IMPORTANT EDIT:
Sorry everyone, i made a big mistake in the structure.
char *name; is meant to be outside of the structure, written to the file after the structure.
This way, you read the structure, find out the size of the name, then read in the string. Also explains why there is no need for a null terminator.
However, i feel somewhere, my actual question has been answered. If someone would like to edit their responses so i can choose one which is the best fitting i'd appreciate it.
Again, the question I was asking is "If you read in a structure, are you also reading in the data it holds, or do you need to access it some other way".
Sorry for the confusion
For an assignment, I've been tasked with a program which writes and reads structures to a disk (using fread and fwrite).
I'm having trouble grasping the concept.
Lets say we have this structure:
typedef struct {
short nameLength;
char* name;
}attendenceList;
attendenceList names;
now assume we give it this data:
names.name = "John Doe\0";
names.nameLength = strlen(names.name); /*potentially -1?*/
and then we use fwrite... given a file pointer fp.
fwrite(&names,sizeof(names),1,fp);
now we close the file, and open it later to read in the structure.
the question is this: when we read in the structure, are we also reading in the variables it stores?
Can we then now do something like:
if(names.nameLength < 10)
{
...
}
Or do we have to fread something more then just the structure, or assign them somehow?
Assuming the fread is:
fread(&names,sizeof(names),1,fp);
Also assuming we've defined the structure in our current function, as above.
Thanks for the help!
You have a problem here:
fwrite(&names,sizeof(names),1,fp);
Since attendenceList saves the name as a char * this will just write out the pointer, not the actual text. When you read that back in, the memory the pointer is referencing will most likely have something else in it.
You have two choices:
Put a character array (char names[MAXSIZE]) in attendenceList.
Don't write the raw data structure, but write the necessary fields.
You're writing the memory layout of the structure, which includes its members.
You'll get them back if you read the structure back in again - atleast if you do it on the same platform, with a program compiled with the same compiler and compiler settings.
Your name member is declared just as a char, so you can't store a string in it.
If name was a pointer like this:
typedef struct {
short nameLength;
char *name;
}attendenceList;
You really should not read/write the struct to a file. You will write the structure as it's laid out in memory, and that includes the value if the name pointer.
fwrite knows nothing about pointers inside your structure, it will not follow pointers and also write whatever they point to.
when you read the structure back again, you'll read in the address in the name pointer, and that might not point to anything sensible anymore.
If you declare name as an array, you'll be ok, as the array and its content is part of the structure.
typedef struct {
short nameLength;
char name[32];
}attendenceList;
As always, make sure you don't try to copy a string - including its nul terminator- to name that's larger than 32. And when you read it back again. set yourstruct.name[31] = 0; so you are sure the buffer is null terminated.
To write a structure, you'd do
attendenceList my_list;
//initialize my_list
if(fwrite(&my_list,sizeof my_list,1,f) != 1) {
//handle error
}
And to read it back again:
attendenceList my_list;
//initialize my_list
if(fread(&my_list,sizeof my_list,1,f) != 1) {
//handle error
}
}
I'm assuming you meant char* name instead of char name.
Also sizeof(name) will return 4 because you are getting the size of a char* not the length of the char array. So you should write strlen(name) not sizeof(name) inside your fwrite.
In your above example I would recommend storing the string exact size without the null termination. You don't need to store the string length as you can get that after.
If you are reading just a string from a file, and you wrote the exact size without the null termination. Then you need to manually null terminate your buffer after you read the data in.
So make sure you allocate at least the size of your data you are reading in plus 1.
Then you can set the last byte of that array to '\0'.
If you write a whole struct at a time to the buffer, you should be careful because of padding. The padding may not always be the same.
when we read in the structure, are we also reading in the variables it stores?
Yes you are, but the problem you have is that as I mentioned above you will be storing the pointer char* (4 bytes) and not the actual char array. I would recommend storing the struct elements individually.
You ask:
now we close the file, and open it later to read in the structure. the question is this: when we read in the structure, are we also reading in the variables it stores?
