Hello!
I have uVision that compiles fine with the C51 v7.03 compiler and the related package, but not complete with the 8.06. I used two different Keil installations. All files are in the same folder.
In the 8.06 I get linker errors like "object does not fit in to pdata page" and "0080H". This looks like the compiler was thinking the PDATA was only 128 bytes, but it is set to 256 bytes in the startup.a51. Any clue what's different in the newer Keil package?
Also there is a warning in 8.06 (which does not show in 7.03) "converting non-pointer to pointer" on this
ptr_xdata = sPtr_obj->Adresse;
while the vars are set like this:
uchar uc_set_obj( uchar pdata *ptr_Set) { uchar i; uchar xdata *ptr_xdata; struct stOBJADR code *sPtr_obj; sPtr_obj=&Obj[*ptr_Set]; . . . ptr_xdata = sPtr_obj->Adresse; }
The struct stOBJADR has a member "uint Adresse;"
I can see no wrong use of the pointers. I just want to be sure that the warning does not affect the code to not work correctly.
As you say, ptr_xdata is a pointer, and sPtr_obj->Adresse isn't - so the warning is perfectly correct!
As usual, an explicit cast should both stop the warning and make your intention clear...
"I just want to be sure that the warning does not affect the code to not work correctly"
There is always a risk in playing fast-and-loose with pointers like this!
I think you should be OK with the XDATA-Specific pointer...
Hello Andy! Thanks for your help.
You are wrong. "sPtr_obj->Adresse" is a struct pointer and gives me a number, an address that "ptr_xdata" is set to.
char buf[1000]; char* p = buf; Now you think that the address of p is a base address. Why? The address of p has nothing to do with buf, and isn't the base of anything meaningful. This is what I consider as pointer address.
But we better stop here. I'm programming since a few years now and never had such problems. I think I don't need to read a bible about pointers, just to write some code. That's not the meaning of a computer language. To be more complicated than its inventor.
My conclusion is: if I set a pointer address in XDATA with a number I read from a struct and as long as this number is 2 bytes long and thus valid, then the pointer has to be set. Period. By assigning a pointer's address I'm not doing a pointer conversion, because the pointer only changes the byte it points to. Nothing more, nothing less.
This is what I consider as pointer address.
We know that by now. And it's still wrong, because this use of the term is guaranteed to confuse everyone exposed to it --- apparently including it's original user, i.e. you.
long as this number is 2 bytes long and thus valid,
That "thus" exposes that you still haven't understood what the issue actually is. No, a 2-byte number is not self-evidently "valid" to hold a pointer. You need 3 bytes to hold an arbitrary pointer in C51.
To put it bluntly: you're deluding yourself. You don't understand pointers anywhere as well as you think you do. Your insistence on ill-fitting terminology may well be part of the problem, preventing true understanding.
then the pointer has to be set.
What exactly made you think it wasn't? You say you've been programming for years, but you still don't recognize the difference between an error and a warning?
"I think I don't need to read a bible about pointers"
The way you write about them suggests otherwise.
"as long as this number is 2 bytes long and thus valid"
That is a non-sequitur; ie, the conclusion does not follow from the reason stated.
A 2-byte number might happen to be valid as an XDATA pointer, or it might not - the fact of its size being 2 bytes has absolutely no relation whatsoever to its validity (or otherwise) as a pointer!
"then the pointer has to be set"
There never was any doubt that the pointer would be set - the question is whether it gets set to anything meaningful...
"By assigning a pointer's address I'm not doing a pointer conversion"
By assinging from an int to a pointer you are doing an implicit conversion!
"the pointer only changes the byte it points to"
Eh??! A pointer of itself changes nothing; only when you dereference the pointer does the pointed-to location get accessed.
You still haven't shown the cast that you used and that still gave the warning. If you want to get back to that, then do it...
"That's not the meaning of a computer language. To be more complicated than its inventor."
It isn't complicated; it is really quite simple. You are the one that is complicating it by messing-up the terminology!
Every program "object" - variables and functions - has an address, and that address is the location at which the object itself is stored. Simple. No exceptions.
