如何有效地将uint16_t [2N]转换为uint32_t [N]？

Question

I've got an array of six uint16_t which is actually three uint32_t , with the bits in the right order and all. How can I cast the former to the latter as effectively as possible?

The number of elements in the array is known at compile-time.

Answer 1

Like this perhaps:

uint16_t arr16[6];
uint32_t *parr32 = (uint32_t*)(&arr16);

And now you can use parr32[i] to refer to elements of the overlayed arr16 array.

Answer 2

On your little-endian machine, your bytes are arranged as

b a d c f e h g

etc., so casting a pointer to the first int16_t, which points at a , to an int32_t* will give the wrong result (or possibly even crash) since it's in the middle of the first int32_t. On a big-endian machine, where the bytes are arranged as

a b c d e f g h

then casting the int16_t* , which points to a , to an int32_t* would yield the right result.