And what's the difference between
void*(*void)(void*)
and
void*(*voi)(void*)
and when to use it?
well void*(void)(void) doesn't compile, because it tries to declare a function that returns a void* named void and you can't use keywords of the language as names.
you can't declare an int named void either (int void=5; doesn't work of course).
Now voi is a valid identifier and you can name a function voi.
Trying to get at what you're actually asking the difference between.
void(*name)(void); and void(name)(void); is that one declares a function and the other declares a function pointer.
In fact the brackets on the second example don't do anything. void(name)(void); is the same as void name(void);.
However when declaring function pointers the brackets are needed or the * will bind left to the return type.
Anyway, I'm not sure what exactly you're asking, so you're best off searching the site for other questions regarding function pointers. Like this one that also explains why using void(name)(void) can be different from void(name)().
Related
I have couple questions regarding some C++ rules.
Why am I able to call a function/method from outside the class in the namespace when I include the return type? (look at the namespace test2::testclass2 in the code below) i.e. this works:
bool b = testclass1::foo<int>(2);
whereas this doesn't: - (it doesn't even compile - compiler throws that this is function redeclaration)
testclass1::foo<int>(2);
C++ complains that it is a function redeclaration. Is that so?
This line:
bool b = testclass1::foo<int>(2);
gets called first before anything else. Is this because static methods get created always first before anything else in C++?
Where can I find those rules? I have a few C++ books at home, so if someone would be kind enough to either point out a book (and chapter or page) or direct me to a website I would greatly appreciate it.
Here below is the sample (partial) code that I tested at home with Visual Studio 2008:
class testclass1
{
public:
testclass1(void);
~testclass1(void);
template<class A> static bool foo(int i)
{
std::cout <<"in static foo";
return true;
}
};
namespace test2
{
class testclass2
{
public:
testclass2(void);
~testclass2(void);
};
bool b = testclass1::foo<int>(2);
}
EDIT:
A few people mentioned that I need to call this inside the function and this will work without any problem.
I understand that; the only reason I asked this question is because I saw this code somewhere (in someone's elses project) and was wondering how and why this works. Since I never really seen anyone doing it before.
Also, this is used (in multiple places) as a way to call and instantiate a large number of classes like this via those function calls (that are outside). They get called first before anything else is instantiated.
C++ is not Python. You write statements in functions and execution starts from the main method. The reason bool b = ... happens to work is that it's defining the global variable b and the function call is merely the initialization expression.
Definitions can exist outside functions while other statements can only exist inside a function body.
Why am I able to call a function/method from outside the class in the namespace when I include the return type? (look at the namespace test2::testclass2)
Your declaration of b is not inside a function, so you are declaring a global variable. If you were inside a function's scope, your second statement would work, but outside a function it makes no sense.
This also answers your second question.
Of course, you wouldn't be allowed to call it this way (i.e. not as a method of an object) if it weren't a static member function.
You can find the rules on e.g. Koenig lookup and template in the standard documentation -- good luck with navigating that! You're not mentioning which compiler you are testing, but I'm not entirely sure it's compliant!
As Mehrdad points out, you're declaring and initializing a global variable within the test2 namespace: this has nothing to do with static methods.
if you write this inside a function like below then it works without a problem. As mentioned above, you need to call these functions from within a function unless you are using the function to initialize a global variable ...
int main()
{
testclass1::foo<int>(2);
return 0;
}
1. First, a helpful correction: you said "...when I include the return type". I think you might be misunderstanding what the <int> part of testclass1::foo<int>(2) does. It doesn't (necessarily) specify the return type, it just provides a value for the template argument "A".
You could have chosen to use A as the return type, but you have the return type hard-coded to "bool".
Basically, for the function as you have written it you will always need to have the <> on it in order to call it. C++ does allow you to omit the <args> off the function when the type can be deduced from the function arguments; in order to get it to do that you have to use the type argument A in your function arguments. For instance if you declared the function this way instead then you could call it without the <>:
template<class A> static bool foo(A i);
In which case it you could call "foo(2)" and it would deduce A to be "int" from the number two.
On the other hand there isn't any way to make it deduce anything based on what you assign the function to. For template argument deduction it only looks at the arguments to the function, not what is done with the result of calling the function. So in:
bool b = testclass1::foo(2);
There is no way to get it to deduce "bool" from that, not even if you made A the return type.
So why doesn't the compiler just tell you "you needed to use <> on the function"? Even though you declared foo once as a template function, you could have also overloaded it with a non-template version too. So the compiler doesn't just automatically assume that you're trying to call a template function when you leave the <> off the call. Unfortunately having NOT assumed you were calling template-foo and not seeing any other declaration for foo, the compiler then falls back on an old C style rule where for a function that takes an int and returns an int, in a very old dialect of C you didn't need to declare that kind of before using it. So the compiler assumed THAT was what you wanted - but then it notices that template-foo and old-crufty-C-foo both take an int parameter, and realizes it wouldn't be able to tell the difference between them. So then it says you can't declare foo. This is why C++ compilers are notorious for giving bad error messages - by the time the error is reported the compiler may have gone completely off the rails and be talking about something that is three or four levels removed from your actual code!