This is the terminology used by the inventors of the language - Messrs Kernighan & Ritchie.
You are the one who is complicating this by using "address of" when applied to a pointer to mean something different from "address of" when applied to any other object!
Because you have thus modified the meaning of "address of" when applied to a pointer, you then also have to invent another term to use when you do want to refer to the location of the pointer itself!
All this is further complicated by the fact that the non-standard terms you have chosen already have well-established and quite different meanings in the more general context!
Note that all variables are not guaranteed to have a memory address - they could be register-based too. On one hand, that regsiter could be seen as an address, but on the other hand, the compiler could move the variable around if it feels like it. For some processors with few GP registers and slow memory access (at least old 16-bit x86 processors) there was some advantages to switching registers.
Seems like we talking past each other.
In other words: I know what a pointer is, but for me the only thing of interest is what it does. That's what I was talking about. An example:
char *ptr; char a; ptr = 7;
is the same principle as
ptr = &a;
where the address of 'a' is not determined. I'm setting the pointer to point to an address, whether this is a number (an absolute address) or a variable. This is all what is done in the code above. Where is the conversion? I can't see it. Because warning C289 says "converting non-pointer to pointer" in v8.06. Actually I'm setting a pointer to an address of a variable.
1. A struct is declared. 2. The struct has a member 'Adresse', uint type. 3. This member is set to the address of a variable, let's say &test. 4. A struct pointer is set to this struct. 5. A char pointer is declared. 6. The struct pointer is set to point to the member 'Adresse'. 7. The value of 'Adresse' is read out and used to set the char pointer to point to 'test'.
Totally OK for me. Where's the problem?
Where is the conversion? I can't see it.
You cannot see it, but the compiler can. Even though C is a fairly weakly-typed language, an integer and a pointer are not the same for the compiler, and need to be treated differently in a number of situations.
Consider the following code snippet:
unsigned int some_int = 0; unsigned int *some_ptr = 0; some_int++; some_ptr++;
Can you tell what the value of some_int will be after the snippet ? (that's trivial, right ?)
Can you tell what the value of some_ptr will be at the end of the code snippet ? (If yes, please state what it will be and why. If no, please state what information you are missing.)
And that is just one example.
Since C is fairly weakly-typed, the compiler might allow a number of so-called implicit type casts. More high-level languages are usually more strongly-typed, and would require an explicit cast in many situations.
Now back to your warning - can you post the line with the explicity type cast you use that still results in a warning ?
The & in 'C' is the "address-of" operator; the result of the expression
&a
gives the "address-of" a - so its type is self-evidently "pointer to a"
So, when you write
p = &a
the compiler can clearly see that you are assigning a pointer-type expression (&a) to a pointer-type variable (p)
However, when you write
p = 7
it is not self-evident that 7 is intended to be a valid address; more particularly, when you write
p = a
there must be some doubt that you might have meant
hence the compiler gives the warning.
To avoid the warning, you can make you intention clear by using an explicit cast:
p = (char*)a
If you're still getting the warning, you must have done it wrong. If you want an explanation of what you've done wrong, you will have to show what you have done!
The problem is that it is not self-evidently correct.
The compiler has noticed the possibility that what you wrote was not what you intended - and so it gives you a Warning.
The Warning is particularly justified as this is such a common mistake - like writing "=" instead of "==" in an 'if' clause...
One more time: you can stop the warning by using an appropriate explicit type-cast.
You are the one who is complicating this by using "address of" when applied to a pointer to mean something different from "address of" when applied to any other object!> I never wrote this. English is not my native language, but I guess I'm not totally wrong when understanding that a pointer address is not the same as the address of a pointer. The address of a pointer is the pointer itself. So how do you call the address the pointer points to in your terminology? I just call it pointer address. Still because I don't need to care where (at which address) the pointer itself is located.
To avoid the warning, you can make you intention clear by using an explicit cast: p = (char*)a
Now that makes sense to me.
"The address of a pointer is the pointer itself."
No: The address of the pointer is the location of the pointer itself.
