How to find the 'sizeof' (a pointer pointing to an array)?

Question

First off, here is some code:

int main() 
{
    int days[] = {1,2,3,4,5};
    int *ptr = days;
    printf("%u\n", sizeof(days));
    printf("%u\n", sizeof(ptr));

    return 0;
}

Is there a way to find out the size of the array that ptr is pointing to (instead of just giving its size, which is four bytes on a 32-bit system)?

Answer 1

No, you can't. The compiler doesn't know what the pointer is pointing to. There are tricks, like ending the array with a known out-of-band value and then counting the size up until that value, but that's not using sizeof().

Another trick is the one mentioned by Zan, which is to stash the size somewhere. For example, if you're dynamically allocating the array, allocate a block one int bigger than the one you need, stash the size in the first int, and return ptr+1 as the pointer to the array. When you need the size, decrement the pointer and peek at the stashed value. Just remember to free the whole block starting from the beginning, and not just the array.

Answer 2

The answer is, "No."

What C programmers do is store the size of the array somewhere. It can be part of a structure, or the programmer can cheat a bit and malloc() more memory than requested in order to store a length value before the start of the array.

Answer 3

For dynamic arrays (malloc or C++ new) you need to store the size of the array as mentioned by others or perhaps build an array manager structure which handles add, remove, count, etc. Unfortunately C doesn't do this nearly as well as C++ since you basically have to build it for each different array type you are storing which is cumbersome if you have multiple types of arrays that you need to manage.

For static arrays, such as the one in your example, there is a common macro used to get the size, but it is not recommended as it does not check if the parameter is really a static array. The macro is used in real code though, e.g. in the Linux kernel headers although it may be slightly different than the one below:

#if !defined(ARRAY_SIZE)
    #define ARRAY_SIZE(x) (sizeof((x)) / sizeof((x)[0]))
#endif

int main()
{
    int days[] = {1,2,3,4,5};
    int *ptr = days;
    printf("%u\n", ARRAY_SIZE(days));
    printf("%u\n", sizeof(ptr));
    return 0;
}

You can google for reasons to be wary of macros like this. Be careful.

If possible, the C++ stdlib such as vector which is much safer and easier to use.

Answer 4

There is a clean solution with C++ templates, without using sizeof(). The following getSize() function returns the size of any static array:

#include <cstddef>

template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
    return SIZE;
}

Here is an example with a foo_t structure:

#include <cstddef>

template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
    return SIZE;
}

struct foo_t {
    int ball;
};

int main()
{
    foo_t foos3[] = {{1},{2},{3}};
    foo_t foos5[] = {{1},{2},{3},{4},{5}};
    printf("%u\n", getSize(foos3));
    printf("%u\n", getSize(foos5));

    return 0;
}

Output:

3
5

Answer 5

As all the correct answers have stated, you cannot get this information from the decayed pointer value of the array alone. If the decayed pointer is the argument received by the function, then the size of the originating array has to be provided in some other way for the function to come to know that size.

Here's a suggestion different from what has been provided thus far,that will work: Pass a pointer to the array instead. This suggestion is similar to the C++ style suggestions, except that C does not support templates or references:

#define ARRAY_SZ 10

void foo (int (*arr)[ARRAY_SZ]) {
    printf("%u\n", (unsigned)sizeof(*arr)/sizeof(**arr));
}

But, this suggestion is kind of silly for your problem, since the function is defined to know exactly the size of the array that is passed in (hence, there is little need to use sizeof at all on the array). What it does do, though, is offer some type safety. It will prohibit you from passing in an array of an unwanted size.

int x[20];
int y[10];
foo(&x); /* error */
foo(&y); /* ok */

If the function is supposed to be able to operate on any size of array, then you will have to provide the size to the function as additional information.

Answer 6

For this specific example, yes, there is, IF you use typedefs (see below). Of course, if you do it this way, you're just as well off to use SIZEOF_DAYS, since you know what the pointer is pointing to.

If you have a (void *) pointer, as is returned by malloc() or the like, then, no, there is no way to determine what data structure the pointer is pointing to and thus, no way to determine its size.

#include <stdio.h>

#define NUM_DAYS 5
typedef int days_t[ NUM_DAYS ];
#define SIZEOF_DAYS ( sizeof( days_t ) )

int main() {
    days_t  days;
    days_t *ptr = &days; 

    printf( "SIZEOF_DAYS:  %u\n", SIZEOF_DAYS  );
    printf( "sizeof(days): %u\n", sizeof(days) );
    printf( "sizeof(*ptr): %u\n", sizeof(*ptr) );
    printf( "sizeof(ptr):  %u\n", sizeof(ptr)  );

    return 0;
} 

Output:

SIZEOF_DAYS:  20
sizeof(days): 20
sizeof(*ptr): 20
sizeof(ptr):  4

Answer 7

There is no magic solution. C is not a reflective language. Objects don't automatically know what they are.

But you have many choices:

1. Obviously, add a parameter
2. Wrap the call in a macro and automatically add a parameter
3. Use a more complex object. Define a structure which contains the dynamic array and also the size of the array. Then, pass the address of the structure.

Answer 8

You can do something like this:

int days[] = { /*length:*/5, /*values:*/ 1,2,3,4,5 };
int *ptr = days + 1;
printf("array length: %u\n", ptr[-1]);
return 0;

Answer 9

My solution to this problem is to save the length of the array into a struct Array as a meta-information about the array.

