling an instance method/variable
Some coding standards use approach (2) as they claim it makes the code easier to read.
Example:
Assume MyClass has a member variable called 'count'
void MyClass::DoSomeStuff(void)
{
int count = 0;
.....
count++;
this->count = count;
}
One other case is when invoking operators. E.g. instead of
bool Type::operator!=(const Type& rhs)
{
return !operator==(rhs);
}
you can say
bool Type::operator!=(const Type& rhs)
{
return !(*this == rhs);
}
Which might be more readable. Another example is the copy-and-swap:
Type& Type::operator=(const Type& rhs)
{
Type temp(rhs);
temp.swap(*this);
}
I don't know why it's not written swap(temp) but this seems to be common.
The other uses for this (as I thought when I read the summary and half the question... .), disregarding (bad) naming disambiguation in other answers, are if you want to cast the current object, bind it in a function object or use it with a pointer-to-member.
Casts
void Foo::bar() {
misc_nonconst_stuff();
const Foo* const_this = this;
const_this->bar(); // calls const version
dynamic_cast<Bar*>(this)->bar(); // calls specific virtual function in case of multi-inheritance
}
void Foo::bar() const {}
Binding
void Foo::baz() {
for_each(m_stuff.begin(), m_stuff.end(), bind(&Foo:framboozle, this, _1));
for_each(m_stuff.begin(), m_stuff.end(), [this](StuffUnit& s) { framboozle(s); });
}
void Foo::framboozle(StuffUnit& su) {}
std::vector<StuffUnit> m_stuff;
ptr-to-member
void Foo::boz() {
bez(&Foo::bar);
bez(&Foo::baz);
}
void Foo::bez(void (Foo::*func_ptr)()) {
for (int i=0; i<3; ++i) {
(this->*func_ptr)();
}
}
Hope it helps to show other uses of this than just this->member.
You only have to use this-> if you have a symbol with the same name in two potential namespaces. Take for example:
class A {
public:
void setMyVar(int);
void doStuff();
private:
int myVar;
}
void A::setMyVar(int myVar)
{
this->myVar = myVar; // <- Interesting point in the code
}
void A::doStuff()
{
int myVar = ::calculateSomething();
this->myVar = myVar; // <- Interesting point in the code
}
At the interesting points in the code, referring to myVar will refer to the local (parameter or variable) myVar. In order to access the class member also called myVar, you need to explicitly use "this->".
You need to use this to disambiguate between a parameters/local variables and member variables.
class Foo
{
protected:
int myX;
public:
Foo(int myX)
{
this->myX = myX;
}
};
The main (or I can say, the only) purpose of this pointer is that it points to the object used to invoke a member function.
Base on this purpose, we can have some cases that only using this pointer can solve the problem.
For example, we have to return the invoking object in a member function with argument is an same class object:
class human {
...
human & human::compare(human & h){
if (condition)
return h; // argument object
else
return *this; // invoking object
}
};
I found another interesting case of explicit usage of the "this" pointer in the Effective C++ book.
For example, say you have a const function like
unsigned String::length() const
You don't want to calculate String's length for each call, hence you want to cache it doing something like
unsigned String::length() const
{
if(!lengthInitialized)
{
length = strlen(data);
lengthInitialized = 1;
}
}
But this won't compile - you are changing the object in a const function.
The trick to solve this requires casting this to a non-const this:
String* const nonConstThis = (String* const) this;
Then, you'll be able to do in above
nonConstThis->lengthInitialized = 1;
While I would assume that in VC++ this would be a no brainer, it's still worth asking.
When creating a getter method for a class that only returns the value of a protected/private member, does the compiler optimize this call so it's the equivalent of referencing that member without having to friend the class and without the overhead of a full method call?
Yes. Both variants compile to the same thing:
struct test
{
int x;
int get() const { return x; }
};
__declspec(noinline) int use_x(const test& t)
{
return t.x;
}
__declspec(noinline) int use_get(const test& t)
{
return t.get();
}
int main()
{
test t = { 111605 };
// pick one:
return use_x(t);
//return use_get(t);
}
Note that it's not as simple as always replacing t.get() with t.x, for the compiler. Consider something like this:
t.get() += 5;
This shouldn't compile, because the result of the function call is an rvalue and += (for primitives) requires an lvalue. The compiler will check for things like that.
When should I explicitly write this->member in a method of
a class?
Usually, you do not have to, this-> is implied.
Sometimes, there is a name ambiguity, where it can be used to disambiguate class members and local variables. However, here is a completely different case where this-> is explicitly required.
Consider the following code:
template<class T>
struct A {
T i;
};
template<class T>
struct B : A<T> {
T foo() {
return this->i; //standard accepted by all compilers
//return i; //clang and gcc will fail
//clang 13.1.6: use of undeclared identifier 'i'
//gcc 11.3.0: 'i' was not declared in this scope
//Microsoft C++ Compiler 2019 will accept it
}
};
int main() {
B<int> b;
b.foo();
}
If you omit this->, some compilers do not know how to treat i. In order to tell it that i is indeed a member of A<T>, for any T, the this-> prefix is required.
Note: it is possible to still omit this-> prefix by using:
template<class T>
struct B : A<T> {
int foo() {
return A<T>::i; // explicitly refer to a variable in the base class
//where 'i' is now known to exist
}
};
If you declare a local variable in a method with the same name as an existing member, you will have to use this->var to access the class member instead of the local variable.
#include <iostream>
using namespace std;
class A
{
public:
int a;
void f() {
a = 4;
int a = 5;
cout << a << endl;
cout << this->a << endl;
}
};
int main()
{
A a;
a.f();
}
prints:
5
4
There are several reasons why you might need to use this pointer explicitly.
