For a char []
, I can easily get its length by:
char a[] = \"aaaaa\";
int length = sizeof(a)/sizeof(char); // length=6
However,
If the char *
is 0-terminated, you can use strlen
Otherwise, there is no way to determine that information
Just use std::vector<char>
which keep the (dynamic) size for you. (Bonus, memory management for free).
Or std::array<char, 10>
which keep the (static) size.
Create a structure to keep the info, something like:
typedef struct {
char* ptr;
int size;
} my_array;
my_array malloc_array(int size)
{
my_array res;
res.ptr = (char*) malloc(size);
res.size = size;
return res;
}
void free_array(my_array array)
{
free(array.ptr);
}
when new allocates an array, depending on the compiler (i use gnu c++), the word in front of the array contains information about the number of bytes allocated.
The test code:
#include <stdio.h>
#include <stdlib.h>
int
main ()
{
int arraySz;
char *a;
unsigned int *q;
for (arraySz = 5; arraySz <= 64; arraySz++) {
printf ("%02d - ", arraySz);
a = new char[arraySz];
unsigned char *p = (unsigned char *) a;
q = (unsigned int *) (a - 4);
printf ("%02d\n", (*q));
delete[] (a);
}
}
on my machine dumps out:
05 - 19
06 - 19
07 - 19
08 - 19
09 - 19
10 - 19
11 - 19
12 - 19
13 - 27
14 - 27
15 - 27
16 - 27
17 - 27
18 - 27
19 - 27
20 - 27
21 - 35
22 - 35
23 - 35
24 - 35
25 - 35
26 - 35
27 - 35
28 - 35
29 - 43
30 - 43
31 - 43
32 - 43
33 - 43
34 - 43
35 - 43
36 - 43
37 - 51
38 - 51
39 - 51
40 - 51
41 - 51
42 - 51
43 - 51
44 - 51
45 - 59
46 - 59
47 - 59
48 - 59
49 - 59
50 - 59
51 - 59
52 - 59
53 - 67
54 - 67
55 - 67
56 - 67
57 - 67
58 - 67
59 - 67
60 - 67
61 - 75
62 - 75
63 - 75
64 - 75
I would not recommend this solution (vector is better), but if you are really desperate, you could find a relationship and be able to conclude the number of bytes allocated from the heap.