No. sizeof(names) is a constant value defined at compile time. It will be the same as
sizeof(short) + sizeof(void*) + some_amount_of_padding_to_align_things
it will NOT include the size of what names.name points to, it will only include the size of the pointer itself.
So you have two problems when writing this to a file.
you aren't actually writing the name string to the file
you are writing a pointer value to the file that will have no meaning when you read it back.
As your code is currently written, When you read back the names, names.name will point to somewhere, but it won't point to "John Doe\0".
What you need to do is to write the string pointed to by names.name instead of the pointer value.
What you need to do is sometimes called "flattening" the structure, You make a structure in memory that contains no pointers, but holds the same data as the structure you want to use, then you write the flattened structure to disk. This is one way to do that.
typedef struct {
short nameLength;
char name[1]; // this will be variable sized at runtime.
}attendenceListFlat;
int cbFlat = sizeof(attendenceListFlat) + strlen(names.name);
attendenceListFlat * pflat = malloc(cbFlat);
pflat->nameLength = names.nameLength;
strcpy(pflat->name, names.name);
fwrite(pflat, cbFlat, 1, fp);
The flattened structure ends with an array that has a minimum size of 1, but when we malloc, we add strlen(names.name) so we can treat that as an array of strlen(names.name)+1 size.
A few things.
Structures are just chunks of memory. It's just taking a bunch of bytes and drawing boundaries on them. Accessing structure elements is just a convenient way of getting a particular memory offset cast as a particular type of data
You are attempting to assign a string to a char type. This will not work. In C, strings are arrays of characters with a NULL byte at the end of them. The easiest way to get this to work is to set a side a fixed buffer for the name. When you create your structure you'll have to copy the name into the buffer (being very careful not to write more bytes than the buffer contains). You can then write/read the buffer from the file in one step.
struct attendanceList {
int namelen;
char name[256]; //fixed size buffer for name
}
Another way you could do it is by having the name be a pointer to a string. This makes what you're trying to do more complicated, because in order to write/read the struct to/from a file, you will have to take into account that the name is stored in a different place in memory. This means two writes and two reads (depending on how you do it) as well as correctly assigning the name pointer to wherever you read the data for the name.
struct attendanceList {
int namelen;
char* name; //the * means "this is a pointer to a char somewhere else in memory"
}
There's a third way you could do it, with a dynamically sized struct using a trick with a zero length array at the end of a struct. Once you know how long the name is, you allocate the correct amount (sizeof(struct attendanceList) + length of string). Then you have it in one contiguous buffer. You just need to remember that sizeof(struct attendanceList) is not the size you need to write/read. This might be a little confusing as a beginning. It is also kind of a hack that's not supported under all compilers.
struct attendanceList {
int namelen;
char name[0]; //this just allows easy access to the data following the struct. Be careful!
}
Related
I'm currently having an issue with the following struct:
typedef struct __attribute__((__packed__)) rungInput{
operation inputOperation;
inputType type;
char* name;
char numeroInput;
u8 is_not;
} rungInput;
I create multiple structs like above inside a for loop, and then fill in their fields according to my program logic:
while (a < 5){
rungInput input;
(...)
Then when I'm done filling the struct's fields appropriately, I then attempt to copy the completed struct to an array as such:
rungArray[a] = input; //memcpy here instead?
And then I iterate again through my loop. I'm having a problem where my structs seem to all have their name value be the same, despite clearly having gone through different segments of code and assigning different values to that field for every loop iteration.
For example, if I have three structs with the following names: "SW1" "SW2" SW3", after I am done adding them to my array I seem to have all three structs point me to the value "SW3" instead. Does this mean I should call malloc() to allocate manually each pointer inside each struct to ensure that I do not have multiple structs that point to the same value or am I doing something else wrong?
When you write rungArray[i] = input;, you are copying the pointer that is in the input structure into the rungArray[i] structure. If you subsequently overwrite the data that the input structure is pointing at, then you also overwrite the data that the rungArray[i] structure is pointing at. Using memcpy() instead of assignment won't change this at all.