When you want to pass a reference to your object to some function.
When there is a locally declared object with the same name as the member object.
When you're trying to access members of dependent base classes.
Some people prefer the notation to visually disambiguate member accesses in their code.
Although I usually don't particular like it, I've seen others use this-> simply to get help from intellisense!
There are few cases where using this must be used, and there are others where using the this pointer is one way to solve a problem.
1) Alternatives Available: To resolve ambiguity between local variables and class members, as illustrated by #ASk.
2) No Alternative: To return a pointer or reference to this from a member function. This is frequently done (and should be done) when overloading operator+, operator-, operator=, etc:
class Foo
{
Foo& operator=(const Foo& rhs)
{
return * this;
}
};
Doing this permits an idiom known as "method chaining", where you perform several operations on an object in one line of code. Such as:
Student st;
st.SetAge (21).SetGender (male).SetClass ("C++ 101");
Some consider this consise, others consider it an abomination. Count me in the latter group.
3) No Alternative: To resolve names in dependant types. This comes up when using templates, as in this example:
#include <iostream>
template <typename Val>
class ValHolder
{
private:
Val mVal;
public:
ValHolder (const Val& val)
:
mVal (val)
{
}
Val& GetVal() { return mVal; }
};
template <typename Val>
class ValProcessor
:
public ValHolder <Val>
{
public:
ValProcessor (const Val& val)
:
ValHolder <Val> (val)
{
}
Val ComputeValue()
{
// int ret = 2 * GetVal(); // ERROR: No member 'GetVal'
int ret = 4 * this->GetVal(); // OK -- this tells compiler to examine dependant type (ValHolder)
return ret;
}
};
int main()
{
ValProcessor <int> proc (42);
const int val = proc.ComputeValue();
std::cout << val << "\n";
}
4) Alternatives Available: As a part of coding style, to document which variables are member variables as opposed to local variables. I prefer a different naming scheme where member varibales can never have the same name as locals. Currently I'm using mName for members and name for locals.
Where a member variable would be hidden by
a local variable
If you just want
to make it explictly clear that you
are calling an instance method/variable
Some coding standards use approach (2) as they claim it makes the code easier to read.
Example:
Assume MyClass has a member variable called 'count'
void MyClass::DoSomeStuff(void)
{
int count = 0;
.....
count++;
this->count = count;
}
One other case is when invoking operators. E.g. instead of
bool Type::operator!=(const Type& rhs)
{
return !operator==(rhs);
}
you can say
bool Type::operator!=(const Type& rhs)
{
return !(*this == rhs);
}
Which might be more readable. Another example is the copy-and-swap:
Type& Type::operator=(const Type& rhs)
{
Type temp(rhs);
temp.swap(*this);
}
I don't know why it's not written swap(temp) but this seems to be common.
The other uses for this (as I thought when I read the summary and half the question... .), disregarding (bad) naming disambiguation in other answers, are if you want to cast the current object, bind it in a function object or use it with a pointer-to-member.
Casts
void Foo::bar() {
misc_nonconst_stuff();
const Foo* const_this = this;
const_this->bar(); // calls const version
dynamic_cast<Bar*>(this)->bar(); // calls specific virtual function in case of multi-inheritance
}
void Foo::bar() const {}
Binding
void Foo::baz() {
for_each(m_stuff.begin(), m_stuff.end(), bind(&Foo:framboozle, this, _1));
for_each(m_stuff.begin(), m_stuff.end(), [this](StuffUnit& s) { framboozle(s); });
}
void Foo::framboozle(StuffUnit& su) {}
std::vector<StuffUnit> m_stuff;
ptr-to-member
void Foo::boz() {
bez(&Foo::bar);
bez(&Foo::baz);
}
void Foo::bez(void (Foo::*func_ptr)()) {
for (int i=0; i<3; ++i) {
(this->*func_ptr)();
}
}
Hope it helps to show other uses of this than just this->member.
You only have to use this-> if you have a symbol with the same name in two potential namespaces. Take for example:
class A {
public:
void setMyVar(int);
void doStuff();
private:
int myVar;
}
void A::setMyVar(int myVar)
{
this->myVar = myVar; // <- Interesting point in the code
}
void A::doStuff()
{
int myVar = ::calculateSomething();
this->myVar = myVar; // <- Interesting point in the code
}
At the interesting points in the code, referring to myVar will refer to the local (parameter or variable) myVar. In order to access the class member also called myVar, you need to explicitly use "this->".
You need to use this to disambiguate between a parameters/local variables and member variables.
class Foo
{
protected:
int myX;
public:
Foo(int myX)
{
this->myX = myX;
}
};
The main (or I can say, the only) purpose of this pointer is that it points to the object used to invoke a member function.
Base on this purpose, we can have some cases that only using this pointer can solve the problem.
For example, we have to return the invoking object in a member function with argument is an same class object:
class human {
...
human & human::compare(human & h){
if (condition)
return h; // argument object
else
return *this; // invoking object
}
};
I found another interesting case of explicit usage of the "this" pointer in the Effective C++ book.
For example, say you have a const function like
unsigned String::length() const
You don't want to calculate String's length for each call, hence you want to cache it doing something like
unsigned String::length() const
{
if(!lengthInitialized)
{
length = strlen(data);
lengthInitialized = 1;
}
}
But this won't compile - you are changing the object in a const function.
The trick to solve this requires casting this to a non-const this:
String* const nonConstThis = (String* const) this;
Then, you'll be able to do in above
nonConstThis->lengthInitialized = 1;