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.
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!
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.
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.
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.
I repeat for the xth time that I absolutely KNOW what a pointer is and I also KNOW the terminology,
Repeating it doesn't help when your explanations indicate the opposite.
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.
A pointer is not uint or whatever. A pointer is a pointer Also, you haven't even shown your definition of uint yet. It could be anything.
A pointer is not an integer value to the compiler. It cannot and must not be, since it needs to be treated completely differently than an integer value in operations. Going back to my earlier example (did you read it, and think about it?):
unsigned int some_int = 0; unsigned int *some_ptr = 0; some_int++; some_ptr++;
What is the value of the two variables after the increment ? The correct answer should also explain why pointers and unsigned integers are two completely different data types. That they take up the same amount of memory is pure coincidence.
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.
What is a "normal variable" ? An integer is "normal", and not a pointer, and you are trying to convert it implicitly to a pointer. The compiler is totally correct when it issues a warning. If you want the warning to disappear, you need to do an explicit type cast (which you said you did, but you did not post the code you used, so we cannot tell if it is correct).
You need to grab a C textbook and read up on 1) pointers and b) operators that work on pointers, especially the "member reference from pointer" operator "->".
some_struct_ptr->some_int_member
evaluates to the value of some_int_member, not to its address. If you want the address, you need to use
&some_struct_ptr->some_int_member
And for the x:th time: you can not absolutely know about pointers and how it is used if you think that the address of a pointer is less important (or more or less irrelevant) than the address of the variable it points to.
Basically, you are ignoring arithmetic on pointers, which is something that is very commonly done.
When you do choose to see a pointer's address, a pointer's value and the value it points to as three separate things, your life will be a lot easier.
char buf[100]; char *p = buf+1; char **pp = &p; *buf = 'H'; *p++ = 'e'; p[0] = 'l'; p[1] = 'l'; buf[4] = 'o'; *(buf+5) = '\0'; *pp = strchr(buf,'\0'); printf("Number of characters in string is %ld\n",(long)(p-buf));
So when reading this value I actually do the same as setting the pointer value directly to the address of the variable And for that, you should tell your friendly compiler that the number you have really is an address - the address of an object of the same type that your left-hand-object (the pointer) is expected to point at.
You are assuming that a pointer is a number and because of this, the compiler should always accept the assign of any number to any pointer without getting upset.
Note that pound and dollar are two currencies. The amount of money are always numeric. But you can't simply assign a dollar value to a pound purchase. If you try to do that, the shop will issue a format conversion warning. The conversion from one currency to a different currency requres a type cast. For currencies, this is normally done by checking the current sell price for dollar and the current buy price for pound. In the case of converting from one currency to another, there is a change to the numeric value. The same thing happens when performing an assign from an integer to a floating point number, or the reverse.
In some special cases, there is no actual conversion performed. But the cast should be there anyway. Not because a change is needed to the numeric value, but because a change is needed to the data type, i.e. to tell the compiler that the object on the left and the object on the right are both of the same data type (or at least of compatible data types) - even if a quick look says something else.
Adding a type cast when assigning a number to a pointer tells the compiler that the numeric source value is a proper value for storage in the pointer. In this case, that the number really is a pointer value and suitable for the pointer you want to assign to.
This is no different from the following:
int a = 5; unsigned b = 5; if (a == b) ...
A lot of compilers will notice that a is a signed integer, and b is an unsigned integer, and will issue a warning about a comparison between signed and unsigned data. The warning is there because there is a reasonable chance that the signed/unsigned mismatch is caused by an invalid assumption somewhere.
When you add a typecast, the compiler will check if it is technically possible to treat the right-hand value as compatible with the left-hand value. If you don't write an explicit typecast, the compiler will check it's available automagic data type rewrite rules (as specified by the standard) to see if it somehow (with or without data loss) can convert the right-hand value into something suitable for the left-hand value. However, unless the compiler is really, really happy about any conversion rules it knows about it will complain by issuing a warning. Some compilers have a larger set of warnings than others. But when they do warn, you should care, because that means you are living on the edge.
