Initialize the structure by another structure, is this possible? - c

I would like to know, is that possible to initialize the structure1 by structure2.
I am new to the concept of casting too.
The output of this code has to be zero. Please guide. Thanks!
#include<stdio.h>
typedef struct student
{
int roll_id[10];
int name_id[10];
int postn;
} student;
typedef struct exams
{
int subject[10];
int area;
}exams;
int main()
{
exams e= { {0} };
student *pptr= (student*)&e;
printf (" %d\n", pptr->name_id[9]);
return 0;
}

Analogy is simple:
You buy an Apple and pretend it is an Orange.
As long as you eat it as something eatable you can eat it, but If you bite in to it expecting to get orange juice you will end up disappointed.
Replace Apple & Orange by your two structures and you by compiler.
Structure is nothing but a block of memory which is usually occupied by different data types.
The compiler implementation may add padding bytes between these types except for the first type in the structure.
Since the first type of your both structures is same(an array of 10 integers). Pretending the structure exam as of other type, student will work but if you try to access any other data type other than the first type it will result in Undefined Behavior.
Undefined Behavior is what you are getting in your code.
Bottom line:
You cannot do this.

First of all, in your code you are not initializing structure1 by structure2, but merely disguising pointer to structure2 as a pointer to structure1.
exams structure instance apparently (let alone weird alignment settings) takes less memory than student instance. Accessing pptr->name_id[9] can very well result in reading past the area dedicated to e. Now it all depends upon your compiler, your computer and so on.. meaning that it is better avoid delving into such details for one's sake.
The output of this code has to be zero
Perhaps it is if you are reading from unallocated stack area (past e) and in your setup stack is filled with zeroes before handing it over to the running thread.
Please read more in your C book about casting and automatic memories and instances and pointers. There is really so much to tell to right your code. You would be better off asking more specific questions after you've studied on the topic more. Good luck

Related

Trying To Learn Dynamic Memory Allocation in C

I am trying to learn dynamic memory allocation and structures and I have some questions.
First of all
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int *number;
number = malloc(1*sizeof(int));
int a;
for(a=0;a<150;a++)
{
number[a]=a;
printf("%d ",number[a]);
}
return 0;
}
In this sample I planned it to give me error. Because I allocated it to size of 1 integer, then I wrote it way too much integers. Shouldn't it give me an error? Can you explain this with detail please?
struct people
{
char *name;
int age;
char *personalInfo;
} human[3];
When I define a structure like this, how can I allocate it to keep more than 3 human? How can I change it to something like human[20] or more? In case if the answer is writing *human instead of human[3], how should I allocate it? Like malloc(number*sizeof(char)*sizeof(int)*sizeof(char))?
And one more thing, in second example do I need to allocate name and personalInfo pointers?
In this sample i planned it to give me error.
You can't "plan to give an error". You're correct to say that the code is wrong because you are writing past the end of the array, but C has no bounds checking -- accessing the array is simply undefined behavior, which means that it can do literally anything -- it can pretend to work fine or it can crash, or it can cause demons to fly out of your nose.
How can i change it to something like human[20] or more?
Well... that's exactly how:
struct people {
// members here
} human[20];
I don't see why the number "3" is so special. If you want to allocate memory dynamically, you would do it the way you would do in the case of every other data type: use the sizeof operator:
struct people *human = malloc(sizeof(human[0]) * number_of_people);
or
struct people *human = malloc(sizeof(struct people) * number_of_people);
C does not provide compile-time or run-time bounds checking of arrays. The program must perform its own bound checking. If you write beyond the end of an array, your program will have undefined behavior. In practice it means that you're corrupting your process' memory by writing to uninitialized areas or areas that are used by allocators for bookkeeping. This can cause the program to crash immediately, crash on exit, or can overwrite the values of other (possibly unrelated) variables in the heap.
To allocate a variable number of people structures, you can do this:
struct people *humans;
humans = malloc(N * sizeof(struct people));
// humans now points to an array of N `people`
For your last question, yes, you must allocate space for those strings, unless you're simply using those pointers to point to strings defined/allocated elsewhere. Note this means that even in the code above where I allocate N objects, I still haven't allocated any space for the strings, and can't write to them until I do so.
struct people
{
char *name;
int age;
char *personalInfo;
};
struct people *human;
human = malloc(sizeof(*human) * 20);
In this sample i planned it to give me error. because i allocated it to size of 1 integer, then i wrote it way too much integers. Shouldn't it give me an error? Can you explain this with detail please?
No, it shouldn't. This is undefined behavior, which may lead to a crash at runtime, but the compiler will neither warn you about it nor produce any errors. You can detect errors like this by running your program in a memory profiler, such as valgrind.
When i define a structure like this, How can i allocate it to keep more than 3 human? How can i change it to something like human[20] or more?
Like this:
struct people *human = malloc(20 * sizeof(struct people));
Like malloc( number*sizeof(char)*sizeof(int)*sizeof(char) ) ?
Besides being too inconvenient, adding up sizes of individual fields may not produce the correct size, because the compiler is allowed to pad your struct to optimize access to its fields. That is why you need to use sizeof(struct people).
The first example invokes undefined behavior.
ยง6.5.6/8: If both the pointer operand and the result point to elements of the same array
object, or one past the last element of the array object, the evaluation shall not
produce an overflow; otherwise, the behavior is undefined
If you want to allocate for 20 humans, you can do...
typedef struct people
{
char *name;
int age;
char *personalInfo;
}People;
People *human;
human = malloc(sizeof(*human) * 20);

