Hi All
I have an array like this
Now if I want to determine its length likeCode:cl_device_id *devices ; devices = (cl_device_id*)malloc(sizeof(cl_device_id) * 2)
and I get 1 (it should be 2). So I do something wrong, any suggestions ?Code:int len = sizeof( devices ) / sizeof( cl_device_id ) ;
Cheers
LuCa
If you use sizeof on a pointer, it will return the size of the pointer, not the size of whatever the pointer points to. You're going to just have to keep track of how many devices you allocated memory for.
If you understand what you're doing, you're not learning anything.
but why does this work:
array is a pointer too!Code:int array[6]= { 1, 2, 3, 4, 5, 6 }; int len=sizeof(array)/sizeof(int);
Arrays decay to pointer, so....
How to find the last value in an array
int len = *((*(&array+1))-1);
Last edited by slingerland3g; 12-23-2009 at 04:32 PM.
No, array is not a pointer! It's an array.Originally Posted by jeanluca
Check this out
If you understand what you're doing, you're not learning anything.
An array is not a pointer.
The wrinkle is that, in some contexts, the name of an array can be used as if it is a pointer.
If you do this;
The size of a pointer (which is simply a variable that contains a memory address) is compiler-dependent. The size of a pointer doesn't change if the pointer happens to contain the address of the first element of an array.Code:#include <stdio.h> int main() { int array[] ={ 1, 2, 3, 4, 5, 6 }; int *pointer = array; /* in this context array is a pointer */ fprintf(stdout, "sizeof(array)/sizeof(int) %u\n", sizeof(array)/sizeof(int)); /* will print 6 */ fprintf(stdout, "sizeof(pointer)/sizeof(int) %u\n", sizeof(pointer)/sizeof(int)); /* will print something else */ }
Anywhere a pointer to T can be used, an array of T can be used. The result is a pointer to the first element of the array. This doesn't mean that an array is a pointer.
The original question is, given a pointer to some malloc'd array, how do you tell how long it is? The answer is you can't. Unless you stash the information away yourself, it's lost.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
So in a few words the
doesn't work when you dynammically allocate arrays. For the reasons already said.Code:length = sizeof(array) / sizeof(type);
In C you just have to be a bit more creative. With proper naming you can solve this kind of things. Just add the line:
and you the _len moto whenever you want the length of a variable.Code:cl_device_id *devices ; devices = (cl_device_id*)malloc(sizeof(cl_device_id) * 2) int devices_len = 2;
If you want an alternative solution, you can combin the devices and devices_len into a struct. Then you simply make your malloc for the struct like:
where cl_device_o your new struct. With a few "replace-all"s you could modify your existing code and be able to find the length of your dynamically allocated pointerish arrays.Code:cl_device_o* createDev(int length) { dev* = malloc(lenght); dev->len = length; return dev; }
Of course the first solution is easier and more safe.
ok, thats all clear!. Thanks a lot for your answers!!
cheers
LuCa