"So how do you call the address the pointer points to in your terminology?"
That is the value of the pointer.
"I just call it pointer address"
Don't call it that - it's just confusing!
"Still because I don't need to care where (at which address) the pointer itself is located"
You would care if you needed a pointer to the pointer...
The address of the pointer (or variable) is the same as the pointer's (or variable's) address, i.e. the location where the pointer (or variable) is stored.
The value of the pointer is an address - the address that the pointer points to.
"The address of a pointer is the pointer itself." should be rewritten as: "The address of a pointer is the location of the pointer itself".
The address the pointer points to? That's the value of the pointer. Just as the value of variable my_int below is 5.
int my_int = 5;
Variables stores values. It doesn't matter what data type the variable has (pointer, pointer, struct, ...), it still stores one or more values.
As soon as you write more complex programs, you regularly do have to care about the address of the pointer, i.e. where the pointer is stored. Or, more specifically, your application will need the address of your pointer, so that it will be able to indirectly (through a pointer) modify the address stored in the pointer (the pointer's value).
Look at strtol(const char* str,char **endptr,int base) which takes a pointer to a pointer as second parameter.
char input_string[] = "12345broken chars"; char *end; long val; val = strtol(input_string,&end,10); if (*end != '\0') printf("Invalid number string\n"); else printf("The value was %ld\n",val);
Now the pointer end (not what it points to) will be modified to point at the first non-valid character in the input string.
This is a simple case where you use the address of a pointer to change the pointer instead of changing what it points to. There are many more - many of them way more complex - that are commonly used in standard C/C++ programs. Some RTOS may have half their API centering around pointers to pointers... Array manipulation functions are a different area where multiple levels of indirection is often needed.
Yes, you did: in your post of 5-Feb-2008 08:25, you wrote,
"A pointer's address for me is the address it points to."
For any other variable, "the variable's address" means the address of the variable - but you said you're using "the pointer's address" to mean the value of the pointer.
I did not!
I can only repeat, that for me a pointer's address is NOT the same as the address of a pointer. Address is defined as a target, at least in my language. So a pointer points to a target (address) and this is the pointer's address (or target). The address of the pointer, where it is located in RAM, should be pronounced as the location of the pointer.
A pointer's value for me is the value it points to. Like the address (number) in the register DPTR is the address it points to. Anything else does not make sense, since no one cares for the pointer itself, except for its data type, even if it is a pointer to a pointer.
You all may call me unteachable, but this is my way of understanding this confusing topic. Confusing because when you hear the first time about pointers (some years ago in my case), people like to confuse with the fact that a pointer points to a certain address/value, but also has an address/value, as you call it. Whereas I don't see any sense in concerning the address of the pointer itself. It just confuses...
"I did not!"
You most certainly did. The quote came from your post. It's right up there in your post that Andy referenced, recorded for history and for all to read.
"A pointer's value for me is the value it points to."
A pointer's value is an address. A dereferenced pointer yields the value addressed by the pointer.
"a pointer's address is NOT the same as the address of a pointer."
OK, maybe that's just a problem with English not being your first language - but, in English, the phrase "a pointer's address" has exactly the same meaning as the phrase "the address of a pointer"
That's nothing to do with 'C' or programming - that's the way English works.
Just as "a dog's breakfast" has the same meaning as "the breakfast of a dog"
Or "a variable's address" has the same meaning as "the address of a variable"
Or "a variable's value" has the same meaning as "the value of a variable"
Or "a pointer's value" has the same meaning as "the value of a pointer"
In general, "a thing's property" has the same meaning as "the property of a thing"
"Address is defined as a target, at least in my language"
Again, not in English.
In English, an "address" identifies a location - whether it's a postal address, that identifies the location of a house, or a memory address, that identifies the location of a particular cell within the memory array.
"So a pointer points to a target (address)"
Correct
"and this is the pointer's address (or target)."
Wrong! That doesn't even make sense by your own reasoning!
If the pointer points to any address, then that address cannot be the pointer's own address - unless the pointer is pointing to itself!
"The address of the pointer, where it is located in RAM, should be pronounced as the location of the pointer."
