In one function I have written:
char *ab;
ab=malloc(10);
Then in another function I want to know the size of memory pointed by the ab pointer.
Is there any way that I can know that ab is pointing to 10 chars of memory?
No, you don't have a standard way to do this. You have to pass the size of the pointed-to memory along with the pointer, it's a common solution.
I.e. instead of
void f(char* x)
{
//...
}
use
void f(char *x, size_t length)
{
//....
}
and in your code
char *ab = malloc( 10 );
f( ab, 10 );
It's a deep secret that only free() knows for sure. It's likely in your system, but in a totally implementation dependent manner.
A bit awkward, but if you want to keep everything together:
typedef struct
{ // size of data followed by data (C only trick! NOT for C++)
int dimension; // number of data elements
int data[1]; // variable number of data elements
} malloc_int_t;
malloc_int_t *ab;
int dimension = 10;
ab = malloc( sizeof(*ab) + (dimension-1)*sizeof(int) );
ab->dimension = dimension;
ab->data[n] // data access
I've changed the data type to int to make the code a more generic template.
You can't (portably anyway). You have to keep track of the size yourself.
Some implementations of malloc could give you an API to access that information, but there is no provisions in the standard for this.
The size is what you passed into malloc, you can use a global variable or macro to remember it.
There is no way, you have to store the size of the allocated memory in another variable.
No, unfortunately.
You need to pass the size of the block along with the pointer.
No.
Now, that being said, there are non-portable hacks to do this, but it is not safe to rely upon them.
If you know with 100% certainty that the memory was allocated via malloc(), you may be able to rewind the pointer a few bytes and inspect the 'malloc node' that is used to track which parts of memory have been allocated and which have not. However, I can not stress this enough--do not ever depend upon this.
There is no way to deduce the size of allocated memory from the pointer itself. Since ab is a char *, sizeof(ab) is the same as sizeof(char *), which obviously is not the same as the size of the allocated chunk of memory.
Since you called malloc with the required size, you know what the size is. Pass this number along with the pointer to the function that needs to know the size.
I had a structure and a char pointer pointing to its memory address. So relating it to your question, I wanted to find the size of the memory location it was pointing to i.e. the size of the structure. So logically what you do is, find the size of the object the pointer creates to. This worked for me:
unsigned char * buffer= Library1Structure;
int x=sizeof(Library1Structure);
So the value of x tells me the size of the memory location the pointer buffer points to.
Related
I have this code
int main(int argc, char *argv[])
{
int i=1;
char **m=malloc(sizeof(char *)*i);
printf("%zu\n",sizeof *m);
m[0]=malloc(strlen("hello")+1);
strcpy(m[0],"hello");
printf("%s\n", m[0]);
i=2;
m=(char **)realloc(m,sizeof (char *)*i);
m[1]=malloc(strlen("hi")+1);
strcpy(m[1],"hi");
printf("%s %s \n",m[0],m[1] );
// TODO: write proper cleanup code just for good habits.
return 0;
}
this is how I am allocating pointer char **m 8 byte single char pointer
int i=1;
char **m=malloc(sizeof(char *)*i);
and this is how I am allocating area of space whose address will be kept in m[0]
m[0]=malloc(strlen("hello")+1);
strcpy(m[0],"hello");
printf("%s\n", m[0]);
I like to know is this normally how its done. I mean allocating space for pointer and then allocating space in memory that the pointer will hold.
Does m[0]=malloc(strlen("hello")+1); is same as this *(m+0)=malloc(strlen("hello")+1); and does this m[1]=malloc(strlen("hi")+1); this *(m+1)=malloc(strlen("hi")+1);
And I am increasing pointer to pointer numbers like this in allocation m=(char **)realloc(m,sizeof (char *)*i); before m[1]=malloc(strlen("hi")+1);
is there anything wrong with above code. I seen similar code on this Dynamic memory/realloc string array
can anyone please explain with this statement char **m=malloc(sizeof(char *)*i); I am allocating 8 byte single pointer of type char but with this statement m=(char **)realloc(m,sizeof (char *)*i); why I am not getting stack smaching detected error. How exactly realloc works. can anyone give me the link of realloc function or explain a bit on this please
I like to know is this normally how its done. I mean allocating space for pointer and then allocating space in memory that the pointer will hold.
It depends on what you are trying to achieve. If you wish to allocate an unspecified amount of strings with individual lengths, then your code is pretty much the correct way to do it.
