Printable View
This is from an OpenGL tutorial on NeHe's site.
It's commented as "Set The Pointer To NULL" but what does it exactly do? MSDN lists memset as "void *memset( void *dest, int c, size_t count );". I understand it as: memset fills the memory at the target of pointer <dest> for the size of <count> bytes with the value <c>.Code:memset(TextureImage, 0, sizeof(void *)*1);
That means <count> is sizeof(void *)*1 but what does THAT mean?
sizeof(void *) on a 32 bit system is 4
if they really meant to multiply it by 1 then you are setting 4 bytes of TextureImage to 0. Simple as that.
What exactly he was doing in this example baffles me! Wont a simple assignment to zero do exactly the same - but more clear and a little more efficiently???:D
I notice that he has abandoned his old ways of coding and has now put some efforts into building a C++ library for his tuts.....and that is better than the old stuff
*(unsigned long *)TextureImage = 0;
would have made more sense probably.
I was never big on NeHe's stuff for newbies, although its getting better and he does have some nice examples.
I like his OpenGL content.....lot's of nice examplesQuote:
Originally posted by RoD
I was never big on NeHe's stuff for newbies, although its getting better and he does have some nice examples.
Quote:
memset(TextureImage, 0, sizeof(void *)*1);
Yuck.
So, basically what you are saying is, you have no clue either, right ? :D
sizeof(void *)*1
Sizeof gets the size of something. Void means no return value. Is it a real data type? That would mean a void func would return something which is .. nothing ?
What does void * mean ? * Has something to do with pointers, but behind void ? Pointer of what ?
*1 means multiply with 1 ? Why would he do that ?
This C stuff is makeing me feel more and more dizzy...
>>Why would he do that ?
Have u tried asking him?
Yea he has some decent OpenGL content but i always found his tutorials a little hard to follow for newbies. Mine arent perfect in any way, but 90% of the emails i get are a combination of praise and coder-error, not RoD error. Not to say i havent messed up, i have lol.
I tried your tutorials, but they are not really a prime example of good portability by using "stdafx". DevC++ also produced a few warnings about non standard code inQuote:
Originally posted by RoD
>>Why would he do that ?
Have u tried asking him?
Yea he has some decent OpenGL content but i always found his tutorials a little hard to follow for newbies. Mine arent perfect in any way, but 90% of the emails i get are a combination of praise and coder-error, not RoD error. Not to say i havent messed up, i have lol.
. Basically, I didn't get past the 5th tutorial because of functions like SetTextColor etc.Code:int APIENTRY WinMain
Don't get me wrong, I'm sure your tutorials are great. It's just that I use a poor compiler, so they didn't help me. A skilled C++ coder could probably easily work around this, but then, a skilled coder wouldn't need a tutorial.
The tutorials I found on NeHe's site worked on DevC++ after I found out that I have to link libraries manually.
Anyway, as for that specific piece of code: If nobody understands it's meaning, it's probably just messy code. I'm fine with not understanding it as long as nobody else does :)
No, they told you, it sets the first 4 bytes pointed at by TextureImage to 0. It's not actually setting TextureImage itself to a NULL pointer; it's setting the memory pointed at to 0.Quote:
Originally posted by darksaidin
So, basically what you are saying is, you have no clue either, right ? :D
sizeof(void *)*1
Sizeof gets the size of something. Void means no return value. Is it a real data type? That would mean a void func would return something which is .. nothing ?
What does void * mean ? * Has something to do with pointers, but behind void ? Pointer of what ?
*1 means multiply with 1 ? Why would he do that ?
This C stuff is makeing me feel more and more dizzy...
People may not be sure what it's accomplishing, but it's certain what the command does.
a void * is a variable which stores the address of something. A void pointer can point to anything, be it an int, a struct, whatever. sizeof(void *) is the size of a generic pointer, and *1 just means multiplied by one. This syntax is common... say you wanted to set an array of 32 ints to zero, you could use:
memset(array,0,sizeof(int)*32);
On a 32 bit OS, the line is equivalent to:
memset(TextureImage,0,4)
although it would be better to write the whole command as:
* (int *) TextureImage = 0;
as was mentioned earlier.
Yea theres a few downsides to my tutorials. Mainly the speed at which i put them out (been rlly busy recently) and that i only use VC++ and that causes some problems for those who dont. Eventually i will solve this problem, but at the present time i dont have the time to do both.
Now, that makes a lot of sense.
My main problem with that piece of code was (void *). The * is listed behind "void", but thats obviously just the way it's done in C. It even makes sense when used in void *varname.
Thanks cat!
darksaidin you're confusing. you don't understand void *?
* after a type is pointer notation. It means that you have a pointer to that type. in the case of void * it basically means it can be a pointer to anything really. sizeof(void *) will return 4. sizeof(int *) will also return 4. It's just the size that a pointer on a 32 bit system is.
Wow RoD, you learn fast.
http://cboard.cprogramming.com/showt...threadid=28479
I already understood that, FillYourBrain.Quote:
Originally posted by FillYourBrain
darksaidin you're confusing. you don't understand void *?
* after a type is pointer notation. It means that you have a pointer to that type. in the case of void * it basically means it can be a pointer to anything really. sizeof(void *) will return 4. sizeof(int *) will also return 4. It's just the size that a pointer on a 32 bit system is.