There are a variety of ways around this. The simplest is to change the structure so that you allocate a big enough array in the structure to hold the name:
enum { MAX_NAME_SIZE = 32 };
…
char name[MAX_NAME_SIZE];
…
However, if the extreme size of a name is large but the average size is small, then this may waste too much space. In that case, you continue using a char *, but you do indeed have to modify the copying process to duplicate the string with dynamically allocated memory:
rungArray[i] = input;
rungArray[i].name = strdup(input.name);
Remember to free the memory when you discard the rungArray. Yes, this code copies the pointer and then overwrites it, but it is more resilient to change because all the fields are copied, even if you add some extra (non-pointer) fields, and then the pointer fields are handled specially. If you write the assignments to each member in turn, you have to remember to track all the places where you do this (that would be a single assignment function, wouldn't it?) and add the new assignments there. With the code shown, that mostly happens automatically.
You should malloc memory for your struct and then store the pointers to the structs inside your array. You could also turn your structs into a linked list by adding a pointer to each struct that points to the next instance of your struct.
http://www.cprogramming.com/tutorial/c/lesson15.html
Hello I am new to this site, and I require some help with understanding what would be considered the "norm" while coding structures in C that require a string. Basically I am wondering which of the following ways would be considered the "industry standard" while using structures in C to keep track of ALL of the memory the structure requires:
1) Fixed Size String:
typedef struct
{
int damage;
char name[40];
} Item;
I can now get the size using sizeof(Item)
2) Character Array Pointer
typedef struct
{
int damage;
char *name;
} Item;
I know I can store the size of name using a second variable, but is there another way?
i) is there any other advantage to using the fixed size (1)
char name[40];
versus doing the following and using a pointer to a char array (2)?
char *name;
and if so, what is the advantage?
ii) Also, is the string using a pointer to a char array (2) going to be stored sequentially and immediately after the structure (immediately after the pointer to the string) or will it be stored somewhere else in memory?
iii) I wish to know how one can find the length of a char * string variable (without using a size_t, or integer value to store the length)
There are basically 3 common conventions for strings. All three are found in the wild, both for in-memory representation and storage/transmission.
Fixed size. Access is very efficient, but if the actual length varies you both waste space and need one of the below methods to determine the end of the "real" content.
Length prefixed. Extra space is included in the dynamically allocation, to hold the length. From the pointer you can find both the character content and the length immediately preceding it. Example: BSTR Sometimes the length is encoded to be more space efficient for short strings. Example: ASN-1
Terminated. The string extends until the first occurrence of the termination character (typically NUL), and the content cannot contain that character. Variations made the termination two NUL in sequence, to allow individual NUL characters to exist in the string, which is then often treated as a packed list of strings. Other variations use an encoding such as byte stuffing (UTF-8 would also work) to guarantee that there exists some code reserved for termination that can't ever appear in the encoded version of the content.
In the third case, there's a function such as strlen to search for the terminator and find the length.
Both cases which use pointers can point to data immediately following the fixed portion of the structure, if you carefully allocate it that way. If you want to force this, then use a flexible array on the end of your structure (no pointer needed). Like this:
typedef struct
{
int damage;
char name[]; // terminated
} Item;
or
typedef struct
{
int damage;
int length_of_name;
char name[];
} Item;
1) is there any other advantage to using the fixed size (1)
char name[40];
versus doing the following and using a pointer to a char array (2)?
char *name;
and if so, what is the advantage?
With your array declared as char name[40]; space for name is already allocated and you are free to copy information into name from name[0] through name[39]. However, in the case of char *name;, it is simply a character pointer and can be used to point to an existing string in memory, but, on its own, cannot be used to copy information to until you allocate memory to hold that information. So say you have a 30 character string you want to copy to name declared as char *name;, you must first allocate with malloc 30 characters plus an additional character to hold the null-terminating character:
char *name;
name = malloc (sizeof (char) * (30 + 1));
Then you are free to copy information to/from name. An advantage of dynamically allocating is that you can realloc memory for name if the information you are storing in name grows. beyond 30 characters. An additional requirement after allocating memory for name, you are responsible for freeing the memory you have allocated when it is no longer needed. That's a rough outline of the pros/cons/requirements for using one as opposed to the other.
