I am trying to port a code base from iar to avr-gcc. Amongst other things that have to replaced, the iar eeprom memory attribute __eeprom has to replaced with a avr-gcc friendly attribute. AFAIK the replacement for that is EEMEM, but the usage differs and I am not able to figure out how to replace __eeprom in the cleanest manner.
../src/myfunc.h:35:46: error: section attribute not allowed for 'src'
UBYTE *strcpye(UBYTE *dest, UBYTE EEMEM *src);
This error is not limited to pointers, but to all variables in general. IMO the usage of EEMEM is correct, where am I going wrong?
In the avr-gcc toolchain, avr-libc defines macro EEMEM in avr/eeprom.h:
#define EEMEM __attribute__((section(".eeprom")))
This means it's just an attribute that determines the section in which an object with this attribute will be located. In particular, EEMEM only makes sense for variables in static storage. Moreover, accesses to objects located in EEMEM have to be done by hand using functions / macros supplied by avr/eeprom.h like
void eeprom_read_block (void *dst, const void *src, size_t n);
void eeprom_write_byte (uint8_t *p, uint8_t value);
void eeprom_update_word (uint16_t *p, uint16_t value);
etc. Also notice that EEMEM is just an attribute and not a qualifier (like __flash for example). This means that even though you can tag a pointer (target) using attributes, that won't change the access in any way. To be more specific, any access through a pointer that's attributed EEMEM will be to RAM and not to eeprom.
In your case, the prototype of strcpye would read
char* strcpye (char *dest, const char *src);
and the implementation of that function would apply eeprom_read_byte on src++ and write to dest++ until it reads a terminal \0. Notice that you might need an explicit pointer cast as eeprom_read_byte expects [const] uint8_t*, and that char, signed char and unsigned char are 3 distinct types in C.
Using C++11, g++ (GCC) 4.4.7 20120313 (Red Hat 4.4.7-18).
Lets pretend I have a templated function (pardon my terminology if it isn't quite right).
I want to perform a "general" algorithm based on what was supposed to be compile-time instances of "field". Where the only things that really changed are these constants which I moved into trait classes (only added one here but imagine there are more). Originally I was declaring it as
constexpr field FIELD1{1};
However in C++11, non-type template params need to have external linkage (unlike C++14 which can have internal and external linkage?). So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr and it seems that losing that constexpr constructor this field is no longer a valid constant expression to qualify as a non-type template param.
Any suggestions if there is some way I can get around this? Open to a new method of doing things. Below is a simplified (incomplete, and non-compiling version to get the gist of the organization).
So the error I am seeing is along the lines of
error: the value of ‘FIELD1’ is not usable in a constant expression
note: ‘FIELD1’ was not declared ‘constexpr’
extern const field FIELD1;
Not quite sure what could be a best alternative.
I can get rid of the second error by removing the constexpr from the constructor. But then I don't know how to approach the constant expression issue.
field.H
struct field
{
int thingone;
constexpr field(int i):thingone(i){}
};
extern const field FIELD1;
field.C
#include "field.H"
const field FIELD1{0};
field_traits.H
#include "field.H"
template< const field& T >
class fieldTraits;
template< >
class fieldTraits<FIELD1>
{
public:
// Let's say I have common field names
// with different constants that I want to plug
// into the "function_name" algorithm
static constexpr size_t field_val = 1;
};
function.H
#include "field.H"
template< const field& T, typename TT = fieldTraits<T> >
void function_name()
{
// Let's pretend I'm doing something useful with that data
std::cout << T.thingone << std::endl;
std::cout << TT::field_val << std::endl;
}
So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr [...]
Per my comment, you can. It wouldn't work for you, but it's a step that helps in coming up with something that would work:
extern constexpr int i = 10;
This is perfectly valid, gives i external linkage, and makes i usable in constant expressions.
But it doesn't allow multiple definitions, so it can't work in a header file which is included in multiple translation units.
Ordinarily, the way around that is with inline:
extern inline constexpr int i = 10;
But variables cannot be declared inline in C++11.
Except... when they don't need to be declared inline because the effect has already been achieved implicitly:
struct S {
static constexpr int i = 10;
};
Now, S::i has external linkage and is usable in constant expressions!
You may not even need to define your own class for this, depending on the constant's type: consider std::integral_constant. You can write
using i = std::integral_constant<int, 10>;
and now i::value will do exactly what you want.