My problem at first was that I thought the * was always in front of the variable (and not behind the type) because there is no such thing as a pointer to a type but only pointers that can be used like variables of specific type.
At least in OP. C obviously defines it the other way round:
You define a new variable (which will be a pointer) of a type that [meaning "the type"] can be used to create pointers that can only point on a variable of that given type. ... which is a strange way of looking at it, but accurate. Only that this way of looking at it allows to have the * behind the type (which otherwise would make no sense) and not in front of the variable.
If you never coded in Delphi you probably won't understand my problem here.
Anyway, my problem has been solved, thanks to Cat explanation.
Dude that was 9 months ago....Quote:
Originally posted by golfinguy4
Wow RoD, you learn fast.
http://cboard.cprogramming.com/showt...threadid=28479
To expand, pointers ARE typed. For example:Quote:
Originally posted by darksaidin
I already understood that, FillYourBrain.
My problem at first was that I thought the * was always in front of the variable (and not behind the type) because there is no such thing as a pointer to a type but only pointers that can be used like variables of specific type.
int * i;
declares i, and says "i holds an address, and at that address lives an int". Similarly,
MyClass * c;
declares c, and says "c holds an address, and at that address lives an object of class MyClass".
Of course, it is up to the programmer to make sure the pointer actually has the proper address. E.g. the programmer would have to make sure c held the address of a valid MyClass object before using c to do something with that object.
Now, void * was added to C, it wasn't an original feature. The goal was to create a fully generic pointer type, and "void" was already a keyword. Further, it is not legal to create an instance of type void (i.e. the line "int i;" is OK but "void v;" is a compiler error). Because they already had the word "void" reserved as a keyword, and a "void *" was meaningless at the time, they made "void *" mean a special kind of pointer.
void * v;
declares v, and says "v holds an address, and something lives at that address but I don't know what that something is".
In this way, a void * is the most generic kind of pointer. You can't do a lot with it, but you can convert any other pointer into a void * (automatically), and you can convert a void * to any other pointer (automatically in C, explicitly in C++).
Sure, pointers can have a type, that much was clear :)
I merely said that there are no pointers *to* a type (as said in one of the posts), meaning a pointer that points on a datatype.
If I may add, it would have been a lot easier to read in OP:
int *an_integer;
var an_integer: ^integer;
You can read it the way it is written: Variable an_integer is a pointer to an integer - they even used an arrow :) . Thats were my trouble originated from: Pointer to an integer. There is always a word behind the pointer symbol. void * lacks that - and right so, it's different in C because they used a "different logic". Both can make sense.
pointer to a type? That makes no sense as a type doesn't take up memory, therefore can't have an address.
C/C++ usually works best if you read it backwards.
E.g.:
int * myPointer;
"myPointer points to an int"
This is especially helpful when dealing with constants:
const int * p;
"p points to an int that is constant". It's equivalent to:
int const * p;
"p points to a constant int"
but is not the same as:
int * const p;
"p is a constant pointer to an int"
and there is also:
const int * const p; or int const * const p;
"p is a constant pointer to an int which is const"/ "p is a constant pointer to a constant int"
Get used to reading pointer stuff backwards, and you'll be fine :D
And "pointer to a type", as was said, isn't technically correct. But a lot of people use it; it's just short for "pointer to an instance of a type" which would be more correct. It's like saying "pointer to int" -- it really means a "pointer to an instance of type int"
Cat, that ROCKED!
-LC
That's what I was referring to. pointer that points to a datatype. Darksaidin seemed to be suggesting that OP (Object Pascal?) has pointers to datatypes themselves.Quote:
Originally posted by darksaidin
Sure, pointers can have a type, that much was clear :)
I merely said that there are no pointers *to* a type (as said in one of the posts), meaning a pointer that points on a datatype.
And how long has your first OpenGL tut been up there?Quote:
Originally posted by RoD
Dude that was 9 months ago....
Nope I wasn't. At least I don't think I was.Quote:
Originally posted by FillYourBrain
That's what I was referring to. pointer that points to a datatype. Darksaidin seemed to be suggesting that OP (Object Pascal?) has pointers to datatypes themselves.
(I would skip the next paragraph if I were you. I just had to mention it :) )
However, it's actually true. You can have pointers to certain types in Delphi: Classes ("object types"). Normally there isn't much use in pointing at them. You could do it in a dynamically linked DLL to import a class from the executable that linked the DLL, create new derived(w?) classes with that classpointer and than even "reimport" that new class back into the executable. It's a bit tricky and a few other conditions have to be met, i.e. you can't just go and use methods of the reimported class like you'd use normal methods, because they were not known at linktime. Either you use a baseclass in both files and define abstact methods to be overwritten by derived classes or you got to get the pointers to the new methods from the dlls classtable (which can get very messy)... ...
Anyway, that wasn't the topic. There is no need to keep it bumped anymore. Thanks for all the replies, really helped me a lot!
Note sure exactly, it was back in january or febuary at least because i was out snowboarding the one day and that postponed the day they came out. Webmaster would know, but i wouldnt waste his time to make you happy.Quote:
Originally posted by golfinguy4
And how long has your first OpenGL tut been up there?
I dont get why you make a life out of attacking me dude.