2. Yes you're exactly right.
3. I find that the C++ references and whitepapers that IBM makes available online are the most informative. Here's a link to the section about templates: C++ Templates
I have a question about changing parameter name of a member function.
I have function foo(type iA) then I change to foo(type iB), I think it is bin comp, but I am not sure.
Could you help me to explain the reason? Thanks a lot!
Of course it will be binary compatible because you do not change the type of object. Hence the size stays the same. Name of variable is only for human.
You can actually declare function not giving any names, like:
int funct(void*, int, Object)
When you define it you can use whatever you like
int funct(void* ptr, int something , Object object){return 42;};
It is fine. It also works for class members.
The (I would say main) reason for binary compatibility or incompatibility are the sizes of passed objects. So when you operate on actual binary data, on assembler level no offsets change. If you changed types, arguments would lie in different places in memory, and offsets for them would need to be recalculated. What is important are addresses not aliases or names.
Edits:
For the sake of the completeness.
As #Lightness pointed out, you can also skip names in definition, so the very example I gave could look: int funct(void*, int, Object){return 42;};.
Also regarding #James' comment. I would say that when two objects have the same size, it gets murky. Of course it makes no sense to make such wild conversion from one type to another. However, considering previous example, if you do not use operand it would probably work ... on the assembler level. If sizes differed it would corrupt stack.
However, when I thought how to comment on this problem, I told myself that you ask about c++. So let's look at it from this point of view. I would say that functions are binary compatible when their signature does not change. And names of operands are not part of it in any definition. Function signature is a basis for name mangling and overloading, and the symbol name in the final binary object.
The names of the arguments is just for you to distinguish the arguments. You can have different names in a function declaration and the functions definition.
So this is okay:
void foo(int bar); // Declare function
...
// Define function
void foo(int bibibibibi)
{
...
}
The declaration and definition above is for the same function. C++ does not override or overload functions based on argument names, only argument types.
The compiler does not store the names of arguments (or other variable) except as debug info. The actual generated code have no information about the variable names.
I'm receiving the following warning:
warning: converting from 'void (MyClass::*)(byte)' to 'void (*)(byte)'
This is because I need to pass as argument a member function instead of an ordinary function. But the program is running correctly.
I'd like to disable this warning (Wno-bad-function-cast doesn't work for C++) or to implement a different way to pass a member function.
No. Take this warning seriously. You should rather change your code to handle this scenario.
Pointer to member function(void (MyClass::*)(byte)) and normal function pointer (void (*)(byte)) are entirely different. See this link. You cannot cast them just like that. It results in undefined behavior or crash.
See here, how they are different:
void foo (byte); // normal function
struct MyClass {
void foo (byte); // member function
}
Now you may feel that, foo(byte) and MyClass::foo(byte) have same signature, then why their function pointers are NOT same. It's because, MyClass::foo(byte) is internally resolved somewhat as,
void foo(MyClass* const this, byte);
Now you can smell the difference between them.
Declare pointer to member function as,
void (MyClass::*ptr)(byte) = &MyClass::foo;
You have to use this ptr with the object of MyClass, such as:
MyClass obj;
obj.*ptr('a');
You can't pass a function that takes two arguments to a place that expects a function that takes one. Can't be done, forget about it, period, end of story. The caller passes one argument to your function. It doesn't know about the second argument, it doesn't pass it to your function, you can't make it do what you want however hard you try.
For the very same reason you can't pass a non-static member function where a regular function is expected. A member function needs an object to operate on. Whatever code calls your function doesn't know about the object, there's no way to pass it the object, and there's no way to make it use the right calling sequence that takes the object into account.
Interfaces that take user's functions, without taking additional data that the user might want to pass to his function, are inherently evil. Look at the qsort() function from the C standard library. That's an example of an evil interface. Suppose you want to sort an array of string according to some collation scheme defined externally. But all it accepts is a comparison function that takes two values. How do you pass that collation scheme to your comparison function? You can't, and so if you want it working, you must use an evil global variable, with all the strings attached to it.
That's why C++ has moved away from passing function pointers around, and towards function objects. Function objects can encapsulate whatever data you want.
Also, this may be helpful
union FuncPtr
{
void (* func)(MyClass* ptr, byte);
void (MyClass::* mem_func)(byte);
};
I'm trying to fix something in some Objective C++ (?!) code. I don't know either of those languages, or any of the relevant APIs or the codebase, so I'm getting stymied left and right.
Say I have:
Vector<char, sizeof 'a'>& sourceData();
sourceData->append('f');
When i try to compile that, I get:
error: request for member 'append' in 'WebCore::sourceData', which is of non-class type 'WTF::Vector<char, 1ul >& ()();
In this case, Vector is WTF::Vector (from WebKit or KDE or something), not STD::Vector. append() very much is supposed to be a member of class generated from this template, as seen in this documentation. It's a Vector. It takes the type the template is templated on.
