查找动态分配数组的大小

Question

Let's say we have an array of strings initialized like this:

char **a = malloc(3 * sizeof(char*)); //edited

a[0]="asd";
a[1]="fghj";
a[2]="klzxc";

How can we print the first dimension of this 2D array (3)?

Answer 1

How can we print the first dimension of this 2d array (3)?

You have to keep track of the number of elements you allocated in a separate variable:

size_t num_elements = 3;
char **a = malloc( sizeof *a * num_elements );
if ( a )
{
  a[0] = "asd";    // NOTE: these lines store the addresses of the string
  a[1] = "fghj";   // literals in a[0], a[1], and a[2] - you are not copying
  a[2] = "klzxc";  // the *contents* of each string, just its address
}
...
for ( size_t i; i < num_elements; i++ )
  printf( "a[%zu] = %s\n", i, a[i] );

A pointer, regardless of type, points to (stores the address of) a single object. That single object may be the first in a larger sequence of objects, but there is no way to determine that from pointer value itself. If you have the following:

  char **    char *            char
  +---+      +---+             +---+---+---+---+
a:|   | ---> |   | a[0] -----> |'a'|'s'|'d'| 0 |
  +---+      +---+             +---+---+---+---+
             |   | a[1] ---+
             +---+         |   +---+---+---+---+---+
             |   | a[2] -+ +-> |'f'|'g'|'h'|'j'| 0 |
             +---+       |     +---+---+---+---+---+
                         |
                         |     +---+---+---+---+---+---+
                         +---> |'k'|'l'|'z'|'x'|'c'| 0 |
                               +---+---+---+---+---+---+

You cannot know from a itself that it points to the first of 3 objects; you cannot know from each a[i] itself that it points to a sequence of char objects. That information must be tracked separately. In the case of a we have a separate variable, num_elements , that keeps track of how may elements are in a . In the case of each a[i] , we have the string terminator to tell us how long each string is.

Answer 2

How can we print the first dimension of this 2D array (3)?

---

A couple of previous points:

To be accurate char **a is not a 2D array, it's a pointer to pointer to char .

A safer way to use this memory allocation would by to use the derefenced pointer instead of the type, this method facilitates the maintenace of the code:

char **a = malloc(sizeof *a * 3);

Another thing to note is that the string literals assigned to those pointers are read only and cannot be changed.

---

To answer your question:

To know the first dimension, ie the number of assigned pointers, you need to keep it in mind, possibly store it, there is no portable way to retrieve the number of pointers you allocated after the fact.

However, there are non-portable methods to do this:

size_t size = _msize(a)/sizeof *a;  //Windows

size_t size = malloc_usable_size(a)/sizeof *a; //Linux

size_t size = malloc_size(a)/sizeof *a; //Mac OS

As the functions return the number of allocated bytes, dividing their return value by the size of the pointer retrieves the actual number of pointers.

---

If you need actual editable char arrays:

If you want to have an array of chars, that you can actually edit and change, you must also allocate memory for each of those three pointers.

eg

for(int i = 0; i < 3; i++)
    a[i] = malloc(4) //for char arrays of 3 chars + null byte

After this, to assign a string you'll need to use something like strcpy or preferably memcpy .

eg

memcpy(a[0], "asd", 4); //copies "asd" to a[0]