Swap two arrays (using pointers, not copy)

Question

I've got two arrays, defined as:

uint8_t array1[10];
uint8_t array2[10];

I'd like to swap the arrays (actually the pointers to each array) using code like this:

uint8_t *tmp = array1;
array1 = array2;
array2 = tmp;

However, I keep getting errors like:

invalid array assignment

for the last two lines. I don't recall this being such a problem in traditional C++. I've tried all sorts of casting to no avail. Is there something I'm missing?

I can get the code to run properly if I switch the declarations for the arrays to use new and have them just be more-or-less pointers rather than arrays, but I'd rather not have to do that:

uint8_t *array1 = new uint8_t[10];
uint8_t *array2 = new uint8_t[10];

Code is running on an ARM-based SAMD51 (Cortex M4 processor).

Thanks for any tips!

Andrew M. · Accepted Answer · 2020-05-14 03:50:16Z

2

Figured it out. Though using new works, I wanted to keep the buffers out of the heap. The following declarations work with the swap code:

uint8_t a1[10];
uint8_t a2[10];
uint8_t *array1 = a1;
uint8_t *array2 = a2;

uint8_t *tmp = array1;
array1 = array2;
array2 = tmp;

answered May 14, 2020 at 3:50

Andrew M.

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