Dereference FFI pointer in Python to get underlying array

Dereference FFI pointer in Python to get underlying array

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Dereference FFI pointer in Python to get underlying array
Tag : python , By : simonth
Date : December 05 2020, 12:23 PM

this one helps. The cast should be the first warning sign. Why do you have to cast from the type to what should be the same type? This is because there are simple typos:
lib.rle_values.restype = POINTER(c_int32)    
lib.rle_values_size.restype = c_int32
def __str__(self):
    values_size = lib.rle_values_size(self.obj)
    print(values_size, "values_size")

    values_pointer = lib.rle_values(self.obj)
    print("values_pointer:", values_pointer)

    thing = values_pointer[:values_size]
    return str(thing)
pub extern "C" fn rle_values(rle: *mut Rle) -> *mut int32_t {
    let mut rle = unsafe { &mut *rle };
$ LD_LIBRARY_PATH=$PWD/target/debug/ python3 main.py
30 values_size
values_pointer: <__main__.LP_c_int object at 0x10f124048>
[1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2]
pub extern "C" fn rle_values_size(rle: *mut Rle) -> int32_t {
    match unsafe { rle.as_ref() } {
        Some(rle) => rle.values.len() as i32,
        None => 0,

pub extern "C" fn rle_values(rle: *mut Rle) -> *mut int32_t {
    match unsafe { rle.as_mut() } {
        Some(mut rle) => rle.values.as_mut_ptr(),
        None => ptr::null_mut(),

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Why can't I dereference a pointer to multidimensional array?

Tag : c , By : Jim F.
Date : March 29 2020, 07:55 AM
I wish this help you You can dereference it, it is just that the result is not going to be what you'd expect: *arr is not an int, it's a pointer to an int (OK, a one-dimensional array). If you want to see 1 printed, add another star:
printf("%d ", **arr);

Dereference 2D Array Pointer In C

Tag : objective-c , By : Joshua Johnson
Date : March 29 2020, 07:55 AM
Hope this helps In this case, the answer is no -- since you simply defined array (didn't use something like malloc to allocate it) you don't have to do anything to free it either. If it was local (defined inside a function) it'll be freed automatically when you exit the function. If you defined it outside any function, it's a global, so it'll exist the entire time the program runs. Either way, you don't have to d any explicit memory management.

Assign and dereference void * to an array pointer in C

Tag : c , By : nabbed
Date : March 29 2020, 07:55 AM
like below fixes the issue I have a pointer to array of fixed size integer elements. After populating that array, I assigned it to void *pBuff. Later on, I need to access array elements through void pointer which I failed in doing so. , Apart from the fact that you need to use:
pBuff = pPB;

Why in C++, I don't need to dereference a pointer to an array in order to access items in the array

Tag : cpp , By : user184975
Date : March 29 2020, 07:55 AM
it helps some times I'm currently learning pointer. And when I create a pointer to an array with int type, I found that I can directly index the pointer without deferencing the pointer and the compiler still output exact items in my array. I do not understand why this works and why we don't need to first deference the pointer. ,
Code that without dereference
[ code ]
 theArray[i] = i;
 *(theArray+i) = i;
 *p = q;
 p[0] = q;

how to understand that dereference a pointer to array get a pointer to element?

Tag : cpp , By : Revision17
Date : March 29 2020, 07:55 AM
will be helpful for those in need Firstly, you should understand array-to-pointer decay,
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