If you wish to have a fixed amount of strings with individual lengths, you could just do char* arr [n]; and then only malloc each arr[i].
Or if you wish to have a fixed amount of strings with a fixed maximum length, you could use a 2D array of characters, char arr [x][y];, and no malloc at all.
Does m[0]=malloc(strlen("hello")+1); is same as this *(m+0)=malloc(strlen("hello")+1);
Yes, m[0] is 100% equivalent to *((m)+(0)). See Do pointers support "array style indexing"?
is there anything wrong with above code
Not really, except stylistic and performance issues. It could optionally be rewritten like this:
char** m = malloc(sizeof(*m) * i); // subjective style change
m[0]=malloc(sizeof("hello")); // compile-time calculation, better performance
why I am not getting stack smaching detected error
Why would you get that? The only thing stored on the stack here is the char** itself. The rest is stored on the heap.
How exactly realloc works. can anyone give me the link of realloc function or explain a bit on this please
It works pretty much as you've used it, though pedantically you should not store the result in the same pointer as the one passed, in case realloc fails and you wish to continue using the old data. That's a very minor remark though, since in case realloc fails, it either means that you made an unrealistic request for memory, or that the RAM on your system is toast and you will unlikely be able to continue execution anyway.
The canonical documentation for realloc would be the C standard C17 7.22.3.5:
#include <stdlib.h>
void *realloc(void *ptr, size_t size);
The realloc function deallocates the old object pointed to by ptr and returns a
pointer to a new object that has the size specified by size. The contents of the new
object shall be the same as that of the old object prior to deallocation, up to the lesser of
the new and old sizes. Any bytes in the new object beyond the size of the old object have
indeterminate values.
If ptr is a null pointer, the realloc function behaves like the malloc function for the
specified size. Otherwise, if ptr does not match a pointer earlier returned by a memory
management function, or if the space has been deallocated by a call to the free or
realloc function, the behavior is undefined. If memory for the new object cannot be
allocated, the old object is not deallocated and its value is unchanged.
Returns
The realloc function returns a pointer to the new object (which may have the same value as a pointer to the old object), or a null pointer if the new object could not be allocated.
Notably there is no guarantee that the returned pointer always has the same value as the old pointer, so correct use would be:
char* tmp = realloc(arr, size);
if(tmp == NULL)
{
/* error handling */
}
arr = tmp;
(Where tmp has the same type as arr.)
Your code looks fine to me. Yes, if you are storing an array of strings, and you don't know how many strings will be in the array in advance, then it is perfectly fine to allocate space for an array of pointers with malloc. You also need to somehow get memory for the strings themselves, and it is perfectly fine for each string to be allocated with its own malloc call.
The line you wrote to use realloc is fine; it expands the memory area you've allocated for pointers so that it now has the capacity to hold 2 pointers, instead of just 1. When the realloc function does this, it might need to move the memory allocation to a different address, so that is why you have to overwrite m as you did. There is no stack smashing going on here. Also, please note that pointers are not 8 bytes on every platform; that's why it was wise of you to write sizeof(char *) instead of 8.
To find more documentation about realloc, you can look in the C++ standard, or the POSIX standard, but perhaps the most appropriate place for this question is the C standard, which documents realloc on page 314.
I am pretty new to C and I am trying to read the user input (some sentence or string in general) and then I want to create an array based on the input lenght. Is there a reasonable way to do it? Thanks for answers
Just for an overview of why all the answers are suggesting pointers instead of arrays:
When I was learning C one thing that helped was to understand arrays and pointers and how similar they are.
For the most part, they can have the same syntax, you can use * syntax with either or you can use [] syntax with either.
The differences are:
1) Arrays have memory allocated for them by the system and pointers don't, you have to "set" a pointer to some memory that you have allocated.
2) I don't think arrays can change where arrays point, they always point at their pre-allocated spot.
Since arrays are pre-allocated and can't be repointed, you want a pointer. You can treat it exactly as an array (You can use [] syntax) but you have to allocate memory for it first.
So for example, if a array with and p is a pointer, a[0]=1, *a=1, p[0]=1 and *p=1 are all identical functions, and while *++p=1 is valid, I don't think *++a=1 is valid because you can't change where a points.
So the short version would be, you need a pointer, not an array, and to change how much is allocated, you allocate the new size (With malloc or something similar), copy what you want to retain over and free the old space (Or you might be able to increase the size of the first one--realloc?, not sure, my C is decades old)
malloc/free, in the case of strings a strlen will get you it's length.