Now, because I never write programs in Real Man's programming languages, I'm hella confused about the notations for references and pointers and dereferences and where we need them.
I ultimately want a Vector reference, because I want to pass it to another function with the signature:
void foobar(const Vector<char>& in, Vector<char>& out)
I'm guessing the const in the foobar() sig is something I can ignore, meaning 'dont worry, this won't be mangled if you pass it in here'.
I've also tried using .append rather than -> because isn't one of the things of C++ references that you can treat them more like they aren't pointers? Either way, its the same error.
I can't quite follow the error message: it makes it sound like sourceData is of type WTF:Vector<char, 1ul>&, which is what I want. It also looks from the those docs of WTF::Vector that when you make a Vector of something, you get an .append(). But I'm not familiar with templates, either, so I can't really tell i I'm reading that right.
EDIT:
(This is a long followup to Pavel Minaev)
WOW THANKS PROBLEM SOLVED!
I was actually just writing an edit to this post that I semi-figured out your first point after coming across a reference on the web that that line tells the compiler your forward declaring a func called sourceData() that takes no params and returns a Vector of chars. so a "non-class type" in this case means a type that is not an instance of a class. I interpreted that as meaning that the type was not a 'klass', i.e. the type of thing you would expect you could call like .addMethod(functionPointer).
Thanks though! Doing what you suggest makes this work I think. Somehow, I'd gotten it into my head (idk from where) that because the func sig was vector&, I needed to declare those as &'s. Like a stack vs. heap pass issue.
Anyway, that was my REAL problem, because I tried what you'd suggested about but that doesn't initialize the reference. You need to explicitly call the constructor, but then when I put anything in the constructor's args to disambiguate from being a forward decl, it failed with some other error about 'temporary's.
So in a sense, I still don't understand what is going on here fully, but I thank you heartily for fixing my problem. if anyone wants to supply some additional elucidation for the benefit of me and future google people, that would be great.
This:
Vector<char, sizeof 'a'>& sourceData();
has declared a global function which takes no arguments and returns a reference to Vector. The name sourceData is therefore of function type. When you try to access a member of that, it rightfully complains that it's not a class/struct/union, and operator-> is simply inapplicable.
To create an object instead, you should omit the parentheses (they are only required when you have any arguments to pass to the constructor, and must be omitted if there are none):
Vector<char, sizeof 'a'> sourceData;
Then you can call append:
sourceData.append('f');
Note that dot is used rather than -> because you have an object, not a pointer to object.
You do not need to do anything special to pass sourceData to a function that wants a Vector&. Just pass the variable - it will be passed by reference automatically:
foobar(sourceData, targetData);
Dipping your toes in C++ is never much fun. In this case, you've run into a couple of classic mistakes. First, you want to create an instance of Vector on the stack. In this case the empty () is interpreted instead as a declaratiton of a function called sourceData that takes no agruments and returns a reference to a Vector. The compiler is complaining that the resulting function is not a class (it's not). To create an instance of Vector instead, declare the instance without the () and remove the &. The parentheses are only required if you are passing arguments to the instance constructor and must be omitted if there are no arguments.
You want
Vector<char, sizeof 'a'> sourceData;
sourceData.append('f');
Vector<char, sizeof 'a'> outData; //if outData is not instantiated already
foobar(sourceData, outData);
This Wikipedia article gives a decent introduction to C++ references.
I was reading litb's question about SFINAE here and I was wondering exactly what his code is declaring. A simpler (without the templates) example is below:
int (&a())[2];
What exactly is that declaring? What is the role of the &? To add to my confusion, if I declare the following instead
int b()[2];
I get an error about declaring a function that returns an array, while the first line has no such error (therefore, one would think the first declaration is not a function). However, if I try to assign a
a = a;
I get an error saying I'm attempting to assign the function a... so now it is a function. What exactly is this thing?
There's these awesome programs called cdecl and c++decl. They're very helpful for figuring out complicated declarations, especially for the byzantine forms that C and C++ use for function pointers.
tyler#kusari ~ $ c++decl
Type `help' or `?' for help
c++decl> explain int (&a())[2]
declare a as function returning reference to array 2 of int
c++decl> explain int b()[2]
declare b as function returning array 2 of int
a returns a reference, b does not.
For future reference, you may find this link helpful when you have a particularly difficult C/C++ declaration to decipher:
How To Read C Declarations
For completeness, I will repeat what others have said to directly answer your question.
int (&a())[2];
...declares a to be a zero-argument function which returns a reference to an integer array of size 2. (Read the basic rules on the link above to have a clear understanding of how I came up with that.)
int b()[2];
...declares b to be a zero-argument function which returns an integer array of size two.
Hope this helps.
int (&a())[2];
It declares a symbol a that is a function that takes no arguments and returns a reference to a two-element array of integers.
int b()[2];
This declares a symbol b that is a function that takes no arguments and returns a two-element array of integers... this is impossible by the design of the language.
It is relatively simple: get an operator precedence chart, start the symbol name (a) and start applying the operators as you see from their precedence. Write down after each operation applied.