How to fix Bad Pointers in C

I am a beginner in C and as I was trying to make a simple snake game, I stumbled upon the problem of value change in pointers when used in another function. I used pointers in order to grow my snake and I used 3 of them. Granted this might be a really noobish algorithm for a snake game but I feel like I am almost there but I cannot figure out what went wrong. I have used 3 pointes for the snake itself, the x coordinate of each part of the snake and also the y. I might as well just post a part of my code.
#include<stdio.h>
#include<dos.h>
#include<conio.h>
printer(int *forgoodness,int *y,char *lang,int tx,int ty, int *x)
{
int h=*forgoodness-1,g=0;
The value changes in here, somehow the x copies 6 values of the y.
/* for(;g<15;g++)
{
printf("%i",x[g]);
}*/
for(;h>=0;h--)
{
gotoxy(x[h],y[h]);
printf("%c",lang[h]);
}
gotoxy(tx,ty);
printf(" ");
}
main()
{
int transferx=1,x=1,transfery=1,ch,game=0,dir;
int *transx, *transy, *numel;
int tempsx,g=0,tempsy,forex,j=0,*totalel;
char *snake;
int *snakey, *snakex;
If I explicitly assign values it works well but I cannot grow my snake using this:
//int snakex[15]={26,27,28,29,30,31,32,33,34,35,36,37,38,39,40};
//int snakey[15]={13,13,13,13,13,13,13,13,13,13,13,13,13,13,13};
clrscr();
*totalel=0;
*numel=14;
forex=26;
snake= "***************";
This is what I did, and I checked the values before the function printer runs and the values are fine.
for(;j<15;j++)
{
snakey[j]=13;
snakex[j]=forex;
*totalel=j;
forex++;
}
printer(numel,snakey,snake,transferx,transfery,snakex);
I hope that you can help me on this. Cheers.
Maybe attempt to rewrite a bit of your code... ideally using less pointers. Based on the code I see, it looks as if you're feeding values into random pointers. That's pretty dangerous.
For instance, doing these two lines in a row is bad:
int *num;
*num = 9;
The reason is because num is just a pointer to memory... and it isn't currently pointing to anything valid yet. The number 9 is attempting to be stored in some random location. You need to allocate some memory for num to point to, or point it to the address of another variable (a non-pointer integer). I think maybe you're not quite grasping the concepts of pointers just yet. But don't worry, it takes a little time.
You should be able to write your game without using any pointers at all, and maybe just use a fixed array for now like someone else had mentioned. Then do a bit more reading up on how pointers work and take another stab at it. I'm sure you'll get it!
But for now, try to revisit the problem with a new set of data and come back to us with what you've done so we can try to help out further. :)
Cool that you are working with pointers! But as you can see, they can be a little tricky. You have to remember that pointers are like the address of a mailbox; they can tell where something is, but they cannot (in of themselves) HOLD ANYTHING. The can just point to something that does. So every pointer you declare has to point to a real thing (like an int or an array element or an array) before it can be used or assigned to. Otherwise to are shoving values into random memory areas which causes crashes. This seems to be the main problem you are having.
int *ptr;
int value;
ptr = &value;
*ptr = 10;
// value now is 10