You can use malloc to allocate new memory, Note that since C's memory isn't managed (contrary to Java, Python or any other high level language), you will have to free the memory once you are done using it.
int arr_size = 0;
int* arr;
printf("Please enter a size to the array:");
scanf("%d", &arr_size);
arr = malloc(arr_size * sizeof(int))
// Use array
free(arr);
void *malloc(size_t size);
The malloc() function allocates size bytes and returns a pointer to
the allocated memory. The memory is not initialized. If size is 0,
then malloc() returns either NULL, or a unique pointer value that can
later be successfully passed to free().
It depends on the standard you're compiling against and your compiler. You can only rely on variable length arrays in C99
The only way to be certain is to use malloc, though you need to ensure you free the memory afterwards:
int length;
// Do something to set the size
// Allocates a contiguous block of memory that is
// (length * size of a char primitive) in length
char *array = (char *)malloc(length * sizeof(char));
// Do whatever you need do to with the array
free(array);
In C, declaring a variable as a pointer (char *a) and as an array (char a[3]) allows you to use that variable in exactly the same way. The only difference is that with a pointer you need to allocate and free the memory yourself, while with the array that block of memory is given to you automatically and it is freed when it goes out of scope.
With the code above, you can still access each individual character via an index like so:
array[0] = 'f';
array[1] = 'o';
array[3] = 'o';
I have an char array of fixed size in C application. I am passing that array to some function and from there I am passing it to multiple functions. So that the array gets filled in some of the functions based on some condition. Since I am sending a fixed size array I am facing problem when I copy data to it if the size is more than the array size. I know that I have to make that char array dynamic but as I said that array gets filled in multiple functions and size will be different. So do I need to dynamically allocate the array wherever it gets filled? Consider the array gets filled in 30+ different functions. Or is there a way to do a minimal modification?
As your question title says C, IMO the best approach will be to decalre a pointer of that particular variable type in your main() function, pass the address of that pointer [essencially a double-pointer] to other functions and allocate memory dynamically.
You can keep on passing the pointer to all other functions. Inside every called functions, measure the amount of memory required to put the data [from that particular function] and use realloc() to increase the available memory.
As mentioned by UncleO, the required pointer should be the pointer to array [i.e, a double pointer]
EDIT
For the very first time allocating memory to that pointer, you can use malloc() or calloc().
From next time onwards, to extend [resize] the amount of memory, you need to use realloc()
You don't pass an array to a function in C, although it appears that way. What gets passed is a pointer to the first element of the array. The pointer is passed by value. That is, the value of the pointer (the memory location) is copied into a local variable. The contents of the array can be changed using this local variable.
If you use malloc() or realloc() with this local variable, then the pointer that was "passed in" won't be affected. realloc() may resize the memory, but it can also free that memory and allocate some new memory to the local variable.
If you want to change the array pointer, then you should pass in a pointer to the pointer. The thing the pointer points to is what is changed. This is a bit more cumbersome. But this way you can allocate more memory is needed.
#include <stdlib.h>
char* arr;
void changeit(char** arrptr)
{
*arrptr = realloc(*arrptr, 20*sizeof(char));
}
void main (void)
{
arr = malloc(10*sizeof(char));
changeit(&arr);
}
To function that do not alter the array, pass the array address and size.
int foo1(const char *array, size_t size, ...)
To each function that does not change the array size, pass array address and array size
int foo2(char *array, size_t size, ...)
To functions that may alter the array size, pass the address of the address of the array and the address of the size.
int foo3(char **array, size_t *size, ...)
Code could put these two variables together
typedef struct {
size_t size;
char array;
} YetAnotherArrayType;
Chux,
There's a little typo at then end of your post. I think you mean:
typedef struct {
size_t size;
char* array;
} YetAnotherArrayType;
You didn't make array a pointer type.
If the poster is handcrafting a container, the classic solution is to track both a size and a capacity.
In that model you allocate the array to some initial a capacity and set size to 0. You then track its population causing it grow by some chunk size or factor each time it fills up.
Frequent reallocation can be a massive performance drain and by the sounds of the program in question such behaviour seems likely.
I need to dynamically allocate an array of type int with size 12MB. I did it as follows:
unsigned int *memory;
memory = (int *) malloc( 3072 * sizeof(int));
How would I iterate through the array without having to use 3072? Is there a way to get the length of the array? Or should I just do for (int i = 0; i < 3072; i++)?
