How can i know the size of the array using a pointer that is allocated using malloc?
#include <stdio.h>
int main(){
int *ptr = (int *)malloc(sizeof(int * 10));
printf("Size:%d",sizeof(ptr));
free(ptr_one);
return 0;
}
I get only the size of the pointer in this case which is 8.How to modify the code to get the size of array which will be 40.
You cannot.
You will need to do the bookkeeping and keep track of it yourself. With new you allocate dynamic memory and while deallocating the memory you just call delete, which knows how much memory it has deallocate because the language takes care of it internally so that users do not need to bother about the bookkeeping. If you still need it explicitly then you need to track it through separate variable.
if your machine is 32 bit you will get pointer size always 4 bytes of any data type
if your machine is 64 bit you will get pointer size always 8 bytes of any data type
if you declare static array you will get size by using sizeof
int a[10];
printf("Size:%lu",sizeof(a));
But you did not get the size of array which is blocked by pointer. where the memory to the block is allocated dynamically using malloc .
see this below code:
#include <stdio.h>
#include<stdlib.h>
int main()
{
int i;
int *ptr = (int *)malloc(sizeof(int) * 10);
// printf("Size:%lu",sizeof(ptr));
// In the above case sizeof operater returns size of pointer only.
for(i=1;ptr && i<13 ;i++,ptr++)
{
printf("Size:%d %p\n",((int)sizeof(i))*i,ptr);
}
return 0;
}
output:
Size:4 0x8ec010
Size:8 0x8ec014
Size:12 0x8ec018
Size:16 0x8ec01c
Size:20 0x8ec020
Size:24 0x8ec024
Size:28 0x8ec028
Size:32 0x8ec02c
Size:36 0x8ec030
Size:40 0x8ec034 //after this there is no indication that block ends.
Size:44 0x8ec038
Size:48 0x8ec03c
Related
I try something like below but all the time I have a segmentation fault.
I don't really want to use (e.g.) #define N 1000 and then declare int buffer[N].
Just in case..I'm not allowed to use any headers except stdio.h as well as dynamic memory.
void input (int *buffer, int *length);
int main()
{
int length, *buffer = NULL, *numbers = NULL;
input(buffer, &length);
}
void input(int *buffer, int *length) {
scanf("%d", length);
if (*length < 0) {
error = 1;
return;
}
for (int i = 0; i < *length; i++) {
scanf("%d", *buffer[i]);
}
}
How to pass an array with unknown 1-d dimension into function
In C, arrays cannot exist until their size is known.
There are other approaches though.
In C, code cannot pass an array to a function. some_function(some_array) converts the array some_array to the address of the first element of the array: &some_array[0]. That is what the function receives, a pointer, not an array. The original size information of the array is not passed, thus also pass the length to the function.
Sample:
Read desired length.
{
int length = 0;
scanf("%d", &length);
Form a variable length array, length >= 1.
if (length <= 0) {
return NULL;
}
int buffer[length];
Now call a function, passing the length and the address of the first element of the array.
// Do stuff with length and buf, like read data
foo1(length, buffer);
// foo1() receives the length & address of the first element of the array as an int *
// Do more stuff with length and buf, like write data
foo2(length, buffer);
}
At the end of the block }, buffer no longer available.
In C, you can't create an array if you can't know its size at compile time (or at least not in certain implementations and standards), so doing something like buffer[length] won't work (again at least not in certain implementations/standards).
What you need to do to make sure this works everywhere is to use a pointer (as I see you're trying to use here). However, what you're doing wrong here that causes your segfault with the pointers is you assign them the value of NULL. This also won't work due to how when you assign a pointer an arbitrary value, there is no memory allocated for the pointer (This applies for everything other than addresses of "regular" variables using the & operator and assigning other pointers that are checked to be OK). Your pointers are just pointing to address 0 and can't be used for anything.