How do I determine the memory layout of a structure?

Suppose I have the following structure (in C):
struct my_struct {
int foo;
float bar;
char *baz;
};
If I now have a variable, say
struct my_struct a_struct;
How can I find out how the fields of that structure are going to be laid out in memory? In other words, I need to know what the address of a_struct.foo, of a_struct.bar and a_struct.baz are going to be. And I cannot do that programatically, because I am actually cross-compiling to another platform.
CLARIFICATION
Thanks the answers so far, but I cannot do this programatically (i.e. with the offsetof macro, or with a small test program) because I am cross-compiling and I need to know how the fields are going to be aligned on the target platform. I know this is implementation-dependent, that's the whole point of my question. I am using GCC to compile, targeting an ARM architecture.
What I need in the end is to be able to dump the memory from the target platform and parse it with other tools, such as Python's struct library. But for that I need to know how the fields were laid out.
In general, this is implementation specific. It depends on things like the compiler, compiler settings, the platform you are compiling on, word-size, etc. Here's a previous SO thread on the topic: C struct memory layout?
If you are cross-compiling, I'd imagine the specific layout will be different depending on which platform you compile for. I'd consult references for your compiler and platform.
There's a program called pahole (Poke-A-Hole) in the dwarves package that will produce a report showing the structures of your program along with markers showing where and how large padding is.
I think you have two options.
The first one is to use __attribute__((packed)) after the struct declaration. This will ensure that each member will be allocated exactly the amount of memory that its type requires.
The other one is to examine your structure and use the alignment rules (n-byte basic type variable has to be n-byte aligned) to figure out the layout.
In your example, in either case each member variable will take 4 bytes and the structure will occupe 12 bytes in memory.
One hacky way to see the memory view of what's inside it would be to cast a struct pointer to a char pointer, then print out all the chars, something like:
struct my_struct s;
s.foo = MAX_INT;
s.bar = 1.0;
s.baz = "hello";
for (int i = 0; i < sizeof(s); i++) {
char *c = ((char*)&s) + i;
printf("byte %d: 0x%02x\n", i, *c);
}
That doesn't explicitly show you the boundaries, you'd have to infer that from the dump.
Comments made by others about packing still apply; you'll also want to use explicitly sized types like uint32 instead of unsigned int

