i have pointer of array(array which is in memory). Can i calculate the size of array from its pointer ? i dont know actually where is the array in memory. i only getting pointer adress(suppose 9001) using that adress i have to calculate array size.
Thanks.
No, you cannot calculate the size of the array. Objects in C do not carry type information, so you must arrange to know the size of the array ahead of time. Something like this:
void my_function(int array[], int size);
int array[] as a function parameter. In principle it might help readers, by telling them that array is a pointer to the first of one or more ints. In practice: (1) size already tells them that; (2) people are used to using char* when passing a pointer to the first character of a string, not char[], so they can cope with int* too; (3) the different syntax frequently misleads newbies into thinking it's somehow different from int *array.
You cannot do this in C. The size of a pointer is the size of a pointer, not the size of any array it may be pointing at.
If you end up with a pointer pointing to an array (either explicitly with something like char *pch = "hello"; or implicitly with array decay such as passing an array to a function), you'll need to hold the size information separately, with something like:
int twisty[] = [3,1,3,1,5,9];
doSomethingWith (twisty, sizeof(twisty)/sizeof(*twisty));
:
void doSomethingWith (int *passages, size_t sz) { ... }
The following code illustrates this:
#include <stdio.h>
static void fn (char plugh[], size_t sz) {
printf ("sizeof(plugh) = %d, sz = %d\n", sizeof(plugh), sz);
}
int main (void) {
char xyzzy[] = "Pax is a serious bloke!";
printf ("sizeof(xyzzy) = %d\n", sizeof(xyzzy));
fn (xyzzy, sizeof(xyzzy)/sizeof(*xyzzy));
return 0;
}
The output on my system is:
sizeof(xyzzy) = 24
sizeof(plugh) = 4, sz = 24
because the 24-byte array is decayed to a 4-byte pointer in the function call.
sz in fn() be 6?
"Pax is a serious bloke!". I suspect you're confused with the integer array above that.xyzzy = [3,1,3,1,5,9] from your previous example. Maybe use the same xyzzy value in both examples?No. arrays in C doesn't have this info. you should hold a variable "next" to the array with this data.
if you have the array it self you can use the sizeof to get his size in compilation stage.
char array1[XXX];
char* pointer1 = array1;
sizeof(array1); // this is XXX
sizeof(pointer1); // this is the size of a pointer in your system
you should do the following and use the variable in you program (or pass it to function when you passing the array to a function)
int array1size = sizeof(array1);
[C]or[C++]but not both. For this question, some answers applicable to one language will not apply to the other.