Ok, muddling though Stack on the particulars about void*, books like The C Programming Language (K&R) and The C++ Programming Language (Stroustrup). What have I learned? That void* is a generic pointer with no type inferred. It requires a cast to any defined type and printing void* just yields the address.
What else do I know? void* can't be dereferenced and thus far remains the one item in C/C++ from which I have discovered much written about but little understanding imparted.
I understand that it must be cast such as *(char*)void* but what makes no sense to me for a generic pointer is that I must somehow already know what type I need in order to grab a value. I'm a Java programmer; I understand generic types but this is something I struggle with.
So I wrote some code
typedef struct node
{
void* data;
node* link;
}Node;
typedef struct list
{
Node* head;
}List;
Node* add_new(void* data, Node* link);
void show(Node* head);
Node* add_new(void* data, Node* link)
{
Node* newNode = new Node();
newNode->data = data;
newNode->link = link;
return newNode;
}
void show(Node* head)
{
while (head != nullptr)
{
std::cout << head->data;
head = head->link;
}
}
int main()
{
List list;
list.head = nullptr;
list.head = add_new("My Name", list.head);
list.head = add_new("Your Name", list.head);
list.head = add_new("Our Name", list.head);
show(list.head);
fgetc(stdin);
return 0;
}
I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out? This implies I already need to know the type, and this reveals nothing about the generic nature of void* while I follow what is here although still no understanding.
Why am I expecting void* to cooperate and the compiler to automatically cast out the type that is hidden internally in some register on the heap or stack?
I'll handle the memory deallocation later. Assuming I have no understanding of the type stored in void*, how do I get the value out?
You can't. You must know the valid types that the pointer can be cast to before you can dereference it.
Here are couple of options for using a generic type:
If you are able to use a C++17 compiler, you may use std::any.
If you are able to use the boost libraries, you may use boost::any.
Unlike Java, you are working with memory pointers in C/C++. There is no encapsulation whatsoever. The void * type means the variable is an address in memory. Anything can be stored there. With a type like int * you tell the compiler what you are referring to. Besides the compiler knows the size of the type (say 4 bytes for int) and the address will be a multiple of 4 in that case (granularity/memory alignment). On top, if you give the compiler the type it will perform consistency checks at compilation time. Not after. This is not happening with void *.
In a nutshell, you are working bare metal. The types are compiler directives and do not hold runtime information. Nor does it track the objects you are dynamically creating. It is merely a segment in memory that is allocated where you can eventually store anything.
The main reason to use void* is that different things may be pointed at. Thus, I may pass in an int* or Node* or anything else. But unless you know either the type or the length, you can't do anything with it.
But if you know the length, you can handle the memory pointed at without knowing the type. Casting it as a char* is used because it is a single byte, so if I have a void* and a number of bytes, I can copy the memory somewhere else, or zero it out.
Additionally, if it is a pointer to a class, but you don't know if it is a parent or inherited class, you may be able to assume one and find out a flag inside the data which tells you which one. But no matter what, when you want to do much beyond passing it to another function, you need to cast it as something. char* is just the easiest single byte value to use.
Your confusion derived from habit to deal with Java programs. Java code is set of instruction for a virtual machine, where function of RAM is given to a sort of database, which stores name, type, size and data of each object. Programming language you're learning now is meant to be compiled into instruction for CPU, with same organization of memory as underlying OS have. Existing model used by C and C++ languages is some abstract built on top of most of popular OSes in way that code would work effectively after being compiled for that platform and OS. Naturally that organization doesn't involve string data about type, except for famous RTTI in C++.
For your case RTTI cannot be used directly, unless you would create a wrapper around your naked pointer, which would store the data.
In fact C++ library contains a vast collection of container class templates that are useable and portable, if they are defined by ISO standard. 3/4 of standard is just description of library often referred as STL. Use of them is preferable over working with naked pointers, unless you mean to create own container for some reason. For particular task only C++17 standard offered std::any class, previously present in boost library. Naturally, it is possible to reimplement it, or, in some cases, to replace by std::variant.
Assuming I have no understanding of the type stored in void*, how do I get the value out
You don't.
What you can do is record the type stored in the void*.
In c, void* is used to pass around a binary chunk of data that points at something through one layer of abstraction, and recieve it at the other end, casting it back to the type that the code knows it will be passed.
void do_callback( void(*pfun)(void*), void* pdata ) {
pfun(pdata);
}
void print_int( void* pint ) {
printf( "%d", *(int*)pint );
}
int main() {
int x = 7;
do_callback( print_int, &x );
}
here, we forget thet ype of &x, pass it through do_callback.