a few beginner C questions

I'm sort of learning C, I'm not a beginner to programming though, I "know" Java and python, and by the way I'm on a mac (leopard).
Firstly,
1: could someone explain when to use a pointer and when not to?
2:
char *fun = malloc(sizeof(char) * 4);
or
char fun[4];
or
char *fun = "fun";
And then all but the last would set indexes 0, 1, 2 and 3 to 'f', 'u', 'n' and '\0' respectively. My question is, why isn't the second one a pointer? Why char fun[4] and not char *fun[4]? And how come it seems that a pointer to a struct or an int is always an array?
3:
I understand this:
typedef struct car
{
...
};
is a shortcut for
struct car
{
...
};
typedef struct car car;
Correct? But something I am really confused about:
typedef struct A
{
...
}B;
What is the difference between A and B? A is the 'tag-name', but what's that? When do I use which? Same thing for enums.
4. I understand what pointers do, but I don't understand what the point of them is (no pun intended). And when does something get allocated on the stack vs. the heap? How do I know where it gets allocated? Do pointers have something to do with it?
5. And lastly, know any good tutorial for C game programming (simple) ? And for mac/OS X, not windows?
PS. Is there any other name people use to refer to just C, not C++? I hate how they're all named almost the same thing, so hard to try to google specifically C and not just get C++ and C# stuff.
Thanks!!
It was hard to pick a best answer, they were all great, but the one I picked was the only one that made me understand my 3rd question, which was the only one I was originally going to ask. Thanks again!
My question is, why isn't the second one a pointer?
Because it declares an array. In the two other cases, you have a pointer that refers to data that lives somewhere else. Your array declaration, however, declares an array of data that lives where it's declared. If you declared it within a function, then data will die when you return from that function. Finally char *fun[4] would be an array of 4 pointers - it wouldn't be a char pointer. In case you just want to point to a block of 4 chars, then char* would fully suffice, no need to tell it that there are exactly 4 chars to be pointed to.
The first way which creates an object on the heap is used if you need data to live from thereon until the matching free call. The data will survive a return from a function.
The last way just creates data that's not intended to be written to. It's a pointer which refers to a string literal - it's often stored in read-only memory. If you write to it, then the behavior is undefined.
I understand what pointers do, but I don't understand what the point of them is (no pun intended).
Pointers are used to point to something (no pun, of course). Look at it like this: If you have a row of items on the table, and your friend says "pick the second item", then the item won't magically walk its way to you. You have to grab it. Your hand acts like a pointer, and when you move your hand back to you, you dereference that pointer and get the item. The row of items can be seen as an array of items:
And how come it seems that a pointer to a struct or an int is always an array?
item row[5];
When you do item i = row[1]; then you first point your hand at the first item (get a pointer to the first one), and then you advance till you are at the second item. Then you take your hand with the item back to you :) So, the row[1] syntax is not something special to arrays, but rather special to pointers - it's equivalent to *(row + 1), and a temporary pointer is made up when you use an array like that.
What is the difference between A and B? A is the 'tag-name', but what's that? When do I use which? Same thing for enums.
typedef struct car
{
...
};
That's not valid code. You basically said "define the type struct car { ... } to be referable by the following ordinary identifier" but you missed to tell it the identifier. The two following snippets are equivalent instead, as far as i can see
1)
struct car
{
...
};
typedef struct car car;
2)
typedef struct car
{
...
} car;
What is the difference between A and B? A is the 'tag-name', but what's that? When do I use which? Same thing for enums.
In our case, the identifier car was declared two times in the same scope. But the declarations won't conflict because each of the identifiers are in a different namespace. The two namespaces involved are the ordinary namespace and the tag namespace. A tag identifier needs to be used after a struct, union or enum keyword, while an ordinary identifier doesn't need anything around it. You may have heard of the POSIX function stat, whose interface looks like the following
struct stat {
...
};
int stat(const char *path, struct stat *buf);
In that code snippet, stat is registered into the two aforementioned namespaces too. struct stat will refer to the struct, and merely stat will refer to the function. Some people don't like to precede identifiers always with struct, union or enum. Those use typedef to introduce an ordinary identifier that will refer to the struct too. The identifier can of course be the same (both times car), or they can differ (one time A the other time B). It doesn't matter.
3) It's bad style to use two different names A and B:
typedef struct A
{
...
} B;
With that definition, you can say
struct A a;
B b;
b.field = 42;
a.field = b.field;
because the variables a and b have the same type. C programmers usually say
typedef struct A
{
...
} A;
so that you can use "A" as a type name, equivalent to "struct A" but it saves you a lot of typing.
Use them when you need to. Read some more examples and tutorials until you understand what pointers are, and this ought to be a lot clearer :)
The second case creates an array in memory, with space for four bytes. When you use that array's name, you magically get back a pointer to the first (index 0) element. And then the [] operator then actually works on a pointer, not an array - x[y] is equivalent to *(x + y). And yes, this means x[y] is the same as y[x]. Sorry.