Thanks
The is no portable or convenient way to find out how large the allocate chunk is. After malloc is done all that is visible is a pointer to an indeterminate amount of memory (*). You must do the bookkeeping yourself.
(*) Actually malloc and friends do know how large the chunk is but there is no standard way for a client to access this information.
Pointers in C have no inherent length hence there is no way to do this. You must keep the length paired with the pointer throughout it's lifetime.
There is no other way. You should put 3072 in a const int, which would make the code better and more maintainable.
With MSVC and MinGW you can use the nonportable _msize function, like
char *c=malloc(1234);
printf("%lu",(unsigned long)_msize(c));
free(c);
This is actually a much more concise, much more clear question than the one I had asked here before(for any who cares): C Language: Why does malloc() return a pointer, and not the value? (Sorry for those who initially think I'm spamming... I hope it's not construed as the same question since I think the way I phrased it there made it unintentionally misleading)
-> Basically what I'm trying to ask is: Why does a C programmer need a pointer to a dynamically-allocated variable/object? (whatever the difference is between variable/object...)
If a C programmer has the option of creating just 'int x' or just 'int *x' (both statically allocated), then why can't he also have the option to JUST initialize his dynamically-allocated variable/object as a variable (and NOT returning a pointer through malloc())?
*If there are some obscure ways to do what I explained above, then, well, why does malloc() seem the way that most textbooks go about dynamic-allocation?
Note: in the following, byte refers to sizeof(char)
Well, for one, malloc returns a void *. It simply can't return a value: that wouldn't be feasible with C's lack of generics. In C, the compiler must know the size of every object at compile time; since the size of the memory being allocated will not be known until run time, then a type that could represent any value must be returned. Since void * can represent any pointer, it is the best choice.
malloc also cannot initialize the block: it has no knowledge of what's being allocated. This is in contrast with C++'s operator new, which does both the allocation and the initialization, as well as being type safe (it still returns a pointer instead of a reference, probably for historical reasons).
Also, malloc allocates a block of memory of a specific size, then returns a pointer to that memory (that's what malloc stands for: memory allocation). You're getting a pointer because that's what you get: an unitialized block of raw memory. When you do, say, malloc(sizeof(int)), you're not creating a int, you're allocating sizeof(int) bytes and getting the address of those bytes. You can then decide to use that block as an int, but you could also technically use that as an array of sizeof(int) chars.
The various alternatives (calloc, realloc) work roughly the same way (calloc is easier to use when dealing with arrays, and zero-fills the data, while realloc is useful when you need to resize a block of memory).
Suppose you create an integer array in a function and want to return it. Said array is a local variable to the function. You can't return a pointer to a local variable.
However, if you use malloc, you create an object on the heap whose scope exceeds the function body. You can return a pointer to that. You just have to destroy it later or you will have a memory leak.
It's because objects allocated with malloc() don't have names, so the only way to reference that object in code is to use a pointer to it.
When you say int x;, that creates an object with the name x, and it is referenceable through that name. When I want to set x to 10, I can just use x = 10;.
I can also set a pointer variable to point to that object with int *p = &x;, and then I can alternatively set the value of x using *p = 10;. Note that this time we can talk about x without specifically naming it (beyond the point where we acquire the reference to it).
When I say malloc(sizeof(int)), that creates an object that has no name. I cannot directly set the value of that object by name, since it just doesn't have one. However, I can set it by using a pointer variable that points at it, since that method doesn't require naming the object: int *p = malloc(sizeof(int)); followed by *p = 10;.
You might now ask: "So, why can't I tell malloc to give the object a name?" - something like malloc(sizeof(int), "x"). The answer to this is twofold:
Firstly, C just doesn't allow variable names to be introduced at runtime. It's just a basic restriction of the language;
Secondly, given the first restriction the name would have to be fixed at compile-time: if this is the case, C already has syntax that does what you want: int x;.
You are thinking about things wrong. It is not that int x is statically allocated and malloc(sizeof(int)) is dynamic. Both are allocated dynamically. That is, they are both allocated at execution time. There is no space reserved for them at the time you compile. The size may be static in one case and dynamic in the other, but the allocation is always dynamic.