It is later passed to code inside do_callback or elsewhere that knows that the void* is actually an int*. So it casts it back and uses it as an int.
The void* and the consumer void(*)(void*) are coupled. The above code is "provably correct", but the proof does not lie in the type system; instead, it depends on the fact we only use that void* in a context that knows it is an int*.
In C++ you can use void* similarly. But you can also get fancy.
Suppose you want a pointer to anything printable. Something is printable if it can be << to a std::ostream.
struct printable {
void const* ptr = 0;
void(*print_f)(std::ostream&, void const*) = 0;
printable() {}
printable(printable&&)=default;
printable(printable const&)=default;
printable& operator=(printable&&)=default;
printable& operator=(printable const&)=default;
template<class T,std::size_t N>
printable( T(&t)[N] ):
ptr( t ),
print_f( []( std::ostream& os, void const* pt) {
T* ptr = (T*)pt;
for (std::size_t i = 0; i < N; ++i)
os << ptr[i];
})
{}
template<std::size_t N>
printable( char(&t)[N] ):
ptr( t ),
print_f( []( std::ostream& os, void const* pt) {
os << (char const*)pt;
})
{}
template<class T,
std::enable_if_t<!std::is_same<std::decay_t<T>, printable>{}, int> =0
>
printable( T&& t ):
ptr( std::addressof(t) ),
print_f( []( std::ostream& os, void const* pt) {
os << *(std::remove_reference_t<T>*)pt;
})
{}
friend
std::ostream& operator<<( std::ostream& os, printable self ) {
self.print_f( os, self.ptr );
return os;
}
explicit operator bool()const{ return print_f; }
};
what I just did is a technique called "type erasure" in C++ (vaguely similar to Java type erasure).
void send_to_log( printable p ) {
std::cerr << p;
}
Live example.
Here we created an ad-hoc "virtual" interface to the concept of printing on a type.
The type need not support any actual interface (no binary layout requirements), it just has to support a certain syntax.
We create our own virtual dispatch table system for an arbitrary type.
This is used in the C++ standard library. In c++11 there is std::function<Signature>, and in c++17 there is std::any.
std::any is void* that knows how to destroy and copy its contents, and if you know the type you can cast it back to the original type. You can also query it and ask it if it a specific type.
Mixing std::any with the above type-erasure techinque lets you create regular types (that behave like values, not references) with arbitrary duck-typed interfaces.
I'm reading through this memory management code overloading operator new. there's expression something like
typedef char *b
and later in the code b was used like this:
b(h); //h is a pointer to some class;
h defined here:
static Head* h= (Head*) HEAP_BASE_ADDRESS;
I'm assuming when b is used it is considered a pointer to a char. But how can a pointer have expressions like b()?? Is there some sort of conversion going on in here? Can I understand it as b now is having the same address as h?
The first code line you posted is a typedef which creates an alias for char* as b. The second code line shows a functional-style type conversion from h to b.
Can I understand it as b now is having the same address as h?
The b is just an alias of char*, so b(h) eventually does nothing unless you store the result of that expresion like:
b b_ptr = b(h); // equivalent to: char* b_ptr = ((char*)h);
The functional-style type conversion works only with single-word type names, so if you want to use this conversion style to e.g. a pointer, you have to typedef it first. (This is the reason of the typedef char *b.) This style of conversion can be used for expressions like int(3.14 + 6.67).
In Linux kernel, structured types are defined like that:
typedef struct _TAG_ { ... };
and then used in routines like that:
struct _TAG_ structured_entity;
struct _TAG_ *pointer_to_structured_entity;
void function(struct _TAG_ *arg, ...);
Why not so:
typedef struct _TAG_ { ... } _typename_;
and then:
_typename_ structured_entity;
_typename_ *pointer_to_structured_entity;
void function(_typename_ *arg, ...);
What is the technical necessity to do so? Or is this simply tradition / style / magic?
The Linux kernel coding style - at kernel.org and in the Documentation directory that comes with the kernel - discourages using typedefs for structs:
Chapter 5: Typedefs
Please don't use things like "vps_t".
It's a _mistake_ to use typedef for structures and pointers. When you see a
vps_t a;
in the source, what does it mean?
In contrast, if it says
struct virtual_container *a;
you can actually tell what "a" is.
The document goes on to list cases where the author believes typedefs are useful - such as an opaque object that can only be accessed using accessor functions - concluding with:
In general, a pointer, or a struct that has elements that can reasonably
be directly accessed should _never_ be a typedef.