What you need to do here to fix the pointers is to use dynamic memory allocation, so you can have a truly variable-sized array. Specifically, you need to use a function like malloc or calloc to allocate memory for the pointers so they are usable. In your case, using calloc and reading its documentation, we see that it takes 2 parameters: The number of elements it should allocate memory for and the size of each element. We also know that it returns a pointer to the starting address of the allocated memory and that in case of failure (which can only happen if you're out of memory), it returns NULL. Using this, we understand that in your case the call to calloc would be like this:
int *buffer = (int *) calloc(length, sizeof(int));
The sizeof() function returns the size of a data type in bytes. Here you allocated enough memory for the pointer to hold length integers (since you'll use it as an array you need enough memory for all the integers, you're not just pointing to 1 integer but storing all of them), and calloc is also noted to initialize every allocated element to 0, so you have an array of integers that are all initialized to 0 (Also note that type casting has been used to make sure the allocated memory block is appropriate for use with an integer array, you can read more about type casting in this small article from Tutorialspoint if you'd like). Then, after this has been allocated, you can start reading your integers into the array. The complete code looks like this:
void input (int *buffer, int *length);
int main() {
// NOTE: I don't see the numbers pointer used here, maybe remove it?
int length, *buffer, *numbers;
input(buffer, &length);
}
void input(int *buffer, int *length) {
scanf("%d", length);
if (*length < 0) {
// Consider printing the exact error here
error = 1;
return;
}
buffer = (int *) calloc(length, sizeof(int));
if (buffer == NULL) {
printf("Couldn't allocate memory for buffer\n");
error = 1;
return;
}
// Accessing the elements of an array doesn't need * and in fact * here can (and probably will) cause terrible things
for (int i = 0; i < *length; i++) {
scanf("%d", buffer[i]);
}
}
Also don't forget to call free() on the pointer after you're done using it to avoid memory leaks (in your case that'd be after the call to input()).
Hope this helped, good luck!
You cannot use arrays because their memory size must be known to the compiler at compile time. Also you can't use Variable Length Arrays because they are allocated at the point of declaration and deallocated when the block scope containing the declaration exits.
The solution to your problem might be to use malloc
I have a char array, we know that that a char size is 1 byte. Now I have to collect some char -> getchar() of course and simultaneously increase the array by 1 byte (without malloc, only library: stdio.h)
My suggestion would be, pointing to the array and somehow increase that array by 1 till there are no more chars to get OR you run out of Memory...
Is it possible to increase char array while using it, WITHOUT malloc?
No.
You cannot increase the size of a fixed size array.
For that you need realloc() from <stdlib.h>, which it seems you are not "allowed" to use.
Is it possible to increase char array while using it, WITHOUT malloc?
Quick answer: No it is not possible to increase the size of an array without reallocating it.
Fun answer: Don't use malloc(), use realloc().
Long answer:
If the char array has static or automatic storage class, it is most likely impossible to increase its size at runtime because keeping it at the same address that would require objects that are present at higher addresses to be moved or reallocated elsewhere.
If the array was obtained by malloc, it might be possible to extend its size if no other objects have been allocated after it in memory. Indeed realloc() to a larger size might return the same address. The problem is it is impossible to predict and if realloc returns a different address, the current space has been freed so pointers to it are now invalid.
The efficient way to proceed with this reallocation is to increase the size geometrically, by a factor at a time, 2x, 1.5x, 1.625x ... to minimize the number of reallocations and keep linear time as the size of the array grows linearly. You would a different variable for the allocated size of the array and the number of characters that you have stored into it.
Here is an example:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char *a = NULL;
size_t size = 0;
size_t count = 0;
int c;
while ((c = getchar()) != EOF && c != '\n') {
if (count >= size) {
/* reallocate the buffer to 1.5x size */
size_t newsize = size + size / 2 + 16;
char *new_a = realloc(a, new_size);
if (new_a == NULL) {
fprintf("out of memory for %zu bytes\n", new_size);
free(a);
return 1;
}
a = new_a;
size = new_size;
}
a[count++] = c;
}
for (i = 0; i < count; i++) {
putchar(a[i]);
}
free(a);
return 0;
}
There are two ways to create space for the string without using dynamic memory allocation(malloc...). You can use a static array or an array with automatic storage duration, you need to specify a maximum amount, you might never reach. But always check against it.
#define BUFFER_SIZE 0x10000
Static
static char buffer[BUFFER_SIZE];
Or automatic (You need to ensure BUFFER_SIZE is smaller than the stack size)
int main() {
char buffer[BUFFER_SIZE];
...