Rather, it is that int x allocates the memory on the stack and malloc(sizeof(int)) allocates the memory on the heap. Memory on the heap requires that you have a pointer in order to access it. Memory on the stack can be referenced directly or with a pointer. Usually you do it directly, but sometimes you want to iterate over it with pointer arithmetic or pass it to a function that needs a pointer to it.
Everything works using pointers. "int x" is just a convenience - someone, somewhere got tired of juggling memory addresses and that's how programming languages with human-readable variable names were born.
Dynamic allocation is... dynamic. You don't have to know how much space you are going to need when the program runs - before the program runs. You choose when to do it and when to undo it. It may fail. It's hard to handle all this using the simple syntax of static allocation.
C was designed with simplicity in mind and compiler simplicity is a part of this. That's why you're exposed to the quirks of the underlying implementations. All systems have storage for statically-sized, local, temporary variables (registers, stack); this is what static allocation uses. Most systems have storage for dynamic, custom-lifetime objects and system calls to manage them; this is what dynamic allocation uses and exposes.
There is a way to do what you're asking and it's called C++. There, "MyInt x = 42;" is a function call or two.
I think your question comes down to this:
If a C programmer has the option of creating just int x or just int *x (both statically allocated)
The first statement allocates memory for an integer. Depending upon the placement of the statement, it might allocate the memory on the stack of a currently executing function or it might allocate memory in the .data or .bss sections of the program (if it is a global variable or static variable, at either file scope or function scope).
The second statement allocates memory for a pointer to an integer -- it hasn't actually allocated memory for the integer itself. If you tried to assign a value using the pointer *x=1, you would either receive a very quick SIGSEGV segmentation violation or corrupt some random piece of memory. C doesn't pre-zero memory allocated on the stack:
$ cat stack.c
#include <stdio.h>
int main(int argc, char *argv[]) {
int i;
int j;
int k;
int *l;
int *m;
int *n;
printf("i: %d\n", i);
printf("j: %d\n", j);
printf("k: %d\n", k);
printf("l: %p\n", l);
printf("m: %p\n", m);
printf("n: %p\n", n);
return 0;
}
$ make stack
cc stack.c -o stack
$ ./stack
i: 0
j: 0
k: 32767
l: 0x400410
m: (nil)
n: 0x4005a0
l and n point to something in memory -- but those values are just garbage, and probably don't belong to the address space of the executable. If we store anything into those pointers, the program would probably die. It might corrupt unrelated structures, though, if they are mapped into the program's address space.
m at least is a NULL pointer -- if you tried to write to it, the program would certainly die on modern hardware.
None of those three pointers actually point to an integer yet. The memory for those integers doesn't exist. The memory for the pointers does exist -- and is initially filled with garbage values, in this case.
The Wikipedia article on L-values -- mostly too obtuse to fully recommend -- makes one point that represented a pretty significant hurdle for me when I was first learning C: In languages with assignable variables it becomes necessary to distinguish between the R-value (or contents) and the L-value (or location) of a variable.
For example, you can write:
int a;
a = 3;
This stores the integer value 3 into whatever memory was allocated to store the contents of variable a.
If you later write:
int b;
b = a;
This takes the value stored in the memory referenced by a and stores it into the memory location allocated for b.
The same operations with pointers might look like this:
int *ap;
ap=malloc(sizeof int);
*ap=3;
The first ap= assignment stores a memory location into the ap pointer. Now ap actually points at some memory. The second assignment, *ap=, stores a value into that memory location. It doesn't update the ap pointer at all; it reads the value stored in the variable named ap to find the memory location for the assignment.
When you later use the pointer, you can choose which of the two values associated with the pointer to use: either the actual contents of the pointer or the value pointed to by the pointer:
int *bp;
bp = ap; /* bp points to the same memory cell as ap */
int *bp;
bp = malloc(sizeof int);
*bp = *ap; /* bp points to new memory and we copy
the value pointed to by ap into the
memory pointed to by bp */
I found assembly far easier than C for years because I found the difference between foo = malloc(); and *foo = value; confusing. I hope I found what was confusing you and seriously hope I didn't make it worse.
Perhaps you misunderstand the difference between declaring 'int x' and 'int *x'. The first allocates storage for an int value; the second doesn't - it just allocates storage for the pointer.
If you were to "dynamically allocate" a variable, there would be no point in the dynamic allocation anyway (unless you then took its address, which would of course yield a pointer) - you may as well declare it statically. Think about how the code would look - why would you bother with:
int x = malloc(sizeof(int)); *x = 0;
When you can just do:
int x = 0;