#include <stdio.h>
#include <stdlib.h>

struct Array
{
    int length;

    double *array;
};

typedef struct Array Array;

Array* NewArray(int length)
{
    /* Allocate the memory for the struct Array */
    Array *newArray = (Array*) malloc(sizeof(Array));

    /* Insert only non-negative length's*/
    newArray->length = (length > 0) ? length : 0;

    newArray->array = (double*) malloc(length*sizeof(double));

    return newArray;
}

void SetArray(Array *structure,int length,double* array)
{
    structure->length = length;
    structure->array = array;
}

void PrintArray(Array *structure)
{       
    if(structure->length > 0)
    {
        int i;
        printf("length: %d\n", structure->length);
        for (i = 0; i < structure->length; i++)
            printf("%g\n", structure->array[i]);
    }
    else
        printf("Empty Array. Length 0\n");
}

int main()
{
    int i;
    Array *negativeTest, *days = NewArray(5);

    double moreDays[] = {1,2,3,4,5,6,7,8,9,10};

    for (i = 0; i < days->length; i++)
        days->array[i] = i+1;

    PrintArray(days);

    SetArray(days,10,moreDays);

    PrintArray(days);

    negativeTest = NewArray(-5);

    PrintArray(negativeTest);

    return 0;
}

But you have to care about set the right length of the array you want to store, because the is no way to check this length, like our friends massively explained.

Answer 10

No, you can't use sizeof(ptr) to find the size of array ptr is pointing to.

Though allocating extra memory(more than the size of array) will be helpful if you want to store the length in extra space.

Answer 11

int main() 
{
    int days[] = {1,2,3,4,5};
    int *ptr = days;
    printf("%u\n", sizeof(days));
    printf("%u\n", sizeof(ptr));

    return 0;
}

Size of days[] is 20 which is no of elements * size of it's data type. While the size of pointer is 4 no matter what it is pointing to. Because a pointer points to other element by storing it's address.

Answer 12

 #define array_size 10

 struct {
     int16 size;
     int16 array[array_size];
     int16 property1[(array_size/16)+1]
     int16 property2[(array_size/16)+1]
 } array1 = {array_size, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9};

 #undef array_size

array_size is passing to the size variable:

#define array_size 30

struct {
    int16 size;
    int16 array[array_size];
    int16 property1[(array_size/16)+1]
    int16 property2[(array_size/16)+1]
} array2 = {array_size};

#undef array_size

Usage is:

void main() {

    int16 size = array1.size;
    for (int i=0; i!=size; i++) {

        array1.array[i] *= 2;
    }
}

Answer 13

In strings there is a '\0' character at the end so the length of the string can be gotten using functions like strlen. The problem with an integer array, for example, is that you can't use any value as an end value so one possible solution is to address the array and use as an end value the NULL pointer.

#include <stdio.h>
/* the following function will produce the warning:
 * ‘sizeof’ on array function parameter ‘a’ will
 * return size of ‘int *’ [-Wsizeof-array-argument]
 */
void foo( int a[] )
{
    printf( "%lu\n", sizeof a );
}
/* so we have to implement something else one possible
 * idea is to use the NULL pointer as a control value
 * the same way '\0' is used in strings but this way
 * the pointer passed to a function should address pointers
 * so the actual implementation of an array type will
 * be a pointer to pointer
 */
typedef char * type_t; /* line 18 */
typedef type_t ** array_t;
int main( void )
{
    array_t initialize( int, ... );
    /* initialize an array with four values "foo", "bar", "baz", "foobar"
     * if one wants to use integers rather than strings than in the typedef
     * declaration at line 18 the char * type should be changed with int
     * and in the format used for printing the array values 
     * at line 45 and 51 "%s" should be changed with "%i"
     */
    array_t array = initialize( 4, "foo", "bar", "baz", "foobar" );

    int size( array_t );
    /* print array size */
    printf( "size %i:\n", size( array ));

    void aprint( char *, array_t );
    /* print array values */
    aprint( "%s\n", array ); /* line 45 */

    type_t getval( array_t, int );
    /* print an indexed value */
    int i = 2;
    type_t val = getval( array, i );
    printf( "%i: %s\n", i, val ); /* line 51 */

    void delete( array_t );
    /* free some space */
    delete( array );

    return 0;
}
/* the output of the program should be:
 * size 4:
 * foo
 * bar
 * baz
 * foobar
 * 2: baz
 */
#include <stdarg.h>
#include <stdlib.h>
array_t initialize( int n, ... )
{
    /* here we store the array values */
    type_t *v = (type_t *) malloc( sizeof( type_t ) * n );
    va_list ap;
    va_start( ap, n );
    int j;
    for ( j = 0; j < n; j++ )
        v[j] = va_arg( ap, type_t );
    va_end( ap );
    /* the actual array will hold the addresses of those
     * values plus a NULL pointer
     */
    array_t a = (array_t) malloc( sizeof( type_t *) * ( n + 1 ));
    a[n] = NULL;
    for ( j = 0; j < n; j++ )
        a[j] = v + j;
    return a;
}
int size( array_t a )
{
    int n = 0;
    while ( *a++ != NULL )
        n++;
    return n;
}
void aprint( char *fmt, array_t a )
{
    while ( *a != NULL )
        printf( fmt, **a++ );   
}
type_t getval( array_t a, int i )
{
    return *a[i];
}
void delete( array_t a )
{
    free( *a );
    free( a );
}

Answer 14

This is how I personally do it in my code. I like to keep it as simple as possible while still able to get values that I need.

typedef struct intArr {
    int size;
    int* arr; 
} intArr_t;

int main() {
    intArr_t arr;
    arr.size = 6;
    arr.arr = (int*)malloc(sizeof(int) * arr.size);

    for (size_t i = 0; i < arr.size; i++) {
        arr.arr[i] = i * 10;
    }

    return 0;
}