Correct - and "the location of the pointer" is synonymous with "the address of the pointer" which is also synonymous with "the pointer's address" which is also synonymous with "the pointer's location"
"Whereas I don't see any sense in concerning the address of the pointer itself."
Just because you don't see a need for it doesn't mean that there is no need for it!
It is also no excuse for allowing your terminology to become misleading.
You have already been given several examples of where there is a need to be concerned with it
You all may call me unteachable, but this is my way of understanding this confusing topic.
You got that backwards. It's exactly your way of thinking that renders this topic so confusing for you. It's getting in the way of your understanding it.
Your private terminology is based on firmly held, yet incorrect beliefs. Mismatch of thought with reality is the consequence. That's pretty much the definition of "confusion".
The address of an ordinary variable is generally every bit as important as its value. The same holds for pointer variables: their address is as important as their value. Pretending otherwise will bite you in the lower back soon enough.
"You all may call me unteachable"
OK then: you are unteachable; that is, in English, at least.
Maybe it'd be better if you could find someone to explain this to you in your own language?
Just think of a pointer to a pointer!
char variable; char *pointer_to_variable; char **pointer_to_pointer_to_variable;
"It just confuses"
Only because you still don't understand it, and still insist on using confusing terminology!
Or "a pointer's value" has the same meaning as "the value of a pointer
Afaik, a pointer points to a value, but it doesn't have one.
You sure? A data packet in the internet is sent to an address, the IP. Network hardware has a MAC address, where data is addressed to. If you send a letter, you send it to an address. Perhaps I should have written destination instead of target. It's sometimes not easy to distinguish between multiple words in English (target, aim, goal, destination) that have only one meaning in German (Ziel).
But still, if a pointer points to a pointer, I don't need to care about both pointer locations. This is C, so I don't care about memory addresses. This is a matter of opinion, but I don't find it very elegant to put a pointer to a pointer...
I'm also not the super programmer like you folks seem to be. I rather do many things in my life than just learning a computer language down to its deepest meaning. What counts is that it has to work. And it does! Even without non-sense warnings of a picky compiler.
You don't seem to get it. I don't understand it, because it's confusing. I can understand anything that is logical for me. Electronics, current, atoms, physics, biology down to molecules, bits and bytes. I can explain you, down to the level of an electron, how a computer works, but this is not logical to me. My terminology helps me to understand it. It's enough to know what a pointer is and does, no need to understand it.
You plainly don't understand it, and your bad terminology may well be a contributing factor to that, if not just a symptom of it.
It's not complicated.
Keep it simple and just use the same terminology as the rest of the world.
"It's enough to know what a pointer is and does, no need to understand it."
If you know what it is and does, then you really should be capable of understanding it!
Afaik, a pointer points to a value, but it doesn't have one.<p>
Well, I'm afraid your knowledge is wrong, then.
A pointer, like any other variable, has an address (where it is stored), and a value (which can be used to access a certain location in memory). A pointer does not need to point to a value.
Example:
void * some_void_ptr
some_void_ptr is a pointer. At least as far as the compiler is concerned, it does not point to a value. Hence, the compiler cannot, for example, increment it:
some_void_ptr++;
will result in an error. It also cannot dereference it:
*some_void_ptr = 0;
will also result in an error.
However, you can take the address of this pointer, for example:
void ** some_ptr_to_void_ptr; some_ptr_to_void_ptr = &some_voidptr;
This is C, so I don't care about memory addresses.
C is still way too close to the actual hardware to ignore the target platform. The C compiler will not save you from alignment faults and access penalties that stem from "not caring about memory addresses".
I rather do many things in my life than just learning a computer language down to its deepest meaning. What counts is that it has to work. And it does! Even without non-sense warnings of a picky compiler.
You are not the only one with that attitude round here.
That attitude is fine for a hobby programmer, but unfortunately is the reason why the world is full of unmaintainable, unreliable software.
Don't you mean: As Far As You Think You Know?
Pointer-to-pointer inellegant? Did you look at my strtol() example? Have you ever used strtol()?