In your case, a proper cast will tell your compiler that you have turned off the power and removed the fuse before doing electrical work in your house. The cast will mute the warning since you are giving the compiler a written contract where you, the developer, dearly promises that the value in your uint really, really do represent a good pointer to a variable of the correct data type.
int a = 5; unsigned b = 5; a = b;
Could generate a warning, because some legal unsigned values won't fit in a signed int
Similarly,
int a = 5; unsigned b = 5; b = a;
Could generate a warning, because some legal signed values won't fit into an unsigned int.
In both cases, you can tell the compiler that it's not an accidental oversight by giving an explicit cast.
Exactly the same applies when assigning a non-pointer to a pointer, as has already been explained.
"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."
Yes correct: The pointer is ptr_xdata; The non-pointer is sPtr_obj->Adresse.
"Since ptr_xdata is declared as a pointer, it will still be a pointer after its value has changed."
Correct - but the warning is not about that: Read it again, the warning is about converting a non-pointer to pointer - that is, the thing being converted is the non-pointer
ptr_xdata = (sPtr_obj->Adresse); ^ ^ | | | | This is a pointer This is not a pointer (it's a uint)
So the above assignment is from a non-pointer (on the right-hand side) to a pointer (on the left-hand side)
And so the warning is correct:
"converting [from] non-pointer to pointer"
You don't seem to get it. I don't understand it, because it's confusing.
In this case, it is actually the reverse. It is confusing because you don't understand it.
A number of million students have thought that pointers are confusing. However, there are quite a lot of programmers in this world that do know about pointers now. And they can testify that as soon as they really did understand pointers, they no longer see them as confusing.
It's just a question of getting the correct internal picture of the pointer concept. Before you have a working mental picture, they will be confusing. With the picture in place, you will think: gosh, that was easy.
The reason so many programs fails because of pointer errors, is not because pointers are hard to grasp but because they are powerful and a more powerful tool is inherently more dangerous. People makes mistakes. When they make mistakes with pointers, bad things happens. So a developer should always be extra careful when working with pointers.
Right now, we are trying to get you to change your internal view of pointers. A number of us believe that the terminology really is crucial to this task. If you see a pointer as a variable, with the normal properties of a variable (location, value, size, ...) then we believe you have a better chance to be able to switch your mental image of pointers, and be able to figure out their full potential. And their full potential really do require you to think about the address of the pointer, i.e. where it is stored. You really do have to understand why several steps of indirection is needed, or why programs must be able to modify the value of the pointer, and not just the value it points to.
I repeat for the xth time that I absolutely KNOW
And everybody else has to keep repeating, that you don't know anywhere as much as you think.
You're contradicting yourself all the time. One moment you claim you understand pointers perfectly, the next you admit you're confused by them.
You've stated painfully obvious nonsense like this, several times by now
this value is an uint value and I put this value to another pointer, also of type uint,
It shouldn't need saying yet another time, but apparently it does: No, that pointer is not of type uint. It's of type pointer-to-uint. A uint is not a pointer any more than a car is a driver's license. Staying in that image, you got a warning because you tried to put the car in your inside pocket.
Everybody, including you, is allowed to be wrong once in a while. But that doesn't mean we'll seeingly tolerate your insisting on staying a confused fool forever.
"One one man stands against the world, the world is often right."
To the insane, the rest of the world is crazy and they are the ones who are sane.
"the struct [element?] was formerly initialised with an address of a variable"
I presume you're talking about sPtr_obj->Adresse - yes?
If this is supposed to hold the address of a variable, why was it not properly defined as a pointer in the first place?!
The only reason why the language allows conversion from integers to pointers is for systems programming (embedded or when you write hardware drivers), i.e. when you need to access a hardware device at a specific location or when you need to locate variables dynamically at absolute locations.
Taking the address of a variable and convert to an integer and later converting the integer back to a pointer is something a program should hardly ever need. Most required address manipulation can be performed on the pointer. If you implement a memory manager, you might like to play with / and % (or more probably with >> and &) to figure out memory page and offset within memory page.