Note also that when you add an integer to a pointer, it's multiplied by the size of the pointed-to elements, so if you do someIntArray[1], you get the second (index 1) element, not somewhere inbetween starting at the first byte.
Also, as a final gotcha - array types in function argument lists - eg, void foo(int bar[4]) - secretly get turned into pointer types - that is, void foo(int *bar). This is only the case in function arguments.
Your third example declares a struct type with two names - struct A and B. In pure C, the struct is mandatory for A - in C++, you can just refer to it as either A or B. Apart from the name change, the two types are completely equivalent, and you can substitute one for the other anywhere, anytime without any change in behavior.
C has three places things can be stored:
The stack - local variables in functions go here. For example:
void foo() {
int x; // on the stack
}
The heap - things go here when you allocate them explicitly with malloc, calloc, or realloc.
void foo() {
int *x; // on the stack
x = malloc(sizeof(*x)); // the value pointed to by x is on the heap
}
Static storage - global variables and static variables, allocated once at program startup.
int x; // static
void foo() {
static int y; // essentially a global that can only be used in foo()
}
No idea. I wish I didn't need to answer all questions at once - this is why you should split them up :)
Note: formatting looks ugly due to some sort of markdown bug, if anyone knows of a workaround please feel free to edit (and remove this note!)
char *fun = malloc(sizeof(char) * 4);
or
char fun[4];
or
char *fun = "fun";
The first one can be set to any size you want at runtime, and be resized later - you can also free the memory when you are done.
The second one is a pointer really 'fun' is the same as char ptr=&fun[0].
I understand what pointers do, but I don't understand what the point of
them is (no pun intended). And when
does something get allocated on the
stack vs. the heap? How do I know
where it gets allocated? Do pointers
have something to do with it?
When you define something in a function like "char fun[4]" it is defined on the stack and the memory isn't available outside the function.
Using malloc (or new in C++) reserves memory on the heap - you can make this data available anywhere in the program by passing it the pointer. This also lets you decide the size of the memory at runtime and finaly the size of the stack is limited (typically 1Mb) while on the heap you can reserve all the memory you have available.
edit 5. Not really - I would say pure C. C++ is (almost) a superset of C so unless you are working on a very limited embedded system it's usualy OK to use C++.
\5. Chipmunk
Fast and lightweight 2D rigid body physics library in C.
Designed with 2D video games in mind.
Lightweight C99 implementation with no external dependencies outside of the Std. C library.
Many language bindings available.
Simple, read the documentation and see!
Unrestrictive MIT license.
Makes you smarter, stronger and more attractive to the opposite gender!
...
In your second question:
char *fun = malloc(sizeof(char) * 4);
vs
char fun[4];
vs
char *fun = "fun";
These all involve an array of 4 chars, but that's where the similarity ends. Where they differ is in the lifetime, modifiability and initialisation of those chars.
The first one creates a single pointer to char object called fun - this pointer variable will live only from when this function starts until the function returns. It also calls the C standard library and asks it to dynamically create a memory block the size of an array of 4 chars, and assigns the location of the first char in the block to fun. This memory block (which you can treat as an array of 4 chars) has a flexible lifetime that's entirely up to the programmer - it lives until you pass that memory location to free(). Note that this means that the memory block created by malloc can live for a longer or shorter time than the pointer variable fun itself does. Note also that the association between fun and that memory block is not fixed - you can change fun so it points to different memory block, or make a different pointer point to that memory block.
One more thing - the array of 4 chars created by malloc is not initialised - it contains garbage values.
The second example creates only one object - an array of 4 chars, called fun. (To test this, change the 4 to 40 and print out sizeof(fun)). This array lives only until the function it's declared in returns (unless it's declared outside of a function, when it lives for as long as the entire program is running). This array of 4 chars isn't initialised either.
The third example creates two objects. The first is a pointer-to-char variable called fun, just like in the first example (and as usual, it lives from the start of this function until it returns). The other object is a bit strange - it's an array of 4 chars, initialised to { 'f', 'u', 'n', 0 }, which has no name and that lives for as long as the entire program is running. It's also not guaranteed to be modifiable (although what happens if you try to modify it is left entirely undefined - it might crash your program, or it might not). The variable fun is initialised with the location of this strange unnamed, unmodifiable, long-lived array (but just like in the first example, this association isn't permanent - you can make fun point to something else).
The reason why there's so many confusing similarities and differences between arrays and pointers is down to two things:
The "array syntax" in C (the [] operator) actually works on pointers, not arrays!
Trying to pin down an array is a bit like catching fog - in almost all cases the array evaporates and is replaced by a pointer to its first element instead.