Please suggest a more elegamnt method of knowing the position of the first invalid character in the input string.
In the general case, a function that needs to send back multiple values need to do one of the following: - Return a struct - Take a pointer to a struct as parameter, and fill in the struct. - Take multiple pointers as parameters, and update the individual parameters. - Return one value and report the other values through one or more pointers sent as parameters.
If you insist that a pointer's value is the value of the variable it points to, what value would you say a NULL pointer has? And what type si that value? If the pointer was of type pointer-to-int, is it's value then 0 or (int)NULL or what?
This is not advanced programming. The distinction between a pointer's address and it's value is what you are expected to learn during the first year if studying at a university.
It is a pre-requisite to later working with red-black trees, hash-tables, abstract data types, multi-dimensional dynamically allocated arrays etc. A real-world application may be performing just as much manipulation of a pointer as it is dereferencing the pointer, i.e. accessing the value (possibly also a pointer) that the pointer points to.
Have you tried to use qsort() to sort an array of pointers to structures? The comparison function really have to care about the level of indirection needed.
On my unix machine, the \usr\include directory contains about 1000 ** (after stripping away ** used in comments). That gives an indication that quite a lot of developers have thought that a pointer-to-a-pointer is an important concept. All these data structures and function calls does not work unless the user do care about the pointer, instead of just thinking about what it points to.
Right now, you are arguing against yourself. You are saying that you are fine with your terminology. You are saying that the important thing is that your programs works. But you are saying that pointers are confusing. If you do spend time learning the correct terminology, you will be less confused. And if you are less confused, your programs will work (even) better.
One one man stands against the world, the world is often right. The alternative is that he is a genious, and have seen the light.
Since the world has not seen the light, and have chosen to implement a huge amount of API with multiple indirection, you have to learn to accept the concept, or you will have to invent your whole world around you.
You sure? A data packet in the internet is sent to an address, the IP. <p>
No, it is not. The packet is sent to a destination. There are _two_ addresses in a packet, a source and a destination.
Network hardware has a MAC address, where data is addressed to.
The MAC of a NIC is a property of the NIC itself. It only becomes a destination/target if it is put in the right field in the network packet.
If you send a letter, you send it to an address.
A house has an address, which describes its location. It becomes a destination/target when it is written on the right part of the envelope.
So, I think that's part of the problem. You're saying "address" when you mean "destination/target", while everyone else (and all available literature) defines "address" as "location/place".
I repeat for the xth time that I absolutely KNOW what a pointer is and I also KNOW the terminology, but this KNOWLEGDE does not help me understanding it. So I just bend the terminology and then I know two terminologies, of which only my one helps me out when using pointers and describing them. Instead of telling me a thousands time, you could also just leave me have my terminology and just try to understand it for the very moment you are thinking about the pointer warning problem I actually had and then turn back to your terminology. Now you could say "why don't you turn to our terminology for a moment?". Well, I could and I did. But it didn't help. My problem is, in your terminology: if I read a value out of a struct via a struct pointer and this value is an uint value and I put this value to another pointer, also of type uint, I simply expect this procedure to be correct. The warning message C289 says "converting non-pointer to pointer". In my understanding and I guess in everyone else too, this means that I was trying to convert something that's not a pointer (a normal variable, for exmple) to a pointer. Since ptr_xdata is declared as a pointer, it will still be a pointer after its value has changed. Because the struct was formerly initialised with an address of a variable and this address turns to an uint value when stored in the location of the struct. So when reading this value I actually do the same as setting the pointer value directly to the address of the variable.
"... another pointer, also of type uint"
No!
A pointer is not of type uint!
In this particular case, it might just happen to be the same size as a uint, but that is a pure coincidence - it does not make it the same type.
Just as the fact that some small dogs are the same size as cats does not make them cats!
A pointer may be of type pointer to uint - but it is not of type uint!
Pointers - and only pointers - have pointer type
Therefore, any thing that is not a pointer does not have pointer type - in other words, they have non-pointer type
Hence the warning is correct - you are assigning from a non-pointer (uint) to a pointer.
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