};
There are also optimizations done by the operating system. It might lazily allocate the whole (static/automatic) buffer, so that only the used part is in the physical memory. (This also applies to the dynamic memory allocation functions.) I found out that calloc (for big chunks) just allocates the virtual memory for the program; memory pages are cleared only, when they are accessed (probably through some interrupts raised by the cpu). I compared it to an allocation with malloc and memset. The memset does unnessecary work, if not all bytes/pages of the buffer are accessed by the program.
If you cannot allocate a buffer with malloc..., create a static/automatic array with enough size and let the operating system allocate it for you. It does not occupy the same space in the binary, because it is just stored as a size.
I'm currently studying variable length array and automatic storage.
I have the following code that allocate memory for an variable length array myArray inside function vla, and return a pointer to the variable length array from the function.
#include <stdio.h>
int * vla(int n){
int myArray[n];
myArray[0] = 10;
myArray[1] = 11;
int * pointerToInt = myArray;
return pointerToInt;
}
int main(void){
int * pointerToInt = vla(10);
printf("%d, %d", pointerToInt[0], pointerToInt[1]); // prints 10, 11
return 0;
}
I thought that variable length array belong to the automatic storage class (i.e. the memory for the variable length array will be allocated when we enter the function containing the variable length array, and the memory is automatically deallocated after the function exit)
So according to this logic, the memory allocated to myArray variable length array is deallocated after we return from vla method, but how come I can still correctly access the first and second element of the variable length array?
Is this behavior defined? or it is undefined behaviour that just happen to work?
myArray is a stack/auto variable created on the stack memory. Remember memory always exists. It is just owned by different pointers based on allocation and deallocation. The reason why you can still access same values is that the same piece of memory has not been assigned to another pointer and not been overwritten.
To evaluate it. Create another function that allocates same amount from stack but puts different values. Or add arguments in the same function and call it twice with different values. You will then see the difference.
#include <stdio.h>
int * vla(int n, int a, int b){
int myArray[n];
myArray[0] = a;
myArray[1] = b;
int * pointerToInt = myArray;
return pointerToInt;
}
int main(void){
int * pointerToInt = vla(10, 10, 11);
vla(10, 20, 21); // over write stack
printf("%d, %d", pointerToInt[0], pointerToInt[1]); // prints 20, 21
return 0;
}
By the way returning stack memory from vla is not a good idea. Dynamic memory is allocated from heap using malloc family of functions.
You can still correctly access the first and second element of the variable length array because you are assigning base address of the myArray to pointerToInt. Auto variables have a life inside the block only, but in this program we are using pointer to access the data in the memory, as long as that part of stack is not allocated to any other program, we can access that part of stack. If that part of stack is allocated to some other process we will get segmentation fault as we are trying to access unauthorized memory
So, to start off I've already looked at a few questions including this one and none of them seem to help.
I'm simply trying to write a function that extends the size of an array using realloc().
My code currently looks like this:
unsigned char *xtnd = malloc(4);
xtndc(&xtnd, 4);
// sizeof(*xtnd) should now be 8
void xtndc ( unsigned char ** bytesRef , uint8_t count ) {
*bytesRef = realloc(*bytesRef, (sizeof(**bytesRef)) + count);
}
But no matter what I do it seems that the size of xtnd is always 4. After running xtndc() on it it should now be 8 bytes long.
Any suggestions?
The type of **bytesRef is unsigned char, so sizeof(**bytesRef) is 1. sizeof doesn't keep track of dynamic allocations, it's a compile time tool that gives you the size of a type, in this case unsigned char.
You have to keep track of the array size manually to calculate the new required size.
Your program does in fact change the size of the memory block. It changes the size of your original memory block from 4 bytes to 5 bytes. It changes to 5 bytes because you are essentially doing sizeof(unsigned char) + 4 which 1 + 4 = 5. If you want to double the size instead, do count*sizeof(unsigned char) + count. There are two points to be noted here:
The sizeof function returns the size of the data type, not the size of the allocated bytes. There is no way to know the size of the dynamically allocated memory.
The function realloc (and malloc and calloc as well) is not always guaranteed to return the requested reallocation. It may or may not succeed all the time.
I fixed the problem with the following code.
typedef struct CArrPtr {
unsigned char* ptr;
size_t size;
} CArrPtr;
void xtndc ( CArrPtr *bytesRef, uint8_t count );
. . .