Resizing a char[x] to char[y] at runtime

OK, I hope I explain this one correctly.
I have a struct:
typedef struct _MyData
{
char Data[256];
int Index;
} MyData;
Now, I run into a problem. Most of the time MyData.Data is OK with 256, but in some cases I need to expand the amount of chars it can hold to different sizes.
I can't use a pointer.
Is there any way to resize Data at run time? How?
Code is appreciated.
EDIT 1:
While I am very thankful for all the comments, the "maybe try this..." or "do that", or "what you are dong is wrong..." comments are not helping. Code is the help here. Please, if you know the answer post the code.
Please note that:
I cannot use pointers. Please don't try to figure out why, I just can't.
The struct is being injected into another program's memory that's why no pointers can be used.
Sorry for being a bit rough here but I asked the question here because I already tried all the different approaches that thought might work.
Again, I am looking for code. At this point I am not interested in "might work..." or " have you considered this..."
Thank you and my apologies again.
EDIT 2
Why was this set as answered?
You can use a flexible array member
typedef struct _MyData
{
int Index;
char Data[];
} MyData;
So that you can then allocate the right amount of space
MyData *d = malloc(sizeof *d + sizeof(char[100]));
d->Data[0..99] = ...;
Later, you can free, and allocate another chunk of memory and make a pointer to MyData point to it, at which time you will have more / less elements in the flexible array member (realloc). Note that you will have to save the length somewhere, too.
In Pre-C99 times, there isn't a flexible array member: char Data[] is simply regarded as an array with incomplete type, and the compiler would moan about that. Here i recommend you two possible ways out there
Using a pointer: char *Data and make it point to the allocated memory. This won't be as convenient as using the embedded array, because you will possibly need to have two allocations: One for the struct, and one for the memory pointed to by the pointer. You can also have the struct allocated on the stack instead, if the situation in your program allows this.
Using a char Data[1] instead, but treat it as if it were bigger, so that it overlays the whole allocated object. This is formally undefined behavior, but is a common technique, so it's probably safe to use with your compiler.
The problem here is your statement "I can't use a pointer". You will have to, and it will make everything much easier. Hey, realloc even copies your existing data, what do you want more?
So why do you think you can't use a pointer? Better try to fix that.
You would re-arrange the structure like that
typedef struct _MyData
{
int Index;
char Data[256];
} MyData;
And allocate instances with malloc/realloc like that:
my_data = (MyData*) malloc ( sizeof(MyData) + extra_space_needed );
This is an ugly approach and I would not recommend it (I would use pointers), but is an answer to your question how to do it without a pointer.
A limitation is that it allows for only one variable size member per struct, and has to be at the end.
Let me sum up two important points I see in this thread:
The structure is used to interact between two programs through some IPC mechanism
The destination program cannot be changed
You cannot therefore change that structure in any way, because the destination program is stuck trying to read it as currently defined. I'm afraid you are stuck.
You can try to find ways to get the equivalent behavior, or find some evil hack to force the destination program to read a new structure (e.g., modifying the binary offsets in the executable). That's all pretty application specific so I can't give much better guidance than that.
You might consider writing a third program to act as an interface between the two. It can take the "long" messages and do something with them, and pass the "short" messages onward to the old program. You can inject that in between the IPC mechanisms fairly easily.
You may be able to do this like this, without allocating a pointer for the array:
