How to get the real and total length of char * (ch

Question

For a char [], I can easily get its length by:

char a[] = "aaaaa";
int length = sizeof(a)/sizeof(char); // length=6

However, I cannot do like this to get the length of a char * by:

char *a = new char[10];
int length = sizeof(a)/sizeof(char);

because, I know, a here is a pointer, such that length here will be always be 4 (or something other in different systems).

My question is that how can I get the length of a char * afterwards? I know someone may challenge me that you already know its 10 because you just created it. I want to know this because this step of getting its length may come long long way from its creation and I don't want to come long long way back to check this number. Moreover, I also want to know its real length.

To be more specific

• how can I get its real length=5?
• how can I get its total length=10?

for the following example:

char *a = new char[10]; 
strcpy(a, "hello");

Answer 1

You can implement your own new and delete functions, as well as an additional get-size function:

#define CEIL_DIV(x,y) (((x)-1)/(y)+1)

void* my_new(int size)
{
    if (size > 0)
    {
        int* ptr = new int[1+CEIL_DIV(size,sizeof(int))];
        if (ptr)
        {
            ptr[0] = size;
            return ptr+1;
        }
    }
    return 0;
}

void my_delete(void* mem)
{
    int* ptr = (int*)mem-1;
    delete ptr;
}

int my_size(void* mem)
{
    int* ptr = (int*)mem-1;
    return ptr[0];
}

Alternatively, you can override the new and delete operators in a similar manner.

Answer 2

You can't. Not with 100% accuracy, anyway. The pointer has no length/size but its own. All it does is point to a particular place in memory that holds a char. If that char is part of a string, then you can use strlen to determine what chars follow the one currently being pointed to, but that doesn't mean the array in your case is that big.
Basically:

A pointer is not an array, so it doesn't need to know what the size of the array is. A pointer can point to a single value, so a pointer can exist without there even being an array. It doesn't even care where the memory it points to is situated (Read only, heap or stack... doesn't matter). A pointer doesn't have a length other than itself. A pointer just is...
Consider this:

char beep = '\a';
void alert_user(const char *msg, char *signal); //for some reason
alert_user("Hear my super-awsome noise!", &beep); //passing pointer to single char!

void alert_user(const char *msg, char *signal)
{
    printf("%s%c\n", msg, *signal);
}

A pointer can be a single char, as well as the beginning, end or middle of an array...
Think of chars as structs. You sometimes allocate a single struct on the heap. That, too, creates a pointer without an array.

Using only a pointer, to determine how big an array it is pointing to is impossible. The closest you can get to it is using calloc and counting the number of consecutive \0 chars you can find through the pointer. Of course, that doesn't work once you've assigned/reassigned stuff to that array's keys and it also fails if the memory just outside of the array happens to hold \0, too. So using this method is unreliable, dangerous and just generally silly. Don't. Do. It.

Another analogy:
Think of a pointer as a road sign, it points to Town X. The sign doesn't know what that town looks like, and it doesn't know or care (or can care) who lives there. It's job is to tell you where to find Town X. It can only tell you how far that town is, but not how big it is. That information is deemed irrelevant for road-signs. That's something that you can only find out by looking at the town itself, not at the road-signs that are pointing you in its direction

So, using a pointer the only thing you can do is:

char a_str[] = "hello";//{h,e,l,l,o,\0}
char *arr_ptr = &a_str[0];
printf("Get length of string -> %d\n", strlen(arr_ptr));

But this, of course, only works if the array/string is \0-terminated.

As an aside:

int length = sizeof(a)/sizeof(char);//sizeof char is guaranteed 1, so sizeof(a) is enough

is actually assigning size_t (the return type of sizeof) to an int, best write:

size_t length = sizeof(a)/sizeof(*a);//best use ptr's type -> good habit

Since size_t is an unsigned type, if sizeof returns bigger values, the value of length might be something you didn't expect...

Answer 3

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.