CArrPtr xtnd = { .ptr = malloc(4), .size = 4 };
xtndc( &xtnd, 4 );
// xtnd.size is now 8 bytes
. . .
void xtndc ( CArrPtr *bytesRef, uint8_t count ) {
unsigned char *nptr;
if((nptr = realloc(bytesRef->ptr, bytesRef->size + count)) != 0)
{
bytesRef->ptr = nptr;
bytesRef->size = bytesRef->size + count;
}
}
As I am somewhat new to C, what I learned from this is that malloc specifically creates a pointer to a memory block, but you have no direct access to information about the memory block. Instead, you must store the size of the array that you created with malloc somewhere as well.
Since in the past I'd been initializing arrays with unsigned char arr[size]; and then using sizeof on it, I was under the impression that sizeof returned the size of the array, which is of course wrong as it gives you the size of a type.
Glad I could learn something from this.
sizeof is used to calculate size of data type or array. Pointer and array are very similar, but they are different things. For int *ap, sizeof(ap) will return 4 on x86, sizeof(*ap) will return 4; for int a[10], sizeof(a) will return 40.
sizeof expression is processed at compile time, so it will be a constant written into the executable file before you run the program.
malloc and realloc don't maintain size.
If realloc succeeds, it will reallocate the requested size. So you don't need to check the size after realloc returns, but you should check the return value of realloc to ensure that realloc succeeds.
I allocate a 2D array of char * and every string length is 12.
50 rows and 2000000 columns.
Lets calculate it:
50*2000000 * (12(length)+8(for pointer)). I use 64 bit.
50*2000000 * 20 =2000000000 bits .. -> 2 GB.
When I check the memory monitor it shows that the process takes 4 GB.
(All that happened after allocation)
This is the code:
int col=2000000,row=50,i=0,j=0;
char *** arr;
arr=(char***)malloc(sizeof(char**)*row);
for(i=0;i<row;i++)
{
arr[i]=(char ** )malloc(sizeof(char*)*col);
for(j=0;j<col;j++)
{
arr[i][j]=(char*)malloc(12);
strcpy(arr[i][j],"12345678901");
arr[i][j][11]='\0';
}
}
May that be from the paging in Linux?
Each call of malloc is taking more memory than you ask. Malloc needs to store somewhere its internal info about allocated place, like size of allocated space, some info about neighbors chunks, etc. Also (very probably) each returned pointer is aligned to 16 bytes. In my estimation each allocation of 12 bytes takes 32 bytes of memory. If you want to save memory allocate all strings in one malloc and split them into sizes per 12 at your own.
Try the following:
int col=2000000,row=50,i=0,j=0;
char *** arr;
arr= malloc(sizeof(*arr)*row);
for(i=0;i<row;i++)
{
arr[i]= malloc(sizeof(*arr[i])*col);
char *colmem = malloc(12 * col);
for(j=0;j<col;j++)
{
arr[i][j] = colmem + j*12;
strcpy(arr[i][j],"12345678901");
}
}
I would re-write the code from scratch. For some reason, around 99% of all C programmers don't know how to correctly allocate true 2D arrays dynamically. I'm not even sure I'm one of the 1% who do, but lets give it a shot:
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
int main()
{
const int COL_N = 2000000;
const int ROW_N = 50;
char (*arr)[ROW_N] = malloc( sizeof(char[COL_N][ROW_N]) );
if(arr == NULL)
{
printf("Out of memory");
return 0;
}
for(int row=0; row<ROW_N; row++)
{
strcpy(arr[row], "12345678901");
puts(arr[row]);
}
free(arr);
return 0;
}
The important parts here are:
You should always allocate multi-dimensional arrays in adjacent memory cells or they are not arrays, but rather pointer-based lookup tables. Thus you only need one single malloc call.
This should save a bit of memory since you only need one pointer and it is allocated on the stack. No pointers are allocated on the heap.
Casting the return value of malloc is pointless (but not dangerous on modern compilers).
Ensure that malloc actually worked, particularly when allocating ridiculous amounts of memory.
strcpy copies the null termination, you don't need to do it manually.
There is no need for nested loops. You want to allocate a 2D array, not a 3D one.
Always clean up your own mess with free(), even though the OS might do it for you.