typedef struct _MyData
{
int Index;
char Data[1];
} MyData;
Later, you allocate like this:
int bcount = 256;
MyData *foo;
foo = (MyData *)malloc(sizeof(*foo) + bcount);
realloc:
int newbcount = 512;
MyData *resized_foo;
resized_foo = realloc((void *)foo, sizeof(*foo) + newbcount);
It looks like from what you're saying that you definitely have to keep MyData as a static block of data. In which case I think the only option open to you is to somehow (optionally) chain these data structures together in a way that can be re-assembled be the other process.
You'd need and additional member in MyData, eg.
typedef struct _MyData
{
int Sequence;
char Data[256];
int Index;
} MyData;
Where Sequence identifies the descending sequence in which to re-assemble the data (a sequence number of zero would indicate the final data buffer).
The problem is in the way you're putting the question. Don't think about C semantics: instead, think like a hacker. Explain exactly how you are currently getting your data into the other process at the right time, and also how the other program knows where the data begins and ends. Is the other program expecting a null-terminated string? If you declare your struct with a char[300] does the other program crash?
You see, when you say "passing data" to the other program, you might be [a] tricking the other process into copying what you put in front of it, [b] tricking the other program into letting you overwrite its normally 'private' memory, or [c] some other approach. No matter which is the case, if the other program can take your larger data, there is a way to get it to them.
I find KIV's trick quite usable. Though, I would suggest investigating the pointer issue first.
If you look at the malloc implementations
(check this IBM article, Listing 5: Pseudo-code for the main allocator),
When you allocate, the memory manager allocates a control header and
then free space following it based on your requested size.
This is very much like saying,
typedef struct _MyData
{
int size;
char Data[1]; // we are going to break the array-bound up-to size length
} MyData;
Now, your problem is,
How do you pass such a (mis-sized?) structure to this other process?
That brings us the the question,
How does the other process figure out the size of this data?
I would expect a length field as part of the communication.
If you have all that, whats wrong with passing a pointer to the other process?
Will the other process identify the difference between a pointer to a
structure and that to a allocated memory?
You cant reacolate manualy.
You can do some tricks wich i was uning when i was working aon simple data holding sistem. (very simple filesystem).
typedef struct
{
int index ;
char x[250];
} data_ztorage_250_char;
typedef struct
{
int index;
char x[1000];
} data_ztorage_1000_char;
int main(void)
{
char just_raw_data[sizeof(data_ztorage_1000_char)];
data_ztorage_1000_char* big_struct;
data_ztorage_250_char* small_struct;
big_struct = (data_ztorage_1000_char*)big_struct; //now you have bigg struct
// notice that upper line is same as writing
// big_struct = (data_ztorage_1000_char*)(&just_raw_data[0]);
small_struct = (data_ztorage_250_char*)just_raw_data;//now you have small struct
//both structs starts at same locations and they share same memory
//addresing data is
small_struct -> index = 250;
}
You don't state what the Index value is for.
As I understand it you are passing data to another program using the structure shown.
Is there a reason why you can't break your data to send into chunks of 256bytes and then set the index value accordingly? e.g.
Data is 512 bytes so you send one struct with the first 256 bytes and index=0, then another with the next 256 bytes in your array and Index=1.
How about a really, really simple solution? Could you do:
typedef struct _MyData
{
char Data[1024];
int Index;
} MyData;
I have a feeling I know your response will be "No, because the other program I don't have control over expects 256 bytes"... And if that is indeed your answer to my answer, then my answer becomes: this is impossible.

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