If you know the maximum length of the string you need, then you can use a character array. It does mean though that you will be using more memory than you'd typically use with dynamically allocated character arrays. Also, take a look at CString if you are using C++. You can find the length of the character array using strlen. In case of static allocation I believe it will be a part of the variable. Dynamic can be anywhere on the heap.
I know that I can copy the structure member by member, instead of that can I do a memcpy on structures?
Is it advisable to do so?
In my structure, I have a string also as member which I have to copy to another structure having the same member. How do I do that?
Copying by plain assignment is best, since it's shorter, easier to read, and has a higher level of abstraction. Instead of saying (to the human reader of the code) "copy these bits from here to there", and requiring the reader to think about the size argument to the copy, you're just doing a plain assignment ("copy this value from here to here"). There can be no hesitation about whether or not the size is correct.
Also, if the structure is heavily padded, assignment might make the compiler emit something more efficient, since it doesn't have to copy the padding (and it knows where it is), but mempcy() doesn't so it will always copy the exact number of bytes you tell it to copy.
If your string is an actual array, i.e.:
struct {
char string[32];
size_t len;
} a, b;
strcpy(a.string, "hello");
a.len = strlen(a.string);
Then you can still use plain assignment:
b = a;
To get a complete copy. For variable-length data modelled like this though, this is not the most efficient way to do the copy since the entire array will always be copied.
Beware though, that copying structs that contain pointers to heap-allocated memory can be a bit dangerous, since by doing so you're aliasing the pointer, and typically making it ambiguous who owns the pointer after the copying operation.
For these situations a "deep copy" is really the only choice, and that needs to go in a function.
Since C90, you can simply use:
dest_struct = source_struct;
as long as the string is memorized inside an array:
struct xxx {
char theString[100];
};
Otherwise, if it's a pointer, you'll need to copy it by hand.
struct xxx {
char* theString;
};
dest_struct = source_struct;
dest_struct.theString = malloc(strlen(source_struct.theString) + 1);
strcpy(dest_struct.theString, source_struct.theString);
If the structures are of compatible types, yes, you can, with something like:
memcpy (dest_struct, source_struct, sizeof (*dest_struct));
The only thing you need to be aware of is that this is a shallow copy. In other words, if you have a char * pointing to a specific string, both structures will point to the same string.
And changing the contents of one of those string fields (the data that the char * points to, not the char * itself) will change the other as well.
If you want a easy copy without having to manually do each field but with the added bonus of non-shallow string copies, use strdup:
memcpy (dest_struct, source_struct, sizeof (*dest_struct));
dest_struct->strptr = strdup (source_struct->strptr);
This will copy the entire contents of the structure, then deep-copy the string, effectively giving a separate string to each structure.
And, if your C implementation doesn't have a strdup (it's not part of the ISO standard), get one from here.
You can memcpy structs, or you can just assign them like any other value.
struct {int a, b;} c, d;
c.a = c.b = 10;
d = c;
In C, memcpy is only foolishly risky. As long as you get all three parameters exactly right, none of the struct members are pointers (or, you explicitly intend to do a shallow copy) and there aren't large alignment gaps in the struct that memcpy is going to waste time looping through (or performance never matters), then by all means, memcpy. You gain nothing except code that is harder to read, fragile to future changes and has to be hand-verified in code reviews (because the compiler can't), but hey yeah sure why not.
In C++, we advance to the ludicrously risky. You may have members of types which are not safely memcpyable, like std::string, which will cause your receiving struct to become a dangerous weapon, randomly corrupting memory whenever used. You may get surprises involving virtual functions when emulating slice-copies. The optimizer, which can do wondrous things for you because it has a guarantee of full type knowledge when it compiles =, can do nothing for your memcpy call.
In C++ there's a rule of thumb - if you see memcpy or memset, something's wrong. There are rare cases when this is not true, but they do not involve structs. You use memcpy when, and only when, you have reason to blindly copy bytes.
Assignment on the other hand is simple to read, checks correctness at compile time and then intelligently moves values at runtime. There is no downside.
