用malloc分配的数组的大小显示比预期少一

Question

I have this:

alloc(Btree *bt, uint8_t *keylen, int16_t n)
{

bt->node[n].key = malloc(sizeof(int16_t)*(*keylen));
{

Where bt->node[n].key is a pointer to int16_t .

With the debugger running, I can verify that keylen is 5 .

But if I put:

int kl = sizeof(bt->node[n].key) / sizeof(bt->node[n].key[0])

kl is 4 .

What did I do wrong?

Answer 1

Look carefully, you are confusing the pointer with the array:

Where bt->node[n].key is a pointer to int16_t.

Thus, bt->node[n].key is a pointer to the allocated memory, not the allocated memory itself, and sizeof bt->node[n].key is sizeof <pointer to ...> which, in your system is 8 (64 bits).

8 / sizeof uint16_t = 8 / 2 = 4

You can not check the size of the allocated memory chunk, you have to trust malloc() to work well or return NULL if it can't.

Answer 2

sizeof operator produces the size of a type , not the amount of memory allocated to a pointer .

In your case, what happenning is

• key is of type int16_t *
• sizeof(bt->node[n].key) gives sizeof(int16_t *) which on 64 bit, 8
• sizeof(bt->node[n].key[0]) which is sizeof(int16_t ) , which is 2

Ultimately, it's 8/2 = 4.

There is absolutely no measurement of the amount of memory returned by malloc() .

Answer 3

sizeof(bt->node[n].key) is sizeof(uint16_t*) which could be 8 (on 64 bits) sizeof(bt->node[n].key[0]) is sizeof(*(uint16_t*)) which is 2

And 8 / 2 equals 4.