You can use the following solution to accomplish your goal:
struct student
{
char name[20];
char country[20];
};
void main()
{
struct student S={"Wolverine","America"};
struct student X;
X=S;
printf("%s%s",X.name,X.country);
}
You can use a struct to read write into a file.
You do not need to cast it as a `char*.
Struct size will also be preserved.
(This point is not closest to the topic but guess it:
behaving on hard memory is often similar to RAM one.)
To move (to & from) a single string field you must use strncpy
and a transient string buffer '\0' terminating.
Somewhere you must remember the length of the record string field.
To move other fields you can use the dot notation, ex.:
NodeB->one=intvar;
floatvar2=(NodeA->insidebisnode_subvar).myfl;
struct mynode {
int one;
int two;
char txt3[3];
struct{char txt2[6];}txt2fi;
struct insidenode{
char txt[8];
long int myl;
void * mypointer;
size_t myst;
long long myll;
} insidenode_subvar;
struct insidebisnode{
float myfl;
} insidebisnode_subvar;
} mynode_subvar;
typedef struct mynode* Node;
...(main)
Node NodeA=malloc...
Node NodeB=malloc...
You can embed each string into a structs that fit it,
to evade point-2 and behave like Cobol:
NodeB->txt2fi=NodeA->txt2fi
...but you will still need of a transient string
plus one strncpy as mentioned at point-2 for scanf, printf
otherwise an operator longer input (shorter),
would have not be truncated (by spaces padded).
(NodeB->insidenode_subvar).mypointer=(NodeA->insidenode_subvar).mypointer
will create a pointer alias.
NodeB.txt3=NodeA.txt3
causes the compiler to reject:
error: incompatible types when assigning to type ‘char[3]’ from type ‘char *’
point-4 works only because NodeB->txt2fi & NodeA->txt2fi belong to the same typedef !!
A correct and simple answer to this topic I found at
In C, why can't I assign a string to a char array after it's declared?
"Arrays (also of chars) are second-class citizens in C"!!!
Ok I have a struct defined as such...
typedef struct
{
enum COMMAND command;
enum CMD_SOURCE source;
CHAR parameters[16];
} focuserCommand;
I am trying to make a function that will add an instance of this type of struct to an array of focuserCommands. The array is defined like this...
extern focuserCommand CommandBuffer[CMD_BUFFER_SIZE];
the function I am trying to write should take a pointer to a focuserCommand and add it to CommandBuffer. I am implementing CommandBuffer as a FIFO ring buffer so I know i need to move the tail forwared so that my other functions can see that the buffer contains data. CmdBuffHead and CmdBuffTail represent the read and write pointers of the buffer. Writes are added to the tail, reads are made from the head.
void AddCmdToBfr( focuserCommand * cmd )
{
// What goes here to add the struct pointed to by cmd to
// element CmdBuffTail of the buffer?
CmdBuffTail++;
}
Given that you have the variable CmdBuffTail pointing at the next element of CommandBuffer to be written to, then:
void AddCmdToBfr(focusCommand *cmd)
{
assert(CmdBuffTail >= 0 && CmdBuffTail < CMD_BUFFER_SIZE);
CommandBuffer[CmdBuffTail++] = *cmd;
if (CmdBuffTail >= CMD_BUFFER_SIZE)
CmdBuffTail = 0;
}
This preserves the invariant that you need - that the index is in the range 0..CMD_BUFFER_SIZE-1.
You may also need a separate index (CmdBuffHead perhaps) to identify where to read from. If so, you would probably need to alter that value in AddCmdToBfr() if CmdBuffTail if catches up with it.
After fixing a goof copying the incoming command, it occurs to me that you need to think carefully about the memory management. In this case, your focususerCommand structure is simple enough (unless the typedef CHAR hides a pointer - in which case, slap wrists; don't hide pointers) that you can simply copy it with impunity. If you can't copy the structure simply (because it contains pointers to allocated memory), then you need to ensure you understand who owns the data to make sure that (a) any allocated memory is released, and (b) any allocated memory is released just once. This would mean, for example, that instead of just copying the new command over the old, you would have to first release the memory allocated to the old entry. Think in terms of C++ copy